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dlinn [17]
3 years ago
8

An object is dropped from rest from a height of 4.4 107 m above the surface of the Earth. If there is no air resistance, what is

its speed when it strikes the Earth?
Physics
1 answer:
kondaur [170]3 years ago
6 0

Answer:

Explanation:

An object is drop from a height, then it is in direction of gravity

g is +ve

When an object drop from a height, the initial velocity is 0,

U=0

Given that h=4.4×10^7m

V=?

g=9.81m/s^2

Then using equation of motion

V^2=U^2+2gh

V^2=0+2×9.81×4.4×10^7

V^2=86.33×10^7m/s

Take square root of both side

V=29381.63m/s

Now to km/s, divide by 1000

Since 1km=1000m

V=29.38km/s

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N76 [4]

Frequency = (speed) / (wavelength)

Frequency = (331 m/s) / (0.6 m) = 551.7 Hz
3 0
3 years ago
A block is released to slide down a frictionless incline of 15∘ and then it encounters a frictional surface with a coefficient o
Elodia [21]

The block's potential energy at the top of the incline (at a height h from the horizontal surface) is equal to its kinetic energy at the bottom of the incline, so that

mgh = 1/2 mv²

where v is its speed at the bottom of the incline. It follows that

v = √(2gh)

If the incline is 20.4 m long, that means the block has a starting height of

sin(15°) = h/(20.4 m)   ⇒   h = (20.4 m) sin(15°) ≈ 5.2799 m

and so the block attains a speed of

v = √(2gh) ≈ 10.1728 m/s

The block then slides to a rest over a distance d. Kinetic friction exerts a magnitude F over this distance and performs an amount of work equal to Fd. By the work-energy theorem, this quantity is equal to the block's change in kinetic energy, so that

Fd = 0 - 1/2 mv²   ⇒   d = (-1293.58 J)/F

By Newton's second law, the net vertical force on the block as it slides is

∑ F [vertical] = n - mg = 0

where n is the magnitude of the normal force, so that

n = mg = (25 kg) g = 245 N

and thus the magnitude of friction is

F = -0.16 (245 N) = -39.2 N

(negative since it opposes the block's motion)

Then the block slides a distance of

d = (-1293.58 J) / (-39.2 N) ≈ 32.9994 m ≈ 33 m

5 0
3 years ago
A vertical wire carries a current straight down. To the east of this wire, the magnetic field points: A) toward the east. B) tow
geniusboy [140]

Answer:

option E

Explanation:

the correct answer is option E

the direction of magnetic field will be found out with the help of right hand rule.

Put you palm in the direction of electric field and curl your finger in the direction of magnetic field which east direction.

 now, the direction shown by the thumb will be the direction of magnetic field  which comes out to be toward South direction.

6 0
3 years ago
Read 2 more answers
A thin film of oil of thickness t is floating on water. The oil has index of refraction no = 1.4. There is air above the oil. Wh
kkurt [141]

Answer:

t = 120.5 nm

Explanation:

given,    

refractive index of the oil = 1.4

wavelength of the red light = 675 nm

minimum thickness of film = ?

formula used for the constructive interference

2 n t = (m+\dfrac{1}{2})\lambda

where n is the refractive index of oil

t is thickness of film

for minimum thickness

m = 0

2 \times 1.4 \times t = (0+\dfrac{1}{2})\times 675

t = \dfrac{0.5\times 675}{2\times 1.4}

        t = 120.5 nm

hence, the thickness of the oil is t = 120.5 nm

7 0
3 years ago
A spring with constant k = 78 N/m is at the base of a frictionless, 30.0°-inclined plane. A 0.50-kg block is pressed against the
olasank [31]

Explanation:

The given data is as follows.

   Spring constant (k) = 78 N/m,     \theta = 30^{o}

 Mass of block (m) = 0.50 kg

According to the formula of energy conservation,

                mgh sin \theta = \frac{1}{2}kx^{2}

       h = \frac{1}{2} \times \frac{kx^{2}}{mg Sin \theta}

          = \frac{78 \times 0.04}{2 \times 0.5 \times 9.8 \times 0.5}

          = 0.64 m

Thus, we can conclude that the distance traveled by the block is 0.64 m.

4 0
3 years ago
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