A massless, hollow sphere of radius R is entirely filled with a fluid such that its density is p. This same hollow sphere is now compressed so that its radius is R/2, and then it is entirely filled with the same fluid as before. As such, what is the density of the compressed sphere?
a. 8p
b. p/8
c. p/4
d. 4p

Answers

Answer 1

Answer:

a. 8p

Explanation:

We are given that

Radius of hollow sphere , R1=R

Density of hollow sphere=[tex]\rho[/tex]

After compress

Radius of hollow sphere, R2=R/2

We have to find density of the compressed sphere.

We know that

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

[tex]Mass=Density\times volume=Constant[/tex]

Therefore,[tex]\rho_1 V_1=\rho_2V_2[/tex]

Volume of sphere=[tex]\frac{4}{3}\pi r^3[/tex]

Using the formula

[tex]\rho\times \frac{4}{3}\pi R^3=\rho_2\times \frac{4}{3}\pi (R/2)^3[/tex]

[tex]\rho R^3=\rho_2\times \frac{R^3}{8}[/tex]

[tex]\rho_2=8\rho[/tex]

Hence, the density of  the compressed sphere=[tex]8\rho[/tex]

Option a is correct.


Related Questions

4. Paper is solid in packets labelled 80 g/m2. This means that a sheet of paper of area
10 000cm? has a mass of 80 g. The thickness of each sheet is 0.11mm. What is the
density of the paper?
A 0.073 g/cm?
B 0.088 g/cm
C 0.73 g/cm3
D 0.88 g/cm
B
с

Answers

Answer:

Option C. 0.73 g/cm³

Explanation:

From the question given above, the following data were obtained:

Mass = 80 g

Area (A) = 10000 cm²

Thickness = 0.11 mm

Density =?

Next, we shall convert 0.11 mm to cm. This can be obtained as follow:

10 mm = 1 cm

Therefore,

0.11 mm = 0.11 mm × 1 cm / 10 mm

0.11 mm = 0.011 cm

Thus, 0.11 mm is equivalent to 0.011 cm.

Next, we shall determine the volume of the paper. This can be obtained as follow:

Area (A) = 10000 cm²

Thickness = 0.011 cm

Volume =?

Volume = Area × Thickness

Volume = 10000 × 0.011

Volume = 110 cm³

Finally, we shall determine the density of the paper. This can be obtained as follow:

Mass = 80 g

Volume = 110 cm³

Density =?

Density = mass / volume

Density = 80 / 110

Density = 0.73 g/cm³

Therefore the density of the paper is 0.73 g/cm³

A fan spins at 6.0 rev/s. You turn it off, and it slows at 1.0 rev/s2. What is the angular displacement before it stops

Answers

Answer:

Angular displacement before it stops = 18 rev

Explanation:

Given:

Speed of fan w(i) = 6 rev/s

Speed of fan (Slow) ∝ = 1 rev/s

Final speed of fan w(f) = 0 rev/s

Find:

Angular displacement before it stops

Computation:

w(f)² = w(i) + 2∝θ

0² = 6² + 2(1)θ

0 = 36 + 2θ

2θ = -36

Angular displacement before it stops = -36 / 2

θ = -18

Angular displacement before it stops = 18 rev

Give reason why a man getting out of moving bus must run in the same direction for a certain distance.​

Answers

Explanation:

Explanation: It's because when he stop down from a moving bus his feet come at rest while the upper portion of his body is still in motion and he falls in the forward direction.

when blueshift occurs,the preceived frequency of the wave would be?​

Answers

Answer:

When blueshift happens, the perceived frequency of the wave would be higher than the actual frequency.

Explanation:

As the name suggests, when blueshift happens to electromagnetic waves, the frequency of the observed wave would shift towards the blue (high-frequency) end of the visible spectrum. Hence, there would be an increase to the apparent frequency of the wave.

Blueshifts happens when the source of the wave and the observer are moving closer towards one another.

Assume that the wave is of frequency [tex]f\; {\rm Hz}[/tex] at the source. In other words, the source of the wave sends out a peak after every [tex](1/f)\; {\text{seconds}}[/tex].

Assume that the distance between the observer and the source of the wave is fixed. It would then take a fixed amount of time for each peak from the source to reach the observer.

