Which of the following careers would physicists be least qualified to perform,
based on their training?
A. Artist
B. Mechanical engineer
C. Laser designer
D. Power plant engineer

Answers

Answer 1
D

Source? Trust Me Bro

Related Questions

A 12.0 g sample of gas occupies 19.2 L at STP. what is the of moles and molecular weight of this gas?​

Answers

At STP, 1 mole of an ideal gas occupies a volume of about 22.4 L. So if n is the number of moles of this gas, then

n / (19.2 L) = (1 mole) / (22.4 L)   ==>   n = (19.2 L•mole) / (22.4 L) ≈ 0.857 mol

If the sample has a mass of 12.0 g, then its molecular weight is

(12.0 g) / n14.0 g/mol

If the Velocity of the body
is increased to 3v, determine the kinetic energy

Answers

ANSWER; KE=5mv^2 so it is proportional to v^2.

Explanation:So if you triple the velocity you are replacing v with 3v. Then you get (3v)^2=9v^2.

A solid conducting sphere of radius ra is placed concentrically inside a conducting spherical shell of inner radius rb1 and outer radius rb2. The inner sphere carries a charge Q while the outer sphere does not carry any net charge. The potential for rb1 < r < rb2 is:________

Answers

Answer:

The right answer is "[tex]\frac{KQ}{r_b_2}[/tex]".

Explanation:

As the outer spherical shell is conducting, so there is no electric field in side from

⇒ [tex]r_b_1 < r < r_b_2[/tex].

So the electric potential at all points inside the conducting shell that from

⇒ [tex]r_b_1<r<r_b_2[/tex]  

and will be similar as well as equivalent to the potential on the outer surface of the shell that will be:

⇒ [tex]v=\frac{KQ}{r_b_2}[/tex]

Thus the above is the right solution.

A rigid, insulated tank whose volume is 10 L is initially evacuated. A pinhole leak develops and air from the surroundings at 1 bar, 25 C enters the tank until the pressure in the tank becomes 1 bar (assume ideal gas model k=1.4 for the air). Find:
A) final temperature in tank.
B) amount of air that leaks into tank in grams.
C) amount of entropy produced in J/K.

Answers

Answer:

The answer is "[tex]143.74^{\circ} \ C , 8.36\ g, and \ 2.77\ \frac{K}{J}[/tex]"

Explanation:

For point a:

Energy balance equation:

[tex]\frac{dU}{dt}= Q-Wm_ih_i-m_eh_e\\\\[/tex]

[tex]W=0\\\\Q=0\\\\m_e=0[/tex]

From the above equation:

[tex]\frac{dU}{dt}=0-0+m_ih_i-0\\\\\Delta U=\int^{2}_{1}m_ih_idt\\\\[/tex]

because the rate of air entering the tank that is [tex]h_i[/tex] constant.

[tex]\Delta U = h_i \int^{2}_{1} m_i dt \\\\= h_i(m_2 -m_1)\\\\m_2u_2-m_1u_2=h_i(M_2-m_1)\\\\[/tex]

Since the tank was initially empty and the inlet is constant hence, [tex]m_2u-0=h_1(m_2-0)\\\\m_2u_2=h_1m_2\\\\u_2=h_1\\\\[/tex]

Interpolate the enthalpy between [tex]T = 300 \ K \ and\ T=295\ K[/tex]. The surrounding air  

temperature:

[tex]T_1= 25^{\circ}\ C\ (298.15 \ K)\\\\\frac{h_{300 \ K}-h_{295\ K}}{300-295}= \frac{h_{300 \ K}-h_{1}}{300-295.15}[/tex]

Substituting the value from ideal gas:

[tex]\frac{300.19-295.17}{300-295}=\frac{300.19-h_{i}}{300-298.15}\\\\h_i= 298.332 \ \frac{kJ}{kg}\\\\Now,\\\\h_i=u_2\\\\u_2=h_i=298.33\ \frac{kJ}{kg}[/tex]

Follow the ideal gas table.

The [tex]u_2= 298.33\ \frac{kJ}{kg}[/tex] and between temperature [tex]T =410 \ K \ and\ T=240\ K.[/tex]

Interpolate

[tex]\frac{420-410}{u_{240\ k} -u_{410\ k}}=\frac{420-T_2}{u_{420 k}-u_2}[/tex]

Substitute values from the table.

