Calculate the density of the following material.

1 kg helium with a volume of 5.587 m³
700 kg/m³
5.587 kg/m³
0.179 kg/m³

Answers

Answer 1

Answer:

[tex]density \: = \frac{mass}{volume} [/tex]

1 / 5.587 is equal to 0.179 kg/m³

Hope it helps:)

Answer 2

Answer:

The answer is

0.179 kg/m³

Explanation:

Density of a substance is given by

[tex]Density \: = \frac{mass}{volume} [/tex]

From the

mass = 1 kg

volume = 5.583 m³

Substitute the values into the above formula

We have

[tex]Density \: = \frac{1 \: kg}{5.583 \: {m}^{3} } [/tex]

We have the final answer as

Density = 0.179 kg/m³

Hope this helps you


Related Questions

A bus carrying 10 people has over turned on a remote hillside during an intense thunderstorm. What three factors could contribute to creating a delay in advanced care

Answers

Answer:

The three factors that can contribute to creating a delay in advanced care for the passengers in the overturned bus include:

1. Lack of communication: Since the accident happened on the remote hillside, there is a possibility that, there would be no communication network which could have afforded them the opportunity to call medical or technical team.

2. Steep Nature of the Hill: This is another factor which will affect the care which they could have received. Steeply area tends to be difficult for climbing in or out from.

3. Thunderstorm: This factor is another reason which could contribute to delay in receiving advance care. Thunderstorm create barriers for location f the area where the bus overturned or in other situation complicate the rescue efforts of the team sent out to rescue.

Explanation:

A metal sphere A of radius a is charged to potential V. What will be its potential if it is enclosed by a spherical conducting shell B of radius b and the two are connected by a wire?

Answers

Answer:

The potential will be Va/b

Explanation:

So Let sphere A charged Q to potential V.

so, V= KQ/a. ....(1

Thus, spherical shell B is connected to the sphere A by a wire, so all charge always reside on the outer surface.

therefore, potential will be ,

V ′ = KQ/b = Va/b... That is from .....(1), KQ=Va]

g Can a rigid body experience any ACCELERATION when the resultant force acting on that rigid body is zero? Explain.Can a rigid body experience any ACCELERATION when the resultant force acting on that rigid body is zero? Explain.

Answers

Answer:

No, a rigid body cannot experience any acceleration when the resultant force acting on the body is zero.

Explanation:

If the net force on a body is zero, then it means that all the forces acting on the body are balanced and cancel out one another. This sate of equilibrium can be static equilibrium (like that of a rigid body), or dynamic equilibrium (that of a car moving with constant velocity)

For a body under this type of equilibrium,

ΣF = 0   ...1

where ΣF is the resultant force (total effective force due to all the forces acting on the body)

For a body to accelerate, there must be a force acting on it. The acceleration of a body is proportional to the force applied, for a constant mass of the body. The relationship between the net force and mass is given as

ΣF = ma   ...2

where m is the mass of the body

a is the acceleration of the body

Substituting equation 2 into equation 1, we have

0 = ma

therefore,

a = 0

this means that if the resultant force acting on a rigid body is zero, then there won't be any force available to produce acceleration on the body.

Which of
of
these
following material is
used as fuse material?
carbon,
silver
Copper
Aluminium

Answers

The provided question is not correct as, there is more than one options are correct, however the explaining every correct option -

Answer:

The correct answer are - silver, copper and aluminium all three used as fuse material.

Explanation:

A safety device in any electric circuit of that prevents the electric system in case of short circuit by breaking the connection of electric system or circuit termed as the Fuse or fuse element. Normally the fuse are made up of wire or element of material that are low in melting point and high in resistance.

Zinc, lead, tin, silver, copper, aluminium, and alloy of tin and alloy are used as fuse element or material for their low melting point and high resistance these are easily breaks the electric path in case of short circuit.

A circuit consists of four 100W lamps
connected in parallel across a 230V supply.
Inadvertently, a voltmeter has been connected
in series with the lamps. The resistance of the
voltmeter is 15000 and that of the lamps
under the conditions stated is six times their
value when burning normally. What will be the
reading of the voltmeter?​

Answers

Complete question is;

A circuit consists of four 100-W lamps connected in parallel across a 230-V supply. Inadvertently, a voltmeter has been connected in series with the lamps. The resistance of the voltmeter is 1500 Ω and that of the lamps under the conditions stated is six times their value then burning normally. What will be the reading of the voltmeter?