The source of this wave sends out a peak after each period of [tex](1/f)\; {\text{seconds}}[/tex]. It would appear to the observer that consecutive peaks arrive every [tex](1/f)\; {\text{seconds}}\![/tex]. That would correspond to a frequency of [tex]f\; {\rm Hz}[/tex].

On the other hand, for a blueshift to be observed, the source of the wave needs to move towards the observer. Assume that the two are moving towards one another at a constant speed of [tex]v \; {\rm m \cdot s^{-1}}[/tex].

Again, the source of this wave would send out a peak after each period of [tex](1/f)\; {\text{seconds}}[/tex]. However, by the time the source sends out the second peak, the source would have been [tex]v \cdot (1 / f) \; { \rm m}= (v / f)\; {\rm m}[/tex] closer to the observer then when the source sent out the first peak.

When compared to the first peak, the second peak would need to travel a slightly shorter distance before it reach the observer. Hence, from the perspective of the observer, the time difference between the first and the second peak would be shorter than [tex](1/f)\; {\text{seconds}}[/tex]. The observed frequency of this wave would be larger than the original [tex]f\; {\rm Hz}[/tex].

•. What is called the error due to the procedure and used apparatuses?
a. Random error
b. Index error
c. Systematic error
d. Parallax error.​

Answers

Answer:

[tex]c.) \: systematic \: error \\ \\ = > it \: is \: the \: error \: caused \: \\ \\ due \: to \: the \: procedure \\ \\ \: and \: used \: apparatuses \\ \\ \huge\mathfrak\red{Hope \: it \: helps}[/tex]

After a laser beam passes through two thin parallel slits, the first completely dark fringes occur at 19.0 with the original direction of the beam, as viewed on a screen far from the slits. (a) What is the ratio of the distance between the slits to the wavelength of the light illuminating the slits

Answers

Answer:

[tex]$\frac{d}{\lambda} = 1.54$[/tex]

Explanation:

Given :

The first dark fringe is for m = 0

[tex]$\theta_1 = \pm 19^\circ$[/tex]

Now we know for a double slit experiments , the position of the dark fringes is give by :

[tex]$d \sin \theta=\left(m+\frac{1}{2}\right) \lambda$[/tex]

The ratio of distance between the two slits, d to the light's wavelength that illuminates the slits, λ :

[tex]$d \sin \theta=\left(\frac{1}{2}\right) \lambda$[/tex]     (since, m = 0)

[tex]$d \sin \theta=\frac{\lambda}{2}$[/tex]

[tex]$\frac{d}{\lambda} = \frac{1}{2 \sin \theta}$[/tex]

[tex]$\frac{d}{\lambda} = \frac{1}{2 \sin 19^\circ}$[/tex]

[tex]$\frac{d}{\lambda} = 1.54$[/tex]

Therefore, the ratio is [tex]$\frac{1}{1.54}$[/tex]  or 1 : 1.54

The voltage across a membrane forming a cell wall is 74.0 mV and the membrane is 9.20 nm thick. What is the electric field strength in volts per meter

Answers

Answer:

7.60× 10^6 V/m

Explanation:

electric field strength can be determined as ratio of potential drop and distance, I.e

E=V/d

Where E= electric field

V= potential drop= 74.0 mV= 0.07 V

d= distance= 9.20 nm = 9.2×10^-9 m

Substitute the values

E= 0.07/ 9.2×10^-9

= 7.60× 10^6 V/m

A 1640 kg merry-go-round with a radius of 7.50 m accelerates from rest to a rate of 1.00 revolution per 8.00 s. Estimate the merry-go-round as a solid cylinder and determine the net work needed for this acceleration.

Answers

Solution :

Given data :

Mass of the merry-go-round, m= 1640 kg

Radius of the merry-go-round, r = 7.50 m

Angular speed, [tex]$\omega = \frac{1}{8}$[/tex]  rev/sec

                             [tex]$=\frac{2 \pi \times 7.5}{8}$[/tex]  rad/sec

                              = 5.89 rad/sec

Therefore, force required,

[tex]$F=m.\omega^2.r$[/tex]

   [tex]$$=1640 \times (5.89)^2 \times 7.5[/tex]  

   = 427126.9 N

Thus, the net work done for the acceleration is given by :

W = F x r

   = 427126.9 x 7.5

   = 3,203,451.75 J

A new car manufacturer advertises that their car can go from zero to sixty mph in 8 [s]. This is a description of

Answers

Answer:

Acceleration

Explanation:

The fact that new can go from zero to 60mph in 8 secs is a description of its pick-up or in physics,  it's called acceleration.