 [tex]\frac{420-410}{300.69-293.43}=\frac{420-T_2}{{u_{420 k}-u_2}}\\\\T_2=416.74\ K\\\\=143.74^{\circ} \ C\\\\[/tex]

For point b:

Consider the ideal gas equation.  therefore, p is pressure, V is the volume, m is mass of gas. [tex]\bar{R} \ is\ \frac{R}{M}[/tex] (M is the molar mass of the  gas that is [tex]28.97 \ \frac{kg}{mol}[/tex] and R is gas constant), and T is the temperature.

[tex]n=\frac{pV}{TR}\\\\[/tex]

[tex]=\frac{(1.01 \times 10^5 \ Pa) \times (10\ L) (\frac{10^{-3} \ m^3}{1\ L})}{(416.74 K) (\frac{8.314 \frac{J}{mol.k} }{2897\ \frac{kg}{mol})}}\\\\=8.36\ g\\\\[/tex]

For point c:

 Entropy is given by the following formula:

[tex]\Delta S = mC_v \In \frac{T_2}{T_1}\\\\=0.00836 \ kg \times 1.005 \times 10^{3} \In (\frac{416.74\ K}{298.15\ K})\\\\=2.77 \ \frac{J}{K}[/tex]

The top of a swimming pool is at ground level. If the pool is 2.60 m deep, how far below ground level does the bottom of the pool appear to be located for the following conditions?

a. The pool is completely filled with water.
______m below ground level

b. The pool is filled halfway with water.
______m below ground level

Answers

Answer:

a)  [tex]d_g=1.95m[/tex]

b) [tex]d_g'=2.3m[/tex]

Explanation:

From the question we are told that:

Depth [tex]d=2.60[/tex]

a)

Generally the equation for distance to ground is mathematically given by

[tex]\frac{d_g}{d}=\frac{1}{n}[/tex]

Where

[tex]n=\frac{4}{3}[/tex]

Therefore

[tex]d_g=\frac{d}{n}[/tex]

[tex]d_g=\frac{2.6}{4/3}[/tex]

[tex]d_g=1.95m[/tex]

b)

For when The pool is filled halfway with water

[tex]\frac{d_g}{d}=\frac{1}{n}[/tex]

[tex]d_g'=\frac{1.3}{4/3}[/tex]

[tex]d_g'=0.98m[/tex]

Therefore

[tex]d_g'=(1.3+0.98)[/tex]

[tex]d_g'=2.3m[/tex]

A bicycle wheel has a diameter of 63.4 cm and a mass of 1.86 kg. Assume that the wheel is a hoop with all of the mass concentrated on the outside radius. The bicycle is placed on a stationary stand and a resistive force of 123 N is applied tangent to the rim of the tire. What force is required if you shift to a 5.60-cm-diameter sprocket?

Answers

Answer:

Njfjrhrjrkrirkehrbrhrrhrhehrhrhejejebrbrhrbrbbbrhje

monochromatic light of wavelength 500 nm is incident normally on a diffraction grating. if the third order maximum is 32. how many total number of maximuima can be seen

Answers

Answer:

The total number of maxima that can be seen is 11

Explanation:

Given the data in the question

wavelength λ = 500 nm = 5 × 10⁻⁷ m

if the third order maximum is 32

i.e m = 3 and θ = 32°

Now, we know that condition for diffraction maximum is as follows;

d × sinθ = m ×  λ

so we substitute in our given values

d × sin( 32° ) = 3 ×  5 × 10⁻⁷ m

d × sin( 32° ) = 1.5  × 10⁻⁶ m

d = [ 1.5  × 10⁻⁶ m ] / sin( 32° )

d = 2.83 × 10⁻⁶ m

Now, maxima n when θ = 90° will be;

sin( 90° ) = nλ / d

1 =  nλ / d

d =  nλ

n = d / λ

we substitute

n = [ 2.83 × 10⁻⁶ m ]  / [ 5 × 10⁻⁷ m ]

n = 5.66

so 5 is the max value

hence, total maxima value is;

⇒ 2n + 1 = 2( 5 ) + 1 = 10 + 1 = 11

Therefore, total number of maxima that can be seen is 11

On the sonometer shown below, a horizontal cord of length 5 m has a mass of 1.45 g. When the cord was plucked the wave produced had a frequency of 120 Hz and wavelength of 6 cm. (a) What was the tension in the cord? (b) How large a mass M must be hung from its end to give it this tension? ​

