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2 years ago

A robot probe drops a camera off the rim of a 239 m high cliff on mars, where the free-fall acceleration is −3.7 m/s2 .

1 answer:

3 0

a. We can find the velocity when the camera hits the ground. v^2 = (v0)^2 + 2ay = 0 + 2ay v = sqrt{ 2ay } v = sqrt{ (2)(3.7 m/s^2)(239 m) } v = 42 m/s The camera hits the ground with a velocity of 42 m/s b. We can find the time it takes for the camera to hit the ground. y = (1/2) a t^2 t^2 = 2y / a t = sqrt{ 2y / a } t = sqrt{ (2)(239 m) / 3.7 m/s^2 } t = 11.4 seconds

It takes 11.4 seconds for the camera to hit the ground.

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A p-type Si sample is used in the Haynes-Shockley experiment. The length of the sample is 2 cm, and two probes are separated by

Airida [17]

Answer:

Mobility of the minority carriers, $\mu_{n} =1184.21 cm^{2} /V-sec$

Diffusion coefficient for minority carriers, $D_{n} = 29.20 cm^2 /s$

Verified from Einstein relation as $\frac{D_{n} }{\mu_{n} } = 25 mV$

Explanation:

Length of sample, $l_{s} = 2 cm$

Separation between the two probes, L = 1.8 cm

Drift time, $t_{d} = 0.608 ms$

Applied voltage, V = 5 V

Mobility of the minority carriers ( electrons), $\mu_{n} = \frac{V_{d} }{E}$

Where the drift velocity, $V_{d} = \frac{L}{t_{d} }$

$V_{d} = \frac{1.8}{0.608 * 10^{-3} } \\V_{d} = 2960.53 cm/s$

and the Electric field strength, $E = \frac{V}{l_{s} }$

E = 5/2

E = 2.5 V/cm

Mobility of the minority carriers:

$\mu_{n} = 2960.53/2.5\\\mu_{n} =1184.21 cm^{2} /V-sec$

The electron diffusion coefficient, $D_{n} = \frac{(\triangle x)^{2} }{16 t_{d} }$

$\triangle x = (\triangle t )V_{d}$ , where Δt = separation of pulse seen in an oscilloscope in time( it should be in micro second range)

$\triangle x = \frac{(\triangle t) L}{t_{d} } \\\triangle x = \frac{180*10^{-6} * 1.8}{0.608*10^{-3} }\\\triangle x =0.533 cm$

$D_{n} = \frac{0.533^{2} }{16 * 0.608 * 10^{-3} }\\D_{n} = 29.20 cm^2 /s$

For the Einstein equation to be satisfied, $\frac{D_{n} }{\mu_{n} } = \frac{KT}{q} = 0.025 V$

$\frac{D_{n} }{\mu_{n} } = \frac{29.20}{1184.21} \\\frac{D_{n} }{\mu_{n} } = 0.025 = 25 mV$

Verified.

4 0

2 years ago

An actor who memorizes his lines word-for-word has demonstrated _____ learning.

Nana76 [90]

Mechanical association learning used by an actor to memorize his lines

8 0

2 years ago

Read 2 more answers

If a 1000-pound capsule weighs only 165 pounds on the moon, how much work is done in propelling this capsule out of the moon's g

yulyashka [42]

Answer:

178200 g mile pounds

Explanation:

Work= Force * Distance= Fh

F=ma=mg where m is mass and g is acceleration due to gravity

Work= 165 pounds *g* 1080 m= 178200 g mile pounds

5 0

2 years ago

A baseball of mass m = 0.49 kg is dropped from a height h1 = 2.25 m. It bounces from the concrete below and returns to a final h

Brilliant_brown [7]

Answer:

Explanation:

Impulse = change in momentum

mv - mu , v and u are final and initial velocity during impact at surface

For downward motion of baseball

v² = u² + 2gh₁

= 2 x 9.8 x 2.25

v = 6.64 m / s

It becomes initial velocity during impact .

For body going upwards

v² = u² - 2gh₂

u² = 2 x 9.8 x 1.38

u = 5.2 m / s

This becomes final velocity after impact

change in momentum

m ( final velocity - initial velocity )

.49 ( 5.2 - 6.64 )

= .7056 N.s.

Impulse by floor in upward direction

= .7056 N.s

6 0

2 years ago

An 1876 N crate is being pushed across a level force at a constant speed by a force of 747 N. What is the coefficient of kinetic

nekit [7.7K]

The crate only moves horizontally, so its net vertical force is 0. The only forces acting in the vertical direction are the crate's weight (pointing downward) and the normal force of the surface on the crate (pointing upward). By Newton's second law, we have

∑ F (vertical) = n - mg = 0 → n = mg = 1876 N

where n is the magnitude of the normal force.

In the horizontal direction, the crate is moving at a constant speed and thus with no acceleration, so it's completely in equilibrium and the net horizontal force is also 0. The only forces acting on it in this direction are the 747 N push (pointing in the direction of the crate's motion) and the kinetic friction opposing it (pointing in the opposite direction). By Newton's second law,

∑ F (horizontal) = 747 N - f = 0 → f = 747 N

The frictional force is proportional to the normal force by a factor of the coefficient of kinetic friction, µ, such that

f = µn → µ = f / n = (747 N) / (1876 N) ≈ 0.398188 ≈ 0.40

8 0

2 years ago