Here initial velocity u= 0

final velocity v = 60 mph = 1m/minute.

or v =1609.344/60 = 26.82m/s

and time taken to do so is 8 sec

Acceleration a = (v-u)/t

a = (26.82-0)/8 = 3.35 m/s^2

Therefore, acceleration of the car a = 3.35 m/s^2.

A boy is playing with a water hose, which has an exit area of
10 cm2 and has water flowing at a rate of 2 m/s. If he covers
the opening of the hose with his thumb so that it now has an
open area of 2 cm2, what will be the new exit velocity of the
water?

Answers

Answer:

The exit velocity of water is  B. 15 m/s.

Explanation:

According to equation of continuity, for a steady flow of water, the volume of liquid entering a pipe in 1 second is equal to the volume that leaves per second.

If the initial exit area of the pipe is A₁ and the speed of exit is v₁ and the final exit area is A₂ and its corresponding exit velocity  is v₂, then,

Rewrite the expression for v₂.

Substitute 10 cm² for A₁, 2 cm² for A₂ and 3 m/s for v₁.

The exit speed of water from the hose is 15 m/s.

a) Viewers of Star Trek hear of an antimatter drive on the Starship Enterprise. One possibility for such a futuristic energy source is to store antimatter charged particles in a vacuum chamber, circulating in a magnetic field, and then extract them as needed. Antimatter annihilates with normal matter, producing pure energy. What strength magnetic field is needed to hold antiprotons, moving at 5.00 x10^7 m/s in a circular path 2.00m in radius? Antiprotons have the same mass as protons but the opposite (negative) charge.b) Is this field strength obtainable with today's technology or is it a futuristic possibility?

Answers

Charge me and do I name for meters

A car is moving north at 5.2 m/s2. Which type of motion do the SI units in this
value express?

Answers

Answer:

the SI unit (meter per second square) indicates a linear type of motion.

Explanation:

Given;

acceleration of the car, a = 5.2 m/s² North

the SI unit of the car, = m/s²

The SI unit of the given value (acceleration), indicates a linear type of motion.

Linear acceleration is the change in linear velocity with time. Also, the northwards direction indicates linear displacement of the car.

Therefore, the SI unit (meter per second square) indicates a linear type of motion.

Answer:

displacement

Explanation:

Two sinusoidal waves have the same frequency and wavelength. The wavelength is 20 cm. The two waves travel from their respective sources and reach the same point in space at the same time, resulting in interference. One wave travels a larger distance than the other. For each of the possible values of that extra distance listed below, identify whether the extra distance results in maximum constructive interference, maximum destructive interference, or something in-between.
a. 10 cm - (A) in-between (2) maximum destructive (3) maximum constructive.
b. 15 cm - (A) in-between (2) maximum destructive (3) maximum constructive.
c. 20 cm - (A) in-between (2) maximum destructive (3) maximum constructive.
d. 30 cm - (A) in-between (2) maximum destructive (3) maximum constructive.
e. 35 cm - (A) in-between (2) maximum destructive (3) maximum constructive.
f. 40 cm - (A) in-between (2) maximum destructive (3) maximum constructive.

Answers

Answer:

Explanation:

When the path difference is equal to wave length or its integral multiple, constructive interference occurs . If it is odd multiple of half wave length , then destructive interference occurs.

For constructive interference , path diff = n λ

For destructive interference path diff = ( 2n+ 1 ) λ /2

where λ is wave length of wave , n is an integer.

a )

path diff = 10 cm which is half the wavelength , so maximum destructive interference will occur.

b )

path diff = 15 cm which is neither  half the wavelength nor full wavelength , so in between is the right option.

c )

path diff = 20 cm which is equal to  the wavelength , so maximum constructive  interference will occur.

d)

path diff = 30 cm which is 3 times half the wavelength , so maximum destructive interference will occur.

e)

path diff = 35 cm which is neither integral multiple of half the wavelength , nor integral multiple of wavelength so in between is th eright answer.

f )

path diff = 40 cm which is 2 times the wavelength , so maximum constructive  interference will occur

Which of the following categories of motion is mutually exclusive with each of the others? A. Translational motion B. Rectilinear motion C. Rotational motion D. Curvilinear motion

Answers

Answer:

C.  Rotational motion

Explanation:

The kinematics of rotational motion describes the relationships between the angle of rotation, angular velocity, angular acceleration, and time. It only describes motion—it does not include any forces or masses that may affect rotation (these are part of dynamics). Recall the kinematics equation for linear motion: v = v+at (constant a).