Answers

Answer:

(a) T = 0.015 N

(b) M = 1.53 x 10⁻³ kg = 1.53 g

Explanation:

(a) T = 0.015 N

First, we will find the speed of waves:

[tex]v =f\lambda[/tex]

where,

v = speed of wave = ?

f = frequency = 120 Hz

λ = wavelength = 6 cm = 0.06 m

Therefore,

v = (120 Hz)(0.06 m)

v = 7.2 m/s

Now, we will find the linear mass density of the coil:

[tex]\mu = \frac{m}{l}[/tex]

where,

μ = linear mass density = ?

m = mass = 1.45 g = 1.45 x 10⁻³ kg

l = length = 5 m

Thereforre,

[tex]\mu = \frac{1.45\ x\ 10^{-3}\ kg}{5\ m}\\\\\mu = 2.9\ x\ 10^{-4}\ kg/m[/tex]

Now, for the tension we use the formula:

[tex]v = \sqrt{\frac{T}{\mu}}\\\\7.2\ m/s = \sqrt{\frac{T}{2.9\ x\ 10^{-4}\ kg/m}}\\\\(51.84\ m^2/s^2)(2.9\ x\ 10^{-4}\ kg/m) = T[/tex]

T = 0.015 N

(b)

The mass to be hung is:

[tex]T = Mg\\\\M = \frac{T}{g}\\\\M = \frac{0.015\ N}{9.8\ m/s^2}\\\\[/tex]

M = 1.53 x 10⁻³ kg = 1.53 g

During a particular thunderstorm, the electric potential difference between a cloud and the ground is Vcloud - Vground = 4.20 108 V, with the cloud being at the higher potential. What is the change in an electron's electric potential energy when the electron moves from the ground to the cloud?

Answers

Answer:

The electric potential energy is 6.72 x 10^-11 J.

Explanation:

Potential difference, V = 4.2 x 10^8 V

charge of electron, q = - 1.6 x 10^-19 C

Let the potential energy is U.

U = q V

U = 1.6 x 10^-19 x 4.2 x 10^8

U = 6.72 x 10^-11 J

In a large chemical factory, a feed pipe carries a liquid at a speed of 5.5 m/s. A pump pushes the liquid along at a gauge pressure of 140,000 Pa. The liquid travels upward 6.0 m and enters a tank at a gauge pressure of 2,000 Pa. The diameter of the pipe remains constant. At what speed does the liquid enter the tank

Answers

Answer:

v₂ = 15.24 m / s

Explanation:

This is an exercise in fluid mechanics

Let's write Bernoulli's equation, where the subscript 1 is for the factory pipe and the subscript 2 is for the tank.

          P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂

They indicate the pressure in the factory P₁ = 140000 Pa, the velocity

v₁ = 5.5 m / s and the initial height is zero y₁ = 0

the tank is at a pressure of P2 = 2000 Pa and a height of y₂ = 6.0 m

          P₁ -P₂ + ρ g (y₁ -y₂) + ½ ρ v₁² = ½ ρ v₂²

           

let's calculate

         140,000 - 2000 + ρ 9.8 (0- 6) + ½ ρ 5.5² = ½ ρ v₂²

         138000 - ρ 58.8 + ρ 15.125 = ½ ρ v2²

         v₂² = 2 (138000 /ρ - 58.8 + 15.125)

         v₂ = [tex]\sqrt{\frac{276000}{\rho } - 43.675 }[/tex]

In the exercise they do not indicate what type of liquid is being used, suppose it is water with

           ρ = 1000 kg / m³

           v₂ = [tex]\sqrt{\frac{276000}{1000} - 43.675}[/tex]

           v₂ = 15.24 m / s

An electric field E⃗ =5.00×105ı^N/C causes the point charge in the figure to hang at an angle. What is θ?