Answer:

150.42 V

Explanation:

We are told that the circuit consists of four 100W lamps.

We know that Power is given by the equation;

P = V²/R

Thus;

R = V²/P

Now, we are told that the four lamps are connected in parallel across a 230V supply.

Thus, V = 230 V

So resistance, R = 230²/100

R = 529 Ω

We are told that the resistance of the lamps under the conditions stated is six times their value when burning normally.

Thus, total resistance of each lamp under the conditions = 529 × 6 = 3174 Ω

So, since they are connected in parallel, equivalent resistance for each lamp = 3174/4 = 793.5 Ω

Now, since this resistance is connected in series with the voltmeter resistance of 1500 Ω

Therefore, total circuit resistance = 1500 + 793.5 = 2293.5 Ω

Thus;

circuit current = 230/2293.5 = 0.100283 A

Now, according to Ohm’s law, voltage drop across the voltmeter = 1500 × 0.100283 ≈ 150.42V

CHECK THE COMPLETE QUESTION BELOW

A circuit consists of four 100-W lamps connected in parallel across a 230-V supply. Inadvertently, a voltmeter has been connected in series with the lamps. The resistance of the voltmeter is 1500 Ω and that of the lamps under the conditions stated is six times their value then burning normally. What will be the reading of the voltmeter?

Answer:

the reading of the voltmeter=150.4V

Explanation:

We can determine the wattage of a lamp using below expression:

: W = I² R....................eqn(1)

But fro ohms law V=IR

then I= V/R

If we substitute I into equation (1)

We have W= V²/R

But W= 100W

V= 230V

Then

W=220²/R

100 = 2302/R

R = 529 Ω

We can as well calculate the Resistance of each lamp under given condition that they are sixtimes their value when burning normally.

R = 6 × 529 = 3174 Ω

We can also calculate quivalent resistance of the abovefour lamps connected in parallel then

R = 3174/4

= 793.5 Ω

total circuit resistance can be calculated since we know that resistance is connected to voltmeter of 1500 Ω resistance in series arrangement

Then

total circuit resistance = 1500 + 793.5

= 2293.5 Ω

Then from ohms law again

I= V/R

circuit current = 230/2293.5 A

The reading of the voltage drop across the voltmeter

= 1500 × 230/2293.5

= 150.4V

Distinguish between physical and chemical changes. Include examples in your explanations.

Answers

Answer:

Chemical changes are recognized when a substance changes its properties permanently and it cannot be the same substance as before.

Instead the physical changes implies that if you can return to the same substance through a reverse process.

Explanation:

A chemical change is, by example, a combustion, if a sheet of paper burns, its result is ashes, the ashes cannot go back to being a sheet of paper because its properties changed, heat energy was generated that changed matter permanently.

A physical change, by example, is that of freezing water, the water becomes ice, but this can easily become water again if the temperature is increased, its properties do not change and the chemistry of the substance does not change.

A chemical change takes place when a chemical reaction takes place, while when a matter changes forms but not the chemical identity then a physical change takes place.  

A product or a new compound formation takes place from a chemical change as the rearrangement of atoms takes place to produce novel chemical bonds.  

• In a chemical change always a chemical reaction takes place.  

• Some of the chemical changes examples are souring milk, burning wood, digesting food, mixing acid and base, cooking food, etc.  

• In a physical change no new chemical species form.  

• The changing of the state of a pure substance between liquid, gas, or solid is a physical change as there is no change in the identity of the matter.  

• Some of the physical changes are melting of ice, tempering of steel, breaking a bottle, crumpling a sheet of aluminum foil, boiling water, and shredding paper.  

Thus, a new substance is formed during a chemical change, while a physical change does not give rise to a new substance.  

To know more about:

https://brainly.com/question/7279398

The resistor used in the procedures has a manufacturer's stated tolerance (percent error) of 5%. Did you results from Data Table agree with the manufacturer's statement? Explain.
Resistor Measured Resistance
100 99.1

Answers

Answer:

     e% = 0.99%   this value is within the 5% tolerance given by the manufacturer

Explanation:

Modern manufacturing methods establish a tolerance in order to guarantee homogeneous characteristics in their products, in the case of resistors the tolerance or error is given by

          e% = | R_nominal - R_measured | / R_nominal 100

where R_nominal is the one written in the resistance in your barcode, R_measured is the real value read with a multimeter and e% is the tolerance also written in the resistors

let's apply this formula to our case

R_nominal = 10 kΩ = 10000 Ω

R_measured = 100 99 Ω

        e% = | 10000 - 10099.1 | / 10000 100

        e% = 0.99%

this value is within the 5% tolerance given by the manufacturer

If the intensity of an electromagnetic wave is 80 MW/m2, what is the amplitude of the magnetic field of this wave