Rotational motion is mutually exclusive with each of the others. Hence, option (C) is correct.

What is  Rotational motion?

"The motion of an object around a circular route, in a fixed orbit, is referred to as rotational motion."

Rotational motion dynamics are identical to linear or translational dynamics in every way. The motion equations for linear motion share many similarities with the equations for the mechanics of rotating objects. Rotational motion only takes stiff bodies into account. A massed object that maintains a rigid shape is referred to as a rigid body.

What is  Curvilinear motion?

Curvilinear motion is the movement of an object along a curved route. Example: A stone hurled at an angle into the air.

The motion of a moving particle that follows a predetermined or known curve is referred to as curvilinear motion. Two coordinate systems—one for planar motion and the other for cylindrical motion—are used to examine this type of motion.

Learn more about motion here:

https://brainly.com/question/22810476

#SPJ2

A star has a declination of approximately -90°. in what direction is the Star located from the celestial equator?
East
North
South
West

Answers

The start is located on theWest

What would the separation between two identical objects, one carrying 4 C of positive charge and the other 4 C of negative charge, have to be if the electrical force on each was precisely 8 N

Answers

Answer:

7.46×10⁻⁶ m

Explanation:

Applying,

F = kqq'/r²............ Equation 1

make r the subject of the equation

r = √(F/kqq').......... Equation 2

From the question,

Given: F = 8 N, q' = q= 4 C

Constant: k = 8.98×10⁹ Nm²/C²

Substitute these values into equation 2

r = √[8/(4×4×8.98×10⁹)]

r = √(55.7×10⁻¹²)

r = 7.46×10⁻⁶ m

What happens to the acceleration if you triple the force that you apply to the painting with your hand? (Use the values from the example given in the previous part of the lecture.) Submit All Answers Answer: Not yet correct, tries 1/5 3. A driver slams on the car brakes, and the car skids to a halt. Which of the free body diagrams below best matches the braking force on the car. (Note: The car is moving in the forward direction to the right.] (A) (B) (C) (D) No more tries. Hint: (Explanation) The answer is A. The car is moving to the right and slowing down, so the acceleration points to the left. The only significant force acting on the car is the braking force, so this must be pointing left because the net force always shares the same direction as the object's acceleration. 4. Suppose that the car comes to a stop from a speed of 40 mi/hr in 24 seconds. What was the car's acceleration rate (assuming it is constant). Answer: Submit Al Answers Last Answer: 55 N Only a number required, Computer reads units of N, tries 0/5. 5. What is the magnitude (or strength) of the braking force acting on the car? [The car's mass is 1200 kg.) Answer: Submit Al Answers Last Answer: 55N Not yet correct, tries 0/5

Answers

Answer:

2) when acceleration triples force triples,  3) a diagram with dynamic friction force in the opposite direction of movement of the car

4)  a = 2.44 ft / s², 5)  fr = 894.3 N

Explanation:

In this exercise you are asked to answer some short questions

2)  Newton's second law is

         F = m a

when acceleration triples force triples

3) Unfortunately, the diagrams are not shown, but the correct one is one where the axis of movement has a friction force in the opposite direction of movement, as well as indicating that the car slips, the friction coefficient of dynamic.

The correct answer is: a diagram with dynamic friction force in the opposite direction of movement of the car

4) let's use the scientific expressions

          v = v₀ - a t

as the car stops v = 0

          a = v₀ / t

let's reduce the magnitudes

          v₀ = 40 mile / h ([tex]\frac{5280 ft}{1 mile}[/tex]) ([tex]\frac{1 h}{3600 s}[/tex]) = 58.667 ft / s

          a = 58.667 / 24

          a = 2.44 ft / s²

5) let's use Newton's second law

           fr = m a

We must be careful not to mix the units, we will reduce the acceleration to the system Yes

           a = 2.44 ft / s² (1 m / 3.28 ft) = 0.745 m / s²

           fr = 1200  0.745

           fr = 894.3 N

93 cm3 liquid has a mass of 77 g. When calculating its density what is the appropriate number of significant figures

Answers

Answer:

828 kg/m³ or 0.828 g/cm³

Explanation:

Applying,

D = m/V............. Equation 1

Where D = density of the liquid, m = mass of the liquid, V = volume of the liquid.