Answers

We have that the angle is

[tex]\theta=32.53[/tex]

From the Question we are told that

E⃗ =5.00×105ı^N/C

Generally the equation for Tension   is mathematically given

[tex]W=Tcos\theta[/tex]

Where

[tex]tan\theta=\frac{2.5*10^{-9}(5*10{5})}{2*10^{-3}(9.8)}[/tex]

[tex]\theta=32.53[/tex]

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Suppose you want to design an air bag system that can protect the driver at a speed 100 km/h (60 mph) if the car hits a brick wall.
Estimate how fast the air bag must inflate to effectively protect the driver. How does the use of a seat belt help the driver?

Answers

When solving question that contains equations and the use mathematical computations, It is always ideal to list the parameters given.

Now, given that:

the speed of the car which is the initial velocity (u) = 100 km/h before it hits the wall.after hitting the wall, the final velocity will be (v) = 0 km/h

Assumptions:

Suppose we make an assumption that the distance traveled during the collision of the car with the brick wall (S) = 1 mThat the car's acceleration is also constant.

For a motion under constant acceleration, we can apply the kinematic equation:

[tex]\mathsf{v^2 = u^2 + 2as}[/tex]

where;

v = final velocity u = initial velocitya = accelerations = distance

From the above equation, making acceleration (a) the subject of the formula:

[tex]\mathsf{v^2 - u^2 =2as }[/tex]

[tex]\mathsf{a = \dfrac{v^2 - u^2 }{2s}}[/tex]

The initial velocity (u) is given in km/h, and we need to convert it to m/s as it has an effect on the unit of the acceleration.

since 1 km/h = 0.2778 m/s

100 km/h = 27.78 m/s

[tex]\mathsf{a = \dfrac{(0)^2 - (27.78)^2 }{2(1)}}[/tex]

[tex]\mathsf{a = \dfrac{- 771.7284 }{2}}[/tex]

a = - 385.86 m/s²

Similarly, from the kinematic equation of motion, the formula showing the relation between time, acceleration, and velocity is;

v = u + at

where;

v = 0

-u = at

[tex]\mathsf{t = \dfrac{-u}{a}}[/tex]

[tex]\mathsf{t = \dfrac{-27.78}{-385.86}}[/tex]

t = 0.07 seconds

An airbag is designed in such a way as to prevent the driver from hitting on the steering wheel or other hard substance that could damage the part of the body. The use of the seat belt is to keep the driver in shape and in a balanced position against the expansion that occurred by the airbag during the collision on the brick wall.

Thus, we can conclude that in order to estimate how fast the airbag must inflate to effectively protect the driver, the airbag must be inflated at 0.07 seconds faster before the collision to effectively protect the driver.

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80 grams of iron at 100°C is dropped into 200 of water at 20°C contained in an iron vessel of mass 50 gram find the resulting temperature.​

Answers

Answer:

the resulting temperature is 23.37 ⁰C

Explanation:

Given;

mass of the iron, m₁ = 80 g = 0.08 kg

mass of the water, m₂ = 200 g = 0.2 kg

mass of the iron vessel, m₃ = 50 g = 0.05 kg

initial temperature of the iron, t₁ = 100 ⁰C

initial temperature of the water, t₂ = 20 ⁰C

specific heat capacity of iron, c₁ = 462 J/kg⁰C

specific heat capacity of water, c₂ = 4,200 J/kg⁰C

let the temperature of the resulting mixture = T

Apply the principle of conservation of energy;

heat lost by the hot iron = heat gained by the water

[tex]m_1c_1 \Delta t_1 = m_2c_2\Delta t_2\\\\m_1c_1 (100 - T) = m_2c_2 (T- 20)\\\\0.08 \times 462 (100-T) = 0.2 \times 4,200 (T-20)\\\\36.96 (100-T) = 840 (T-20) \\\\100 - T = 22.72 (T-20)\\\\100-T = 22.72 T - 454.4 \\\\554.4 = 23.72T\\\\T = \frac{554.4}{23.72} \\\\T = 23.37 \ ^0C[/tex]

Therefore, the resulting temperature is 23.37 ⁰C

A solid object is made of two materials, one material having density of 2 000 kg/m3 and the other having density of 6 000 kg/m3. If the object contains equal masses of the materials, what is its average density

Answers

Answer:

[tex]\rho_{avg}=4000kg/m^3[/tex]

Explanation:

From the question we are told that:

Density of Material 1 [tex]\rho_1=2000kg/m^3[/tex]

Density of Material 2 [tex]\rho_2=6000kg/m^3[/tex]

Generally the equation for Average density is mathematically given by

[tex]\rho_{avg}=frac{\rho _1+rho _2}{2}[/tex]

[tex]\rho_{avg}=\frac{2000+6000}{2}[/tex]

[tex]\rho_{avg}=4000kg/m^3[/tex]

what is the prefix notation of 0.0000738?​

Answers

Answer:

7.38 × 10-5

Explanation:

All numbers in scientific notation or standard form are written in the form m × 10n, where m is a number between 1 and 10 ( 1 ≤ |m| < 10 ) and the exponent n is a positive or negative integer.