Answers

CHECK THE COMPLETE QUESTION BELOW;

the intensity of an electromagnetic wave is 80 MW/m2, what is the amplitude of the magnetic field of this wave? (c=3.0×108m/s, μ0=4π×10−7T⋅m/A, ε0=8.85×10−12C2/N⋅m2)

Answer:

2.4×10^5 N/C

Explanation:

the amplitude can be explained as the maximum field strength of the electric and magnetic fields. Wave energy is proportional to its amplitude squared (E2 or B2).

We were told to calculate the amplitude of the magnetic field, which can be done using expresion below

S=ε²/2cμ

Where S is the intensity intensity of an electromagnetic wave given as 80 MW/m2

ε² is the Amplitude which we are looking for

c= speed of light given as 3×10^8m/s

Substitute the values into above formula we have,

S=ε²/2cμ

Making Amplitude subject of formula

ε²=S×2cμ

ε²=[80×10^6)(2×3×10^8)(4Π×10^-7)

= 245598.44

ε²=2.4×10^5 N/C

Therefore, amplitude of the magnetic field of this wave is S=2.4×10^5 N/C

The rectangular plate is tilted about its lower edge by a cable tensioned at a constant 600 N. Determine and plot the moment of this tension about the lower edge AB of the plate for the range 0 ≤ θ ≤ 90°

Answers

Answer:

Explanation:

From the figure , it is clear that moment of tension is balanced by moment of weight of plate about the line AB which is acting as axis . If W be the weight of plate ,

moment of tension about AB = moment of weight W about line AB

= W x 2.5 cosθ

moment of tension about AB = 2.5 W cosθ

here only variable  is cosθ which changes when θ changes

So, moment of tension about AB varies according to cosθ.

When θ = 0

moment of tension about AB = 2.5 W x cos 0 = 2.5 W . It is the maximum value of moment of tension .

When θ = 90°

moment of tension about AB = 2.5 W cos 90 = 0

moment of tension about AB = 0

So graph of moment of tension about AB will vary according to graph of

cosθ . It has been shown in the file attached .

Question 2.
In the US, lengths are often measured in inches, feet, yards and miles. Let's do
some conversions. The definition of the inch is: 1 inch = 25.4 mm, exactly. A foot is
12 inches and a mile is 5280 ft, exactly. A centimetre is exactly 0.01 m or 10 mm.
Sammy is 5 feet and 5.3 inches tall.
a). What is Sammy's height in Inches? (answer to 3 significant figures)
(3)
b). What is Sammy's height in Feet? (answer to 3 significant figures)
what is Sammy's hight in feet according to this statement ​

Answers

Explanation:

1 inch = 25.4 mm

1 foot = 12 inches

1 mile = 5260 feet

1 cm = 0.01 m or 10 mm

Now Sammy's height is 5 feet and 5.3 inches.

(a) We need to find Sammy's height in inches.

Since, 1 foot = 12 inches

5 feet = 5 × 12 inches = 60 inches

Now, 5 feet and 5.3 inches = 60 inches + 5.3 inches = 65.3 inches

Sammy's height is 65.3 inches.

(b) We need to find Sammy's height in feet.

Since, 1 foot = 12 inches

[tex]1\ \text{inch}=\dfrac{1}{12}\ \text{feet}[/tex]

So,

[tex]5.3\ \text{inch}=\dfrac{5.3}{12}\ \text{feet}=0.4416\ \text{feet}[/tex]

5 feet and 5.3 inches = 5 feet + 0.4416 feet = 5.44 feet

Sammy's height is 5.44 feet.

A square loop 16.0 cm on a side has a resistance of 6.35 Ω . It is initially in a 0.510 T magnetic field, with its plane perpendicular to B , but is removed from the field in 40.5 ms.

Required:
Calculate the electric energy dissipated in this process.

Answers

Answer:

Explanation:

change in magnetic flux = .16 x .16 x .510 - 0

= .013056 weber .

rate of change of flux = change in flux / time

= .013056 / 40.5 x 10⁻³

= .32237

voltage induced = .32237 V

electrical energy dissipated = v² / R where v is voltage , R is resistance

= .32237² / 6.35

= 16.36 x 10⁻³ J .