From the question,

Given: m = 77 g , V = 93 cm³

Substitute these values into equation 1

D = 77/93

D = 0.828 g/cm³

Converting to kg/m³

D = 828 kg/m³

A loop of wire is in a magnetic field such that its axis is parallel with the field direction. Which of the following would result in an induced emf in the loop?
A. Moving the loop outside of the magnetic field region.
B. Change the diameter of the loop.
C. Change the magnitude of the magnetic field.
D. Spin the loop such that its axis does not consistently line up with the magnetic field direction.

Answers

Answer:

All the given options will result in an induced emf in the loop.

Explanation:

The induced emf in a conductor is directly proportional to the rate of change of flux.

[tex]emf = -\frac{d \phi}{dt} \\\\where;\\\\\phi \ is \ magnetic \ flux\\\\\phi = BA\ cos \theta[/tex]

where;

A is the area of the loop

B is the strength of the magnetic field

θ is the angle between the loop and the magnetic field

Considering option A, moving the loop outside the magnetic field will change the strength of the magnetic field and consequently result in an induced emf.

Considering option B, a change in diameter of the loop, will cause a change in the magnetic flux and in turn result in an induced emf.

Option C has a similar effect with option A, thus both will result in an induced emf.

Finally, considering option D, spinning the loop such that its axis does not consistently line up with the magnetic field direction will change the angle between the loop and the magnetic field. This effect will also result in an induced emf.

Therefore, all the given options will result in an induced emf in the loop.

A coin and feather are dropped in a moon. what will fall earlier on ground.give reasons.if they are dropped in the earth,which one will fall faster?

Answers

Answer:

When an object is dropped, the "principal" force that acts on that object is the gravitational force.

Thus, in the absence of air resistance and such, the acceleration of the object will be equal to the gravitational acceleration:

g = 9.8m/s^2

So, when we drop objects in the moon (where there is no air) the acceleration of every object will be exactly the same. (so there is no dependence in the mass or shape of the object)

Thus, if we drop a coin and a feather in the moon, both objects will fall with the same acceleration, and then both objects will hit the ground at the same time.

But if we are in Earth, we can not ignore the air resistance (a force that acts in the opposite direction than the movement of the object)

And this force depends on the shape and mass of the object (for example, something with a really larger surface and really thin, like a sheet of paper will be more affected by this force than a small rock)

Then here, when the air resistance applies, we should expect that the heavier and smaller object (the coin) to be less affected by this force, then the resistance that the coin experiences is smaller, then the coin falls "faster" than the feather.

A ball drops from a height, bounces three times, and then rolls to a stop when it reaches the ground the fourth time.

At what point is its potential energy greatest?

At what points does it have zero kinetic energy?

At what point did it have maximum kinetic energy?

Answers

Answer:

Greatest potential: moment before being dropped

Zero Kinetic: when it comes to rest

Greatest Kinetic: moment before first bounce

Explanation:

Physics help please

Answers

Answer:

i think the answer is 0.001m³

In the figure, particle A moves along the line y = 31 m with a constant velocity v with arrow of magnitude 2.8 m/s and parallel to the x axis. At the instant particle A passes the y axis, particle B leaves the origin with zero initial speed and constant acceleration a with arrow of magnitude 0.35 m/s2. What angle between a with arrow and the positive direction of the y axis would result in a collision?

Answers

Answer:

59.26°

Explanation:

Since a is the acceleration of the particle B, the horizontal component of acceleration is a" = asinθ and the vertical component is a' = acosθ where θ angle between a with arrow and the positive direction of the y axis.

Now, for particle B to collide with particle A, it must move vertically the distance between A and B which is y = 31 m in time, t.

Using y = ut + 1/2a't² where u = initial velocity of particle B = 0 m/s, t = time taken for collision, a' = vertical component of particle B's acceleration =  acosθ.

So, y = ut + 1/2a't²

y = 0 × t + 1/2(acosθ)t²

y = 0 + 1/2(acosθ)t²

y = 1/2(acosθ)t²   (1)

Also, both particles must move the same horizontal distance to collide in time, t.