To convert 0.0000738 into scientific notation, follow these steps:

Move the decimal 5 times to the right in the number so that the resulting number, m = 7.38, is greater than or equal to 1 but less than 10

Since we moved the decimal to the right the exponent n is negative

n = -5

Write in the scientific notation form, m × 10n

= 7.38 × 10-5

Therefore, the decimal number 0.0000738 written in scientific notation is 7.38 × 10-5 and it has 3 significant figures.

Answer = 7.38 × 10-5

The above answer is correct

A penny of mass 3.10 g rests on a small 20.0 g block supported by a spinning disk with radius of 12.0 cm. The coefficients of friction between block and disk are 0.850 (static) and 0.575 (kinetic) while those for the penny and block are 0.395 (kinetic) and 0.495 (static). What is the maximum rate of rotation in revolutions per minute that the disk can have, without the block or penny sliding on the disk

Answers

Answer:

do this Q yourself because i havent read the chapter

The maximum rate of rotation in revolutions per minute that the disk can have, without the block or penny sliding on the disk is 63 rpm.

How to solve

This is calculated using the coefficient of static friction between the penny and block, which is 0.495.

The maximum angular velocity of the disk is when the force of static friction is just sufficient to prevent the penny from sliding.

This force is equal to the mass of the penny multiplied by the acceleration due to gravity, multiplied by the coefficient of static friction.

The angular velocity of the disk is then calculated from this force and the radius of the disk.

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A T-shirt cannon launches a shirt at 5.30 m/s from a platform height of 4.00 m from ground level. How fast (in m/s) will the shirt be traveling if it is caught by someone whose hands are at 5.20 m from ground level (b) 4.00 m from ground level?

Answers

Answer:

(a) the velocity of the shirt is 2.14 m/s

(b) the velocity of the shirt is 5.3 m/s

Explanation:

Given;

initial velocity of the shirt, u = 5.3 m/s

height of the platform above the ground, h = 4.00 m

(a) When the shirt is caught by someone whose hand is 5.20 m from the ground level, the height traveled by the shirt = 5.2 m - 4.0 m = 1.2 m

The velocity at this position is calculated as;

[tex]v^2 = u^2 + 2(-g)h\\\\v^2 = u^2 - 2gh\\\\v^2 = 5.3^2 - (2\times 9.8 \times 1.2)\\\\v^2 = 4.57\\\\v= \sqrt{4.57} \\\\v = 2.14 \ m/s[/tex]

(b) When the shirt is caught by someone whose hand is 4.00 m from the ground level, the height traveled by the shirt = 4.00 m - 4.00 m = 0 m

The velocity at this position is calculated as;

[tex]v^2 = u^2 + 2(-g)h\\\\v^2 = u^2 - 2gh\\\\v^2 = 5.3^2 - (2\times 9.8 \times 0)\\\\v^2 = 28.09\\\\v= \sqrt{28.09} \\\\v = 5.3 \ m/s[/tex]

A technician builds an RLC series circuit which includes an AC source that operates at a fixed frequency and voltage. At the operating frequency, the resistance R is equal to the inductive reactance XL. The technician notices that when the plate separation of the parallel-plate capacitor is reduced to one-half its original value, the current in the circuit doubles. Determine the initial capacitive reactance in terms of the resistance R.

