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3 years ago

A.) If its booster rockets accelerate the space shuttle at 15m/s2, how high will it be one minute after launch?

Physics

1 answer:

poizon [28]3 years ago

5 0

Answer:

27,000 m

450 m/s

Explanation:

Assuming the initial velocity is 0 m/s:

v₀ = 0 m/s

a = 15 m/s²

t = 60 s

A) Find: Δy

Δy = v₀ t + ½ at²

Δy = (0 m/s) (60 s) + ½ (15 m/s²) (60 s)²

Δy = 27,000 m

B) Find: v_avg

v_avg = Δy / t

v_avg = 27,000 m / 60 s

v_avg = 450 m/s

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BEST ANSWER GETS BRAINLIEST!

wel

Answer:

Distance = 16.9 m

Explanation:

We are given;

Power; P = 70 W

Intensity; I = 0.0195 W/m²

Now, for a spherical sound wave, the intensity in the radial direction is expressed as a function of distance r from the center of the sphere and is given by the expression;

I = Power/Unit area = P/(4πr²)

where;

P is the sound power

r is the distance.

Thus;

Making r the subject, we have;

r² = P/4πI

r = √(P/4πI)

r = √(70/(4π*0.0195))

r = √285.6627

r = 16.9 m

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3 years ago

What are the two main<br> categories of particles and<br> antiparticles?

Mice21 [21]

Answer:

Particles can be classified as hadrons – baryons and mesons – and leptons, each with its anti-particle, and they should know that interactions between these particles can be described in terms of transfer of other particles known as vector bosons.

Explanation:

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3 years ago

A girl police light on level ground with a force of 45 N at a 30° angle above the horizontal the girl pulls a sled 7.3 m how muc

Naya [18.7K]

Explanation:

W=Fdcos0

=45×7.3cos30

=284.4J

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3 years ago

A 75-g bullet is fired from a rifle having a barrel 0.540 m long. choose the origin to be at the location where the bullet begin

lyudmila [28]

Part a) The work done by the gas on the bullet is the integral of the force in dx, where x is the distance covered by the bullet inside the barrel with respect to the origin:
$W= \int\limits^{0.540m}_{0} {F} \, dx = \int\limits^{0.540m}_{0} {(16000+10000x-26000x^2)} \, dx =$
$=16000x+10000 \frac{x^2}{2} - 26000 \frac{x^3}{3}$
By substituting the length of the barrel, L=0.540 m, we find the total work done by the gas on the bullet:
$W=16000(0.540m)+10000 \frac{(0.540m)^2}{2} - 26000 \frac{(0.540m)^3}{3} =$
$=8733 J=8.73 kJ$

part b) The resolution of the problem is the same, we just have to use the new length of the barrel (L=0.95 m) inside the final formula, and we find the new value of the work:
$W=16000(0.95m)+10000 \frac{(0.95m)^2}{2} - 26000 \frac{(0.95m)^3}{3} =$
$=12280 J=12.28 kJ$

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3 years ago

A body goes from P to Q with a velocity of 10 m/s and comes back from Q to P with a velocity of 20 m/s. What is the average velo

Pachacha [2.7K]

Let's say the distance is D. Then the time going is D/10 sec. The time returning is D/20 s. The total time is 3D/20 s, and the total distance is 2D. The average speed for the round trip is (total distance)/(total time). That's (2D) ÷ (3D/20). That's (40D/3D) which is 13-1/3 m/s. (I thought it was going to depend on the distance, but it doesn't.)

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3 years ago