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4 years ago

2. An object experiences an acceleration, g, when it is on the surface of a planet of radius R. What will be

Physics

1 answer:

blsea [12.9K]4 years ago

7 0

The new acceleration of gravity is D) 1/16 g

Explanation:

The magnitude of the acceleration of gravity in the gravitational field of a planet is given by

$g=\frac{GM}{r^2}$

where

G is the gravitational constant

M is the mass of the planet

r is the distance of the object from the centre of the planet

In this problem, the acceleration of gravity g on the surface of the planet (when r=R) is

$g=\frac{GM}{R^2}$

Then the object is moved to a distance of

r' = 4R

Substituting into the original equation, we can find what is the new acceleration of gravity:

$g'=\frac{GM}{(4R)^2}=\frac{1}{16}(\frac{GM}{R^2})=\frac{1}{16}g$

So, the acceleration of gravity has decreased by a factor 16.

Learn more about gravity:

brainly.com/question/1724648

brainly.com/question/12785992

#LearnwithBrainly

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An icicle falls off of a skyscraper from rest and falls for 34 seconds. How fast will that icicle be moving after that time? Sho

goblinko [34]

Answer:

<em>The icicle will be moving at 333.54 m/s</em>

Explanation:

<u>Free Fall Motion </u>

A free-falling object falls under the sole influence of gravity. Any object that is being acted upon only by the force of gravity is said to be in a state of free fall. Free-falling objects do not encounter air resistance.

If an object is dropped from rest in a free-falling motion, it falls with a constant acceleration called the acceleration of gravity, which value is $g = 9.81 m/s^2$ .

The final velocity of a free-falling object after a time t is given by:

vf=g.t

The icicle falls from rest for 34 seconds. We need to find the speed after that time:

vf = 9.81*34

vf = 333.54 m/s

The icicle will be moving at 333.54 m/s

7 0

3 years ago

The frequency of the middle c note on a piano is 261.63 hz. What is the wavelength of this note in centimeters? The speed of sou

Sholpan [36]

Answer:

1.31 m

Explanation:

The relationship between frequency and wavelength of a sound wave is

$c=f \lambda$

where

c is the speed of the wave

f is the frequency

$\lambda$ is the wavelenfth

In this problem, we have

c = 343.06 m/s

f = 261.63 Hz

So we can solve the formula for the wavelength:

$\lambda=\frac{c}{f}=\frac{343.06 m/s}{261.63 Hz}=1.31 m$

8 0

3 years ago

It takes about 20 seconds for a jet plane to go from rest to the takeoff speed of 100 mph (44.7 m/s). What is the average horizo

Ksju [112]

Answer:

Correct answer: F = 214.56 N

Explanation:

Given:

V₀ = 0 m/s initial speed

V = 44.7 m/s speed after t = 29 seconds

m = 96 kg

F = ? horizontal force

Taking off an airplane is a uniformly accelerated motion to which the formula applies:

V = V₀ + a t and V₀ = 0 m/s

V = a t ⇒ a = V / t = 44.7 / 20 = 2.235 m/s²

a = 2.235 m/s²

the horizontal force is calculated using the formula:

F = m · a = 96 · 2.235 = 214.56 N

F = 214.56 N

God is with you!!!

4 0

4 years ago

A block of mass m = 2.0 kg lies on a rough ramp that is inclined at an angle θ = 20oto the horizontal. A force F of magnitude 5.

Marina86 [1]

Answer:

a) 0.64 b) 2.17m/s^2 c) 8.668joules

Explanation:

The block was on the ramp, the ramp was inclined at 20degree. A force of 5N was acting horizontal to the but not parallel to the ramp,

Frictional force = horizontal component of the weight of the block along the ramp + the applied force since the block was just about move

Frictional force = mgsin20o + 5N = 6.71+5N = 11.71

The force of normal = the vertical component of the weight of the block =mgcos20o = 18.44

Coefficient of static friction = 11.71/18.44= 0.64

Remember that g = acceleration due to gravity (9.81m/s^2) and m = mass (2kg)

b) coefficient of kinetic friction = frictional force/ normal force

Fr = 0.4* mgcos 20o = 7.375N

F due to motion = ma = total force - frictional force

Ma = 11.71 - 7.375 = 4.335

a= 4.335/2(mass of the block) = 2.17m/s^2

C) work done = net force *distance = 4.335*2= 8.67Joules

8 0

3 years ago

A sun-like star is barely visible to naked-eye observers on earth when it is a distance of 7.0 light years, or 6.6 * 1016 m, awa

igomit [66]

Answer:

At a distance of 1376.49 candle emits 0.2 watt power

Explanation:

Distance between Sun and earth $6.6\times 10^{16}m$

Sun emits a power of $P=3.8\times 10^{26}watt$

Power emitted by candle = 0.20 watt

We know that brightness is given by

$B=\frac{P}{4\pi d^2}$

So $\frac{3.8\times 10^{26}}{4\pi (6\times 10^{16})^2}=\frac{0.20}{4\pi d^2}$

$3.8\times 10^{26}d^2=7.2\times 10^{32}$

$d^2=1.89\times 10^6$

$d=1376.49m$

So at a distance of 1376.49 candle emits 0.2 watt power

3 0

3 years ago