Answers

Answer:

Xc = (0.467 - 0.427j)R

Explanation:

Since the resistance in the circuit is R, the reactance of the inductor is XL and the reactance of the capacitor is XC, then the impedance of the circuit is

Z = √[R² + (XL - XC)²]

Since the inductive reactance XL equals the resistance R, we have that

Z = √[R² + (XL - XC)²]

Z = √[R² + (R - XC)²]

Thus, the current in the circuit is thus I = V/Z = V/√[R² + (R - XC)²]

Now, when the plate separation of the parallel plate capacitor is reduced to one-half its original value, the current doubles. Also, when the plate separation is reduced to half, the capacitance doubles since C ∝ 1/d where C is capacitance and d separation between the plates. Since the capacitance doubles, the new reactance XC' is twice the initial reactance XC. So, XC' = 2XC. Thus the new impedance is thus

Z' = √[R² + (R - XC')²]

Z' = √[R² + (R - 2XC)²]

The new current is I' = V/Z' = V/√[R² + (R - 2XC)²]

Since the current doubles, I' = 2I.

V/√[R² + (R - 2XC)²] = 2V/√[R² + (R - XC)²]

1/√[R² + (R - 2XC)²] = 2/√[R² + (R - XC)²]

√[R² + (R - XC)²] = 2√[R² + (R - 2XC)²]

squaring both sides, we have

[R² + (R - XC)²] = 4[R² + (R - 2XC)²]

expanding the brackets, we have

[R² + R² - 2RXC + XC²] = 4[R² + R² - 4RXC + 4XC²]

[2R² - 2RXC + XC²] = 4[2R² - 4RXC + 4XC²]

2R² - 2RXC + XC² = 8R² - 16RXC + 16XC²

collecting like terms, we have

16RXC - 2RXC + XC² - 16XC² = 8R² - 2R²

14RXC - 15XC² = 6R²

15XC² - 14RXC + 6R² = 0

Using the quadratic formula to find XC, we have

[tex]XC = \frac{-(-14R) +/- \sqrt{(-14R)^{2} - 4 X 15 X 6R^{2} } }{2 X 15}\\= \frac{-(-14R) +/- \sqrt{196R^{2} - 360R^{2} } }{30}\\ \\= \frac{14R +/- \sqrt{- 164R^{2} } }{30}\\ \\= \frac{14R +/- 12.81Ri }{30}\\\\= 0.467R +/- 0.427Ri[/tex]

Since it is capacitive, we take the negative part.

So, Xc = (0.467 - 0.427j)R

Consider a wave along the length of a stretched slinky toy, where the distance between coils increases and decreases. What type of wave is this

Answers

"Longitudinal wave" is the wave where the difference between the coils increases as well as decreases.

Generating waves whenever the form of communication being displaced in a similar direction as well as in the reverse way of the wave's designated points, could be determined as Longitudinal waves.A wave running the length of something like a Slinky stuffed animal, which expands as well as reduces the spacing across spindles, produces a fine image or graphic.

Thus the above answer is correct.

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Consider a neutron star with a mass equal to the sun, a radius of 19 km, and a rotation period of 1.0 s. What is the speed of a point on the equator of the star

Answers

Answer:

120 km/s

Explanation:

Given data :

Radius of the star is r = 19 km

Rotational time period of the star is T = 1 s

Therefore, we know that the velocity of the star is given by :

[tex]$V=\frac{2\pi r}{T}$[/tex]

[tex]$V=\frac{2 \times 3.14 \times 19\times 10^3}{1}$[/tex]

V = 119380.52 m/s

Therefore, the velocity of the point on the equator of the star is = 120 km/s

What are the major sources of energy utilized during a 100 meter race, a 1000 meter race, and a marathon

Answers

Answer:

The energy from food and then from plants and then from sun.

As sun is the ultimate source of energy.

Explanation:

Distance = 100 m, 1000m, marathon

As the distance is covered by the person, so the muscular energy is used and thus the energy comes form out food.

As we know that the energy can neither be created nor be destroyed it can transform from one form to another.

So, the energy form the food which we consume is converted into the kinetic energy as we run.

6. Two astronauts of equal mass step from the top of a rocket on Venus. One slides down a 10- degree ramp while the other slides down a 75-degree ramp. If all friction is ignored, which astronaut reaches the surface of Venus with the lower kinetic energy

Answers

Answer:

Final kinetic energies of both astronauts will be the same.

Explanation:

If we ignore all the friction present between the ramp and the person. Then essentially there is no loss of energy in the system. Hence the initial potential energies of the astronauts must be equal to their final kinetic energies.

Now the potential energy depends upon mass, height and acceleration due to gravity. All these parameters for both the astronauts. Therefore,both astronauts have same initial potential energies.

