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ziro4ka [17]
3 years ago
11

A bullet in a gun is accelerated from the firing chamber to the end of the barrel at an average rate of 5.70 x 10^5 m/s^2 for 9.

60 x 10^−4 s. What is its muzzle velocity (in m/s) (that is, its final velocity)? (Enter the magnitude.)
Physics
1 answer:
s2008m [1.1K]3 years ago
3 0

Answer:

Final speed, v = 547.2 m/s

Explanation:

Initial velocity of the bullet, u = 0

Acceleration of the bullet, a=5.7\times 10^5\ m/s^2

Time taken, t=9.6\times 10^{-4}\ s

Let v is the final velocity of the muzzle. It can be calculated using the first equation of motion as :

v=u+at

v=at      

v=5.7\times 10^5\ m/s^2\times 9.6\times 10^{-4}\ s    

v = 547.2 m/s

So, the final speed of the muzzle is 547.2 m/s. Hence, this is the required solution.                                

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A car is accelerating at 30 m/s2, if the car is 400 kg how much force
Verizon [17]
It would be 12,000 because newton’s third 2nd law states F=ma (force=matter x acceleration) so 30x400 would be your force .

please mark brainliest and i hope this helps!
6 0
3 years ago
A man-made satellite of mass 6105 kg is in orbit around the earth, making one revolution in 430 minutes. What is the magnitude o
blondinia [14]

Answer:

A gravitational force of 6841.905 newtons is exerted on the satellite by the Earth.

Explanation:

At first we assume that Earth is represented by an uniform sphere, such that the man-made satellite rotates in a circular orbit around the planet. Hence, the following condition must be satisfied:

\left(\frac{4\pi^{2}}{T^{2}} \right)\cdot r = \frac{G\cdot M}{r^{2}} (1)

Where:

T - Period of rotation of the satellite, measured in seconds.

r - Distance of the satellite with respect to the center of the planet, measured in meters.

G - Gravitational constant, measured in newton-square meters per square kilogram.

M - Mass of the Earth, measured in kilograms.

Now we clear the distance of the satellite with respect to the center of the planet:

r^{3} = \frac{G\cdot M\cdot T^{2}}{4\pi^{2}}

r = \sqrt[3]{\frac{G\cdot M\cdot T^{2}}{4\pi^{2}} } (2)

If we know that G = 6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}}, M = 6.0\times 10^{24}\,kg and T = 25800\,s, then the distance of the satellite is:

r = \sqrt[3]{\frac{\left(6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}} \right)\cdot (6.0\times 10^{24}\,kg)\cdot (25800\,s)^{2}}{4\pi^{2}} }

r \approx 18.897\times 10^{6}\,m

The gravitational force exerted on the satellite by the Earth is determined by the Newton's Law of Gravitation:

F = \frac{G\cdot m\cdot M}{r^{2}} (3)

Where:

m - Mass of the satellite, measured in kilograms.

F - Force exerted on the satellite by the Earth, measured in newtons.

If we know that G = 6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}}, M = 6.0\times 10^{24}\,kg, m = 6105\,kg and r \approx 18.897\times 10^{6}\,m, then the gravitational force is:

F = \frac{\left(6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}} \right)\cdot (6105\,kg)\cdot (6\times 10^{24}\,kg)}{(18.897\times 10^{6}\,m)^{2}}

F = 6841.905\,N

A gravitational force of 6841.905 newtons is exerted on the satellite by the Earth.

4 0
3 years ago
Please help! Calculate velocity. Show all work!
Eduardwww [97]

Answer:

v = 23.66 m/s

Explanation:

recall that one of the equations of motion may be expressed:

v² = u² + 2as,

Where

v = final velocity (we are asked to find this)

u = initial velocity = 0 m/s since we are told that it starts from rest

a = acceleration = 0.56m/s²

s = distance traveled = given as 500m

Simply substitute the known values into the equation:

v² = u² + 2as

v² = 0 + 2(0.56)(500)

v² = 560

v = √560

v = 23.66 m/s

4 0
3 years ago
The increase in height as a tsunami approaches shore is due to
ryzh [129]

Answer:

C. The decrease in speed as the wave approaches shore.

Explanation:

The waves break when approaching the shore because the depth decreases. Thus, the wave travels more slowly and increases its height. There comes a time when the part of the wave on the surface travels faster than the one that travels under water, the ridge destabilizes and falls against the ground.

5 0
3 years ago
Read 2 more answers
Learning Task 1: Determine what kind of carbon allotropes are the given pictures
kotegsom [21]

Answer:

1.C70

2.Fullerene

3.Lonsdaleite

4.Graphite

5.Diamond

6.Amorphous carbon

I hope this helps. thank you

Explanation:

4 0
3 years ago
Read 2 more answers
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