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Elena-2011 [213]
3 years ago
14

If 2000 kg cannon fires 2 kg projectile having muzzle velocity 200 m/s then the KINETIC ENERGY of the CANNON will be ("E4" means

"*10^4") *
1 point
4 J
40 J
400 J
None of the above
Physics
2 answers:
Alex3 years ago
7 0
40 j would be your answer
lilavasa [31]3 years ago
6 0

Answer:

M V = m v        conservation of momentum (Caps-cannon  Small-projectile)

V = m / M * V = 2 / 2000 * 200 m/s = .2 m/s    recoil velocity of cannon

KE = 1/2 M V^2 = 2000 / 2 kg * (.2 m/s)^2  = 40 kg m^2/s^2 = 40 J

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Answer:

1977.696 J

Explanation:

Given;

Weight of the box = 28.0 kg

Force applied by the boy = 230 N

angle between the horizontal and the force = 35°

Therefore,

the horizontal component of the force = 230 × cosθ

= 230 × cos 35°

= 188.405 N

Coefficient of kinetic friction, μ = 0.24

Force by friction, f = μN

here,

N = Normal force = Mass × acceleration due to gravity

or

N = 28 × 9.81 = 274.68 N

therefore,

f = 0.24 × 274.68

or

f = 65.9232 N

Now,

work done by the boy, W₁ = 188.405 N × Displacement  

= 188.405 N × 30

= 5652.15 J

and,

the

work done by the friction, W₂ = - 65.9232 N × Displacement  

= - 65.9232 N × 30 m

= - 1977.696 J

[ since the friction force acts opposite to the direction of motion, therefore the workdone will be negative]

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3 years ago
A spacecraft in outer space increases velocity by firing its rockets. How can hot gases escaping from its rockets change the vel
Elena-2011 [213]

Answer:

Momentum is conserved.

Explanation:

-Momentum is conserved.

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Let's talk skateboard velocity how much mean when you get my going down a 3 x 3" metal ramp. No posers are allowed to answer th
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Answer:

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Explanation:

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Define rotational speed?- short answer
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Determine the velocity required for a moving object 2.00 x 10^4 m above the surface of Mars to escape from Mars's gravity. The m
Ket [755]

Answer:

Therefore the escape velocity from Mar's gravity is 15.88 \times 10^4 m/s.

Explanation:

Escape velocity: Escape velocity is a the minimum velocity that a object needs to escape from the gravitational field of massive body.

V_{escape}=\sqrt{\frac{2GM}{R}}

V_{escape}= Escape velocity

G=Universal gravitational constant = 6.673×10⁻¹¹N m²/Kg²

M= mass of Mars = 6.42×10²³ kg

R = Radius of the Mars = 3.40×10³m

The escape velocity does not depend on the velocity of a object.

V_{escape}=\sqrt{\frac{2\times6.673\times 10^{-11}\times 6.42\times 10^{23}}{3.40\times10^3}}

           =15.88 \times 10^4 m/s

Therefore the escape velocity from Mar's gravity is 15.88 \times 10^4 m/s.

           

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