Similarly, the final kinetic energies of astronauts will also be the same.

Which two factors does the power of a machine depend on? А. work and distance B.. force and distance C. work and time D. time and distance?​

Answers

[tex]Hello[/tex] [tex]There![/tex]

[tex]AnimeVines[/tex] [tex]is[/tex] [tex]here![/tex]

The answer is...

C. Work and time.

[tex]HopeThisHelps!![/tex]

[tex]AnimeVines[/tex]

When a ray of light passes through a boundary between two transparent media, it changes direction. What is this phenomenon known as?
Choose the best answer from the choices provided.
a. reflection
b. refraction
c. interference
d. diffraction
e. dispersion

Answers

Answer:

I think Refraction

Explanation:

I think refraction is light which passes through a boundary between 2 transparent media.

Refraction would be your answer which would be b

What is the result of (305.120 + 267.443) x 0.50? How many answers can be written based on the principle of significant digits?

Answers

Answer:

The answer is 286.2815.

dujevduxjehhsusheheh
m=100g
F-?​

Answers

Answer:

Force = mass × acceleration

[tex]F =(100 \times 1000) \times 10 \\ = 1 \times {10}^{6} \: newtons[/tex]

A ten loop coil of area 0.23 m2 is in a 0.047 T uniform magnetic field oriented so that the maximum flux goes through the coil. The average emf induced in the coil is

Answers

Answer:

Explanation:

From the question we are told that:

Number of turns [tex]N=10[/tex]

Area [tex]a=0.23m^2[/tex]

Magnetic field [tex]B=0.947T[/tex]

Generally the equation for maximum flux is mathematically given by

 [tex]\phi=NBa[/tex]

 [tex]\phi=10*0.047*0.23[/tex]

 [tex]\phi=0.1081wbi[/tex]

Therefore induced emf

 [tex]e= \frac{d\phi}{dt}[/tex]

Since

 [tex]t=0[/tex]

Therefore

 [tex]e=0[/tex]

At a distance of 14,000 km from the center of Planet Z-99, the acceleration due to gravity is 32 m/s2. What is the acceleration due to gravity at a point 28,000 km from the center of this planet

Answers

A body of mass m feels a gravitational force due to the planet of

F = GmM/R ² = ma

where

• G = 6.67 × 10⁻¹¹ N•m²/kg² is the universal gravitational constant

• M is the mass of the planet

• R is the distance between the body and the planet's center

• a is the acceleration due to gravity

Solving for a gives

a = GM/R ²

Notice that 28,000 km is twice 14,000 km. The equation says that the acceleration varies inversely with the square of the distance. So if R is changed to 2R, we have a new acceleration of

GM/(2R)² = 1/4 × GM/R ² = a/4

so the acceleration of the body at 28,000 km from the planet's center would be (32 m/s²)/4 = 8 m/s².

A negative slope on the velocity vs. time graph indicates a negative
acceleration.
A True
B. False

Answers

A. true

means decelerating

found in brainly itself

shirleywashington

Ambitious

2.3K answers

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Explanation :

We know that the slope of velocity -time graph gives the acceleration. Acceleration of an object is defined as the rate of change of velocity i.e.

Suppose a driver suddenly applies brakes. In this case the initial velocity of his or her vehicle is more and the final velocity is less.

So, the acceleration is negative in this case i.e. the object is decelerating.

A negative slope on the velocity versus time graph indicates that an object is not accelerating. This statement is false as the object is decelerating.

Which quantities below of a solid object on this planet are NOT the same as on Earth?

Choose all
possible answers.

Weight

Mass

Volume

Density

Acceleration when it falls vertically.

Color

Answers

Answer:

Weight, acceleration when it falls vertically, are not same as that of earth.

Explanation:

Weight of the object is given by the product of mass of the object and the acceleration due to gravity of the planet.

So, the weight of object is not same as that on earth.

The mass is defined as the amount of matter contained in the object.

So, the mass of the object is same as that of earth.  

The volume of the object is defined as the space occupied by the object.

So, the volume of the object is same as that of earth.  

The density is defined as the ratio of mass of the object to its volume.

So, the density of the object is same as that of earth.  

The acceleration due to gravity on a planet depends on the mass of planet and radius of planet.

So, the acceleration is not same as that of earth.

The color of the object is its characteristic.

It is same as that of earth.

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