All household circuits are wired in parallel. A 1140-W toaster, a 270-W blender, and a 80-W lamp are plugged into the same outlet. (The three devices are in parallel when plugged into the same outlet.) Assume that this is the standard household 120-V circuit with a 15-A fuse.
a. What current is drawn by each device?
b. To see if this combination will blow the 15-A fuse, find the total current used when all three appliances are on.

Answers

Answer:

total current = 12.417 A

so it will not fuse as current is less than 15 A

Explanation:

given data

toaster = 1140-W

blender = 270-W

lamp = 80-W

voltage = 120 V

solution

we know that current is express as

current = power ÷ voltage   ......................1

here voltage is same in all three device

so

current by toaster is

I = [tex]\frac{1140}{120}[/tex]

I = 9.5 A

and

current by blender

I = [tex]\frac{270}{120}[/tex]

I = 2.25 A

and

current by lamp is

I = [tex]\frac{80}{120}[/tex]

I = 0.667 A

so here device in parallel so

total current is = 9.5 A + 2.25 A + 0.667 A

total current = 12.417 A

so it will not fuse as current is less than 15 A

Without actually calculating any logarithms, determine which of the following intervals the sound intensity level of a sound with intensity 3.66×10^−4W/m^2 falls within?

a. 30 and 40
b. 40 and 50
c. 50 and 60
d. 60 and 70
e. 70 and 80
f. 80 and 90
g. 90 and 100

Answers

Answer:

f. 80 and 90

Explanation:

1 x 10⁻¹² W/m² sound intensity falls within 0 sound level

1 x 10⁻¹¹ W/m² sound intensity falls within 10 sound level

1 x 10⁻¹⁰ W/m² sound intensity falls within 20 sound level

1 x 10⁻⁹ W/m² sound intensity falls within 30 sound level

1 x 10⁻⁸ W/m² sound intensity falls within 40 sound level

1 x 10⁻⁷ W/m² sound intensity falls within 50 sound level

1 x 10⁻⁶ W/m² sound intensity falls within 60 sound level

1 x 10⁻⁵ W/m² sound intensity falls within 70 sound level

1 x 10⁻⁴ W/m² sound intensity falls within 80 sound level

1 x 10⁻³ W/m² sound intensity falls within 90 sound level

Given sound intensity (3.66 x 10⁻⁴ W/m²) falls with 1 x 10⁻⁴ W/m² of intensity which is within 80 and 90 sound level.

f. 80 and 90

Find the rms current delivered by the power supply when the frequency is very large. Answer in units of A.

Answers

Answer:

The rms current is 0.3112 A.

Explanation:

Given that,

Suppose, The capacitance is 170 μF and the inductance is 2.94 mH. The resistance in the top branch is 278 Ohms, and in the bottom branch is 151 Ohms. The potential of the power supply is 47 V .

We know that,

When the frequency is very large then the capacitance can be treated as a short circuit and inductance as open circuit.

So,

We need to calculate the rms current

Using formula of current

[tex]I=\dfrac{V}{R}[/tex]

Where, V = voltage

R = resistance

Put the value into the formula

[tex]I=\dfrac{47}{151}[/tex]

[tex]I= 0.3112 \ A[/tex]

Hence, The rms current is 0.3112 A.

A square coil of wire with 15 turns and an area of 0.40 m2 is placed parallel to a magnetic field of 0.75 T. The coil is flipped so its plane is perpendicular to the magnetic field in 0.050 s. What is the magnitude of the average induced emf

Answers

Answer:

The magnitude of the average induced emf is 90V

Explanation:

Given;

area of the square coil, A = 0.4 m²

number of turns, N = 15 turns

magnitude of the magnetic field, B = 0.75 T

time of change of magnetic field, t = 0.05 s

The magnitude of the average induced emf is given by;

E = -NAB/t

E = -(15 x 0.4 x 0.75) / 0.05

E = -90 V

|E| = 90 V

Therefore, the magnitude of the average induced emf is 90V

When light is either reflected or refracted, the quantity that does not change in either process is its

Answers

Answer:

Frequency

Explanation:

When waves travel from one medium to another, it is only the frequency of the wave that remains constant . when a wave is refracted at the boundary between two media, the wave will slow down and its wavelength decreases. The wave usually bends at the interface between the two media. The wavelength and speed of a wave may change at the boundary between two media but its frequency remains the same.

Hence the frequency of light is its only property that remains constant.