Let x be the horizontal distance,

x = vt (2)where v = velocity of particle A = 2.8 m/s and t = time for collision

Also,  using x = ut + 1/2a"t² where u = initial velocity of particle B = 0 m/s, t = time taken for collision, a" = horizontal component of particle B's acceleration =  asinθ.

So, x = ut + 1/2a"t²

x = 0 × t + 1/2(ainsθ)t²

x = 0 + 1/2(asinθ)t²

x = 1/2(asinθ)t²  (3)

Equating (2) and (3), we have

vt = 1/2(asinθ)t²   (4)

From (1) t = √[2y/(acosθ)]

Substituting t into (4), we have

v√[2y/(acosθ)] = 1/2(asinθ)(√[2y/(acosθ)])²  

v√[2y/(acosθ)] = 1/2(asinθ)(2y/(acosθ)  

v√[2y/(acosθ)] = ytanθ

√[2y/(acosθ)] = ytanθ/v

squaring both sides, we have

(√[2y/(acosθ)])² = (ytanθ/v)²

2y/acosθ = (ytanθ/v)²

2y/acosθ = y²tan²θ/v²

2/acosθ = ytan²θ/v²

1/cosθ = aytan²θ/2v²

Since 1/cosθ = secθ = √(1 + tan²θ) ⇒ sec²θ = 1 + tan²θ ⇒ tan²θ = sec²θ - 1

secθ = ay(sec²θ - 1)/2v²

2v²secθ = aysec²θ - ay

aysec²θ - 2v²secθ - ay = 0

Let secθ = p

ayp² - 2v²p - ay = 0

Substituting the values of a = 0.35 m/s, y = 31 m and v = 2.8 m/s into the equation, we have

ayp² - 2v²p - ay = 0

0.35 × 31p² - 2 × 2.8²p - 0.35 × 31 = 0

10.85p² - 15.68p - 10.85 = 0

dividing through by 10.85, we have

p² - 1.445p - 1 = 0

Using the quadratic formula to find p,

[tex]p = \frac{-(-1.445) +/- \sqrt{(-1.445)^{2} - 4 X 1 X (-1)}}{2 X 1} \\p = \frac{1.445 +/- \sqrt{2.088 + 4}}{2} \\p = \frac{1.445 +/- \sqrt{6.088}}{2} \\p = \frac{1.445 +/- 2.4675}{2} \\p = \frac{1.445 + 2.4675}{2} or p = \frac{1.445 - 2.4675}{2} \\p = \frac{3.9125}{2} or p = \frac{-1.0225}{2} \\p = 1.95625 or -0.51125[/tex]

Since p = secθ

secθ = 1.95625 or secθ = -0.51125

cosθ = 1/1.95625 or cosθ = 1/-0.51125

cosθ = 0.5112 or cosθ = -1.9956

Since -1 ≤ cosθ ≤ 1 we ignore the second value since it is less than -1.

So, cosθ = 0.5112

θ = cos⁻¹(0.5112)

θ = 59.26°

So, the angle between a with arrow and the positive direction of the y axis would result in a collision is 59.26°.

Help me with my physics, please

Answers

The right answer would be

-20t+ 80

if 145kl of energy is added to water, what mass of water can be heated from 35C to 100C then vaporized at 100C

Answers

Answer:

m = 0.057 kg = 57 g

Explanation:

Energy Added to Water = Heat added to raise the temperature of water + Heat used to vaporize water

[tex]E = mC\Delta T + mH\\E = m(C\Delta T + H)[/tex]

where,

E = Energy added to water = 145 KJ

m = mass of water = ?

C = specific heat capacity of water = 4.2 KJ/kg.°C

ΔT = change in temperature = 100°C - 35°C = 65°C

H = Latent heat of vaporization of water = 2260 KJ/kg

Therefore,

[tex]145\ KJ = m[(4.2\ KJ/kg.^oC)(65^oC)+2260\ KJ/kg]\\\\145\ KJ = m(2533\ KJ/kg)\\\\m = \frac{145\ KJ}{2533\ KJ/kg}[/tex]

m = 0.057 kg = 57 g

The mass of water that can be heated is equal to 0.527 kilograms.

Given the following data:

Quantity of energy = 145 kJ = 145,000 Joules.Initial temperature = 35.0°CFinal temperature = 100.0°C

Scientific data:

Specific heat capacity of water = 4200 J/kg°CLatent heat of vaporization of water = 2260 KJ/kg

To calculate the mass of water that can be heated:

The quantity of energy and heat.

