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s2008m [1.1K]
3 years ago
8

Please help me i’ll mark u branliest !!

Physics
1 answer:
9966 [12]3 years ago
8 0

Answer:

sorry but I can understand that

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Which of these is equivalent to the law of conservation of energy?
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Two cars cover the same distance in a straight line. Car a covers the distance at a constant velocity. Car b starts from rest an
Artyom0805 [142]

a) For the motion of car with uniform velocity we have , s = ut+\frac{1}{2}at^2, where s is the displacement, u is the initial velocity, t is the time taken a is the acceleration.

In this case s = 520 m, t = 223 seconds, a =0 m/s^2

Substituting

       520 = u*223\\ \\u = 2.33 m/s

 The constant velocity of car a = 2.33 m/s

b) We have s = ut+\frac{1}{2} at^2

s = 520 m, t = 223 seconds, u =0 m/s

Substituting

      520 = 0*223+\frac{1}{2} *a*223^2\\ \\ a = 0.0209 m/s^2

Now we have v = u+at, where v is the final velocity

Substituting

        v = 0+0.0209*223 = 4.66 m/s

So final velocity of car b = 4.66 m/s

c) Acceleration = 0.0209 m/s^2

7 0
3 years ago
In a shipping company distribution center, an open cart of mass 50.0 kg is rolling to the left at a speed of 5.00 m/s. Ignore fr
spin [16.1K]

Answer:

a) v_p=9.35m/s

Explanation:

From the question we are told that:

Open cart of mass   M_o=50.0 kg

Speed of cart   V=5.00m/s

Mass of package   M_p=15.0kg

Speed of package at end of chute V_c=3.00m/s

Angle of inclination   \angle =37

Distance of chute from bottom of cart   d_x=4.00m

a)

Generally the equation for work energy theory is mathematically given by

  \frac{1}{2}mu^2+mgh=\frac{1}{2}mv_p^2

Therefore

  \frac{1}{2}u^2+gh=\frac{1}{2}v_p^2

  v_p=\sqrt{2(\frac{1}{2}u^2+gh)}

  v_p=\sqrt{2(\frac{1}{2}v_c^2+gd_x)}

  v_p=\sqrt{2(\frac{1}{2}(3)^2+(9.8)(4))}

  v_p=9.35m/s

4 0
2 years ago
A bullet of mass m=26g is fired into a wooden block of mass M=4.7kg. The block is attached to a string of length 1.5m. The bulle
vodka [1.7K]

Answer:

u = 449 m/s

Explanation:

Given,

Mass of the bullet, m = 26 g

Mass of the wooden block,M = 4.7 Kg

height of the block,h = 0.31 m

initial speed of the block, u = ?

Using conservation of energy

(M+ m)gh = \dfrac{1}{2}(M+m)v^2

gh = \dfrac{1}{2}v^2

v=\sqrt{2gh}

v=\sqrt{2\times 9.81\times 0.31}

v = 2.47 m/s

Now, using conservation of momentum to calculate the speed of the bullet.

m u + M u' = (M+m)v

m u  = (M+m)v

0.026 x u  = (4.7+0.026) x 2.47

u = 449 m/s

Hence, the speed of the bullet is equal to 449 m/s.

5 0
3 years ago
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