A spring with spring constant 15 N/m hangs from the ceiling. A ball is attached to the spring and allowed to come to rest. It is then pulled down 6.0 cm and released. If the ball makes 30 oscillations in 20 s, what are its (a) mass and (b) maximum speed?

Answers

Answer:

a

   [tex]m = 0.169 \ kg[/tex]

b

  [tex]|v_{max} |= 0.5653 \ m/s[/tex]

Explanation:

From the question we are told that

    The  spring constant is  [tex]k = 14 \ N/m[/tex]

     The  maximum extension of the spring is  [tex]A = 6.0 \ cm = 0.06 \ m[/tex]

     The number of oscillation is  [tex]n = 30[/tex]

      The  time taken is  [tex]t = 20 \ s[/tex]

Generally the the angular speed of this oscillations is mathematically represented as

           [tex]w = \frac{2 \pi}{T}[/tex]

where T is the period which is mathematically represented as

     [tex]T = \frac{t}{n}[/tex]

substituting values

     [tex]T = \frac{20}{30 }[/tex]

     [tex]T = 0.667 \ s[/tex]

Thus  

       [tex]w = \frac{2 * 3.142 }{ 0.667}[/tex]

       [tex]w = 9.421 \ rad/s[/tex]

this angular speed can also be represented mathematically as

       [tex]w = \sqrt{\frac{k}{m} }[/tex]

=>   [tex]m =\frac{k }{w^2}[/tex]

substituting values

      [tex]m =\frac{ 15 }{(9.421)^2}[/tex]

      [tex]m = 0.169 \ kg[/tex]

In SHM (simple harmonic motion )the equation for velocity is  mathematically represented as

        [tex]v = - Awsin (wt)[/tex]

The  velocity is maximum when  [tex]wt = \(90^o) \ or \ 1.5708\ rad[/tex]

     [tex]v_{max} = - A* w[/tex]

=>   [tex]|v_{max} |= A* w[/tex]

=>    [tex]|v_{max} |= 0.06 * 9.421[/tex]

=>   [tex]|v_{max} |= 0.5653 \ m/s[/tex]

Water flows at 0.00027 m3/s through a 10-m long garden hose lying on the ground, with a radius of 0.01 m. Water has a viscosity of 1 mPa.s What is the magnitude of gauge pressure in Pa of the water entering the hose

Answers

Answer:

The gauge pressure is  [tex]P = 687.4 \ Pa[/tex]

Explanation:

From the question we are told that

    The rate of flow is  [tex]Q = 0.00027 m^3 /s[/tex]

      The height is h  =  10 m

      The radius is  r =  0.01 m

     The  viscosity is  [tex]\eta = 1mPa \cdot s = 1 *10^{-3} \ Pa\cdot s[/tex]

       

Generally the gauge pressure according to Poiseuille's equation  is mathematically represented as  

               [tex]P = 8 \pi \eta * \frac{L * v }{ A}[/tex]

Here v is the velocity of the water which is mathematically represented according to continuity equation as

             [tex]v = \frac{Q}{A }[/tex]

Where A is the cross-sectional area which is mathematically represented as

            [tex]A = \pi r^2[/tex]

substituting values

          [tex]A = 3.142 *(0.01)^2[/tex]

           [tex]A = 3.142 *10^{-4} \ m^2[/tex]

So

      [tex]v = \frac{ 0.00027}{3.142*10^{-4}}[/tex]

       [tex]v = 0.8593 \ m/s[/tex]

So

       [tex]P = 8 * 3.142 * 1.0*10^{-3}* \frac{10 * 0.8593 }{ 3.142*10^{-4}}[/tex]

       [tex]P = 687.4 \ Pa[/tex]

A 1.2-m length of wire centered on the origin carries a 20-A current directed in the positive y direction. Determine the magnetic field at the point x= 5.0m on x-axis.

a. 1.6 nt in the negative z direction
b. 1.6 nt in the positive z direction
c. 2.4 T in the positive z direction
d. 2.4 nt in the negative z direction
e. None of the above

Answers

Answer:

None of the above

Explanation:

The formula of the magnetic field of a point next to a wire with current is:

B = 2×10^(-7) × ( I /d)

I is the intensity of the current.

d is the distance between the wire and the point.

● B = 2*10^(-7) × (20/5) = 8 ×10^(-7) T

Two 1.0 nF capacitors are connected in series to a 1.5 V battery. Calculate the total energy stored by the capacitors.

Answers

Answer:

1.125×10⁻⁹ J

Explanation:

Applying,

E = 1/2CV²................... Equation 1

Where E = Energy stored in the capacitor, C = capacitance of the capacitor, V = Voltage of the battery.

Given; C = 1.0 nF,  = 1.0×10⁻⁹ F, V = 1.5 V

Substitute into equation 1

E = 1/2(1.0×10⁻⁹×1.5²)

E = 1.125×10⁻⁹ J

Hence the energy stored by the capacitor is 1.125×10⁻⁹ J

Question 2
A) A spring is compressed, resulting in its displacement to the right. What happens to the spring when it is released? (1 point)
The spring exerts a restoring force to the right and compresses even further
The spring exerts a restoring force to the left and returns to its equilibrium position
The spring exerts a restoring force to the right and returns to its equilibrium position
The spring exerts a restoring force to the left and stretches beyond its equilibrium position

Answers

1. Which example best describes a restoring force?

B) the force applied to restore a spring to its original length

2. A spring is compressed, resulting in its displacement to the right. What happens to the spring when it is released?

C) The spring exerts a restoring force to the left and returns to its equilibrium position.

3. A 2-N force is applied to a spring, and there is displacement of 0.4 m. How much would the spring be displaced if a 5-N force was applied?

D) 1 m

4. Hooke’s law is described mathematically using the formula Fsp=−kx. Which statement is correct about the spring force, Fsp?

D)It is a vector quantity.

5. What happens to the displacement vector when the spring constant has a higher value and the applied force remains constant?

A) It decreases in magnatude.

Suppose we want to calculate the moment of inertia of a 56.5 kg skater, relative to a vertical axis through their center of mass.

Required:
a. First calculate the moment of inertia (in kg-m^2) when the skater has their arms pulled inward by assuming they are cylinder of radius 0.11 m.
b. Now calculate the moment of inertia of the skater (in kg-m^2) with their arms extended by assuming that each arm is 5% of the mass of their body. Assume the body is a cylinder of the same size, and the arms are 0.875 m long rods extending straight out from their body being rotated at the ends.

Answers

Answer:

a. 0.342 kg-m² b. 2.0728 kg-m²

Explanation:

a. Since the skater is assumed to be a cylinder, the moment of inertia of a cylinder is I = 1/2MR² where M = mass of cylinder and r = radius of cylinder. Now, here, M = 56.5 kg and r = 0.11 m

I = 1/2MR²

= 1/2 × 56.5 kg × (0.11 m)²

= 0.342 kgm²

So the moment of inertia of the skater is

b. Let the moment of inertia of each arm be I'. So the moment of inertia of each arm relative to the axis through the center of mass is (since they are long rods)

I' = 1/12ml² + mh² where m = mass of arm = 0.05M, l = length of arm = 0.875 m and h = distance of center of mass of the arm from the center of mass of the cylindrical body = R/2 + l/2 = (R + l)/2 = (0.11 m + 0.875 m)/2 = 0.985 m/2 = 0.4925 m

I' = 1/12 × 0.05 × 56.5 kg × (0.875 m)² + 0.05 × 56.5 kg × (0.4925 m)²

= 0.1802 kg-m² + 0.6852 kg-m²

= 0.8654 kg-m²

The total moment of inertia from both arms is thus I'' = 2I' = 1.7308 kg-m².

So, the moment of inertia of the skater with the arms extended is thus I₀ = I + I'' = 0.342 kg-m² + 1.7308 kg-m² = 2.0728 kg-m²

a) The moment of inertia as the skater pulled his/her arm inward by assuming he/she is a cylinder is 0.3418kg-m².

b) If the body of the skater is assumed to be a cylinder of the same size, and the arms are rods extending straight out from his/her body being rotated at the ends, the moment of inertia is 1.7495kg-m².

Given the data in the question;

Mass of skater; [tex]M = 56.5kg[/tex]

a)

When the skater has his arms pulled inward by assuming they are cylinder of radius; [tex]R = 0.11 m[/tex]

Moment of inertia; [tex]I = \ ?[/tex]

From Parallel axis theorem; Moment of Inertia for a cylindrical body is expressed:

[tex]I = \frac{1}{2}MR^2[/tex]

Where M is the mass and R is the radius

We substitute our given values into the equation

[tex]I = \frac{1}{2}\ *\ 56.5kg\ *\ (0.11m)^2\\\\I = \frac{1}{2}\ *\ 56.5kg\ *\ 0.0121m^2\\\\I = 0.3418kg.m^2[/tex]

Therefore, the moment of inertia as the skater pulled his/her arm inward by assuming he/she is a cylinder is 0.3418kg-m²

b)

With the skater's arms extended by assuming that each arm is 5% of the mass of their body

Mass of each arm; [tex]M_a = \frac{5}{100} * M = \frac{5}{100} * 56.5kg = 2.825kg[/tex]

Remaining mass; [tex]M_b = M - 2M_a = 56.5kg - 2(2.825kg) = 50.85kg[/tex]

Assume the body is a cylinder of the same size and the arms are 0.875 m long rods extending straight out from their body being rotated at the ends.

