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IgorC [24]
2 years ago
10

A bullet has a mass of 0.06 kg. Starting from rest, after the gun's trigger is pulled, a constant force acts on the bullet for t

he next 0.025 seconds until the bullet leaves the barrel of the gun with a speed