Note: The quantity of energy added to water is equal to the quantity of heat used to vaporize water and the quantity of heat that is added to raise the temperature of water.

Mathematically, this is given by this expression:

[tex]E=mc\theta + mH\\\\E= m(c\theta + H)[/tex]

Making m the subject of formula, we have:

[tex]m=\frac{E}{c\theta + H}[/tex]

Substituting the parameters into the formula, we have;

[tex]m=\frac{145000}{[42000\times (100-35)] + 2260}\\\\m=\frac{145000}{(4200\times 65) + 2260}\\\\m=\frac{145000}{273000 + 2260}\\\\m=\frac{145000}{275260}[/tex]

Mass, m = 0.527 kilograms.

Read more on quantity of energy here: https://brainly.com/question/13439286

Define measurements.​

Answers

Answer:

act or process of measuring

Explanation:

Explanation:

the comparison of an unknown quantity with a known quantity.

Question 7 of 10
A railroad freight car with a mass of 32,000 kg is moving at 2.0 m/s when it
runs into an at-rest freight car with a mass of 28,000 kg. The cars lock
together. What is their final velocity?
A.1.1 m/s
B. 2.2 m/s
C. 60,000 kg•m/s
D. 0.5 m/s

Answers

Answer:

a

Explanation:

you take 32,000kg ÷2.0m

A person with a near point of 85 cm, but excellent distant vision, normally wears corrective glasses. But he loses them while traveling. Fortunately, he has his old pair as a spare.
(a) If the lenses of the old pair have a power of +2.25 diopters, what is his near point (measured from his eye) when he is wearing the old glasses if they rest 2.0 cm in front of his eye?
(b) What would his near point be if his old glasses were contact lenses instead?

Answers

Answer:

a)   p = 95.66 cm, b) p = 93.13 cm

Explanation:

For this problem we use the  constructor equation

         [tex]\frac{1}{f} = \frac{1}{p} + \frac{1}{q}[/tex]

where f is the focal length, p and q are the distances to the object and the image, respectively

the power of the lens is

         P = 1 / f

         f = 1 / P

         f = 1 / 2.25

         f = 0.4444 m

the distance to the object is

         [tex]\frac{1}{p} = \frac{1}{f} -\frac{1}{q}[/tex]

the distance to the image is

          q = 85 -2

           q = 83 cm

we must have all the magnitudes in the same units

           f = 0.4444 m = 44.44 cm

we calculate

           [tex]\frac{1}{p} = \frac{1}{44.44} - \frac{1}{83}[/tex]

           1 / p = 0.010454

            p = 95.66 cm

b) if they were contact lenses

            q = 85 cm

            [tex]\frac{1}{p} = \frac{1}{44.44} - \frac{1}{85}[/tex]

             1 / p = 0.107375

             p = 93.13 cm

Your dog is running around the grass in your back yard. He undergoes successive displacements 3.20 m south, 8.16 m northeast, and 15.6 m west. What is the resultant displacement

Answers

Answer:

D1 = 3.50 m, south; D2 = 8.20 m, northeast; D3 = 15.0 m, west. Converting all these displacements from east where zero degrees is at east or + x-axis, the converted displacements are: D1 = 3.50 m 270°; D2 = 8.20 m 45° and D3 = 15.0 m 180°. We then tabulate these vectors including there x and y components. The x-components are solved by magnitudes * cos of direction angle while the y-components of the three vectors are solved by magnitudes * sin of direction angle.

The resultant is computed by summing the components algebraically. The direction in degrees is the arc tangent of the sum of all y divided by the sum of all x.

Explanation:

Paauto A: Isulat sa papel ang alpabetong Ingles at bilang I hanggang 10 sa istilong
Roman ng pagleletra.​

Answers

Answer:

Explanation:

English alphabets numbered fro 1 to 26

and the numbers 1 to10 so they are written in roman numbers as

1 - I

2 - II

3 - III

4 - IV

5 -V

6 - VI

7 -VII

8 - VIII

9 - IX

10 -X

11 - XI

12 - XII

13 - XIII

14 - XIV

15 - XV

16 - XVI

17 - XVII

18 - XVIII

19 - XIX

20- XX

21 - XXI

22 - XXII

23 - XXIII

24 - XXIV

25 - XXV

26 - XXVI  

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