Length of arm; [tex]L = 0.875 m[/tex]

From Parallel axis theorem; Moment of Inertia about vertical axis is expressed as:

[tex]I = \frac{1}{2}M_bR^2 + \frac{2}{3}M_aL^2[/tex]

We substitute in our values

[tex]I = \frac{1}{2}*50.85kg*(0.11m)^2 + \frac{2}{3}*2.825kg*(0.875m)^2\\\\I = [\frac{1}{2}*50.85kg * 0.0121m^2] + [\frac{2}{3}*2.825kg*0.765625m^2]\\\\I = 0.3076kg.m^2 + 1.4419kg.m^2\\\\I = 1.7495kg.m^2[/tex]

Therefore, if the body of the skater is assumed to be a cylinder of the same size, and the arms are rods extending straight out from his/her body being rotated at the ends, the moment of inertia is 1.7495kg-m².

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How much energy is required to accelerate a spaceship with a rest mass of 121 metric tons to a speed of 0.509 c?

Answers

Answer

1.07E22 Joules

Explanation;

We know that mass expands by a factor

=>>1/√[1-(v/c)²]

But v= 0.509c

So

1/√(1 - 0.509²)

=>>> 1/√(1 - 0.2591)

= >> 1/√(0.7409) = 1.16

But given that 121 tons is rest mass so 121- 1.16= 119.84 tons is kinetic energy

And we know that rest mass-energy equivalence is 9 x 10^19 joules per ton.

So Multiplying by 119.84

Kinetic energy will be 1.07x 10^22 joules

The Moon orbits Earth in a nearly circular orbit (mean distance is 378,000 km ). The moon Charon orbits Pluto in a nearly circular orbit as well (mean distance is 19,600 km ).
Earth Moon Pluto Charon
Mass (kg) 5.97 x 10^24 0.07342 x 10^24 0.0146 x 10^24 0.00162 x 10^24
Equatorial radius (km) 6378.1 1738.1 1185 604
Which object exhibits the longest orbital period? Hint: perform order of magnitude analysis.
a. Moon around Earth
b. Charon around Pluto
c. About the same for both

Answers

Answer:

a. Moon around Earth.

Explanation:

Charon orbit takes around 6.4 earth days to complete its orbit. Charon does not rises or sets, it hovers over same spot around the Pluto. The same side of Charon faces the Pluto, this is called Tidal Locking.

The moon orbit takes around 27 days to complete its orbit. The moon has different sides that are faced with sun which creates light or dark face of moon on the earth. Moon has 384,400 km distance from the earth.

The object that should exhibit the longest orbital period is option a. Moon around Earth.

What is Charon's orbit?

Charon's orbit takes around 6.4 earth days to finish its orbit. Charon does not rise or sets, it hovers over similar spot around Pluto. The same side of Charon faces Pluto, this we called Tidal Locking. Here the moon orbit should take approx 27 days to finish its orbit. The moon has various sides that are faced with the sun which developed the light or dark face of the moon on the earth. Also, Moon has 384,400 km distance from the earth.

learn more about orbit here: https://brainly.com/question/25404554

What did the results of photoelectric-effect experiments establish?

Answers

Answer:

Option A

Electrons are emitted if low intensity, high-frequency light hits a metal surface.

Explanation:

From the experiments conducted to study the photoelectric effect, conclusions were made that the key factor that contributes to the emission of electrons from the surface of the metal is the frequency of the beam of light. This frequency has to be beyond a minimum threshold, if not, there will be no emission of electrons from the metal surface no matter the intensity of the beam of light or the length of time it is incident upon the metal surface.

This makes option A correct because it highlights the contributions made by the threshold frequency to the photoelectric effect.

A pulley 326 mm in diameter and rotating initially at 4.00 revolutions per second receives a constant angular acceleration of 2.25 radians per second squared by a drive belt. What is the linear velocity of the belt after 5.00 seconds

Answers

Answer:

The linear velocity, v = 5.93 m/s

Explanation:

To find the linear velocity after 5 seconds, we find its angular velocity after 5 seconds using

ω' = ω + αt where ω = initial angular speed = 4.00 rev/s = 4.00 × 2π rad/s = 25.13 rad/s, ω' =  = final angular speed, α = angular acceleration = 2.25 rad/s² and t = time = 5.00 s

ω' = ω + αt

= 25.13 rad/s + 2.25 rad/s² × 5.00 s

= 25.13 rad/s + 11.25 rad/s

= 36.38 rad/s

The linear velocity v is gotten from v = rω' where r = radius of pulley = 326 mm/2 = 163 mm = 0.163 m

v = rω'

= 0.163 m × 36.38 rad/s

= 5.93 m/s

So, the linear velocity  v = 5.93 m/s

Find an analytic expression for p(V)p(V)p(V), the pressure as a function of volume, during the adiabatic expansion.

Answers

Answer:

In an adiabatic process we have

pV γ = const..

This explains that the pressure is a function of volume, p ( V ) ,

So can be written as:

p ( V ) × V γ = p 0 V γ 0 ,

or p ( V ) = p 0 V 0 / V γ

= p 0 V 0 / V ^(7 / 5)

The heat capacity of object B is twice that of object A. Initially A is at 300 K and B at 450 K. They are placed in thermal contact and the combination is thermally insulated. The final temperature of both objects is

Answers

Answer:

The final temperature of both objects is 400 K

Explanation:

The quantity of heat transferred per unit mass is given by;

Q = cΔT

where;

c is the specific heat capacity

ΔT is the change in temperature

The heat transferred by the  object A per unit mass is given by;

Q(A) = caΔT

where;

ca is the specific heat capacity of object A

The heat transferred by the  object B per unit mass is given by;

Q(B) = cbΔT

where;

cb is the specific heat capacity of object B

The heat lost by object B is equal to heat gained by object A

Q(A) = -Q(B)

But heat capacity of object B is twice that of object A

The final temperature of the two objects is given by

[tex]T_2 = \frac{C_aT_a + C_bT_b}{C_a + C_b}[/tex]

But heat capacity of object B is twice that of object A

[tex]T_2 = \frac{C_aT_a + C_bT_b}{C_a + C_b} \\\\T_2 = \frac{C_aT_a + 2C_aT_b}{C_a + 2C_a}\\\\T_2 = \frac{c_a(T_a + 2T_b)}{3C_a} \\\\T_2 = \frac{T_a + 2T_b}{3}\\\\T_2 = \frac{300 + (2*450)}{3}\\\\T_2 = 400 \ K[/tex]

Therefore, the final temperature of both objects is 400 K.

Approximating the eye as a single thin lens 2.70 cmcm from the retina, find the focal length of the eye when it is focused on an object at a distance of 265 cmcm

Answers

Answer:

0.37 cm

Explanation:

The image is formed on the retina which is at a constant distance of 2.70 cm to the lens. Therefore, image distance = 2.70 cm.

The object is at a distant of 265 cm to the lens of the eye.

From lens formula,

[tex]\frac{1}{f}[/tex] = [tex]\frac{1}{u}[/tex] + [tex]\frac{1}{v}[/tex]

where: f is the focal length, u is the object distance and v is the image distance.

Thus, u = 265.00 cm and v = 2.70 cm.

[tex]\frac{1}{f}[/tex] = [tex]\frac{1}{265}[/tex] + [tex]\frac{27}{10}[/tex]

  = [tex]\frac{10+7155}{2650}[/tex]

[tex]\frac{1}{f}[/tex]  = [tex]\frac{7165}{2650}[/tex]

⇒ f = [tex]\frac{2650}{7165}[/tex]

      = 0.37

The focal length of the eye is 0.37 cm.

If R = 20 Ω, what is the equivalent resistance between points A and B in the figure?​

Answers

Answer:

c. 70 Ω

Explanation:

The R and R resistors are in parallel.  The 2R and 2R resistors are in parallel.  The 4R and 4R resistors are in parallel.  Each parallel combination is in series with each other.  Therefore, the equivalent resistance is:

Req = 1/(1/R + 1/R) + 1/(1/2R + 1/2R) + 1/(1/4R + 1/4R)

Req = R/2 + 2R/2 + 4R/2

Req = 3.5R

Req = 70Ω

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