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kondaur [170]
4 years ago
11

S the work done on Amanda's car while speeding up (i) greater than, (ii) less than, or (iii) the same as the work done on Bertha

's car while speeding up?
Physics
1 answer:
ankoles [38]4 years ago
7 0

Answer:

Explanation:

Mass of amanda and bertha car = Ma = Mb

Initial velocity of amanda, ua = 10 m/s

Final velocity of amanda, va = 20 m/s

Initial velocity of bertha, ub = 20 m/s

Final velocity of bertha, vb = 30 m/s

Workdone = change in Energy = 1/2 mv^2 - 1/2 mu^2

Ea = Ma × 1/2 × (20^2 - 10^2)

= Ma × 150

= 150 Ma J

Workdone = change in Energy = 1/2 mv^2 - 1/2 mu^2

Eb = Mb × 1/2 × (30^2 - 20^2)

= Mb × 250

= 250 Mb J

Option ii. Less than. Thus this is because the workdone by amanda car (150Ma) is less than the workdone in berthas car (250Mb).

B.

Impulse, p = force × time

p = Mass × change in velocity

pa = Ma × va - Ma × ua

= Ma × (20 - 10)

= 10 × Ma

= 10 Ma

p = Mass × change in velocity

pa = Mb × vb - Mb × ub

= Mb × (30 - 20)

= 10 × Mb

= 10 Mb

Option iii. The same as.

The impulse as seen above is the same in amanda car (10Ma) as the same with bertha (10Mb)

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A tennis ball of mass 44.0 g is held just above a basketball of mass 594 g. With their centers vertically aligned, both are rele
ZanzabumX [31]

Answer:

u = 4.6 m/s

h = 8.01 m

Explanation:

Given:

Mass of the tennis ball, m = 44.0 g

Mass of the basket ball, M = 594 g

Height of fall, h = 1.08m

Now,

we have

u^2-u'^2 = 2as

where, s = distance = h

a = acceleration

u = final speed before the collision

u' = initial speed

since it is free fall case

thus,

a = g = acceleration due to gravity

u' = 0

thus we have

u^2-0^2 = 2\times9.8\tiimes1.08

or

u = \sqrt{21.168}

or

u = 4.6 m/s

b) Now after the bounce, the ball moves with the same velocity

thus, v = v₂

thus,

final speed (v_f) = v = 4.6 m/s

Then conservation of energy says  

\frac{1}{2}mu_1^2+\frac{1}{2}Mu_2^2 = \frac{1}{2}mv_1^2+\frac{1}{2}Mv_2^2  

also

applying the concept of conservation of momentum

we have

mu₁ + Mu₂ = mv₁ + Mv₂

u₁ =velocity of the tennis ball before collision = -4.6 m/s  

u₂ = velocity of the basketball before collision= 4.6 m/s  

v₁ =  velocity of the tennis ball after collision  

v₂ = velocity of the basketball  after collision

substituting the values in the equation, we get

Now,

solving both the equations simultaneously we get

v = (\frac{2M}{m+M})u_1+(\frac{m-M}{m+M})u_2

substituting the values in the above equation we get

v = (\frac{2\times594}{44+594})(-4.6)+(\frac{44-594}{44+594})4.6

or

v = -8.565-3.965

or

v = -12.53m/s

here negative sign depicts the motion of the ball in the upward direction

now the kinetic energy of the tennis ball

K.E = \frac{1}{2}mv^2

or

K.E = \frac{1}{2}44\times 10^{-3}kg\times 12.53^2

or

K.E = 3.45 J

also at the height the K.E will be the potential energy of the tennis ball

thus,

3.45 J = mgh

or

3.45 = 44 × 10⁻³ × 9.8 × h

h = 8.01 m

5 0
3 years ago
Please Help!!
Allushta [10]

Answer:

h = 219.96 m

Explanation:

Speed of the stone with which it was thrown horizontally, v = 10 m/s

We need to find the height of the cliff if the stone’s trajectory time from the top of the hill to the bottom to be 6.7s.

It means we need to find the distance covered by the stone. As the horizontal speed of the stone is given , it means there is no vertical motion in the stone,u'=0

Using second equation of motion,

d=u't+\dfrac{1}{2}at^2

Put u'=0 and a=g

So,

d=\dfrac{1}{2}gt^2\\\\\text{Putting all the values to find d}\\\\d=\dfrac{1}{2}\times 9.8\times 6.7^2\\\\d=219.96\ m

So, the height of the cliff is 219.96 m.

4 0
3 years ago
Can someone help me please? thank you
stepan [7]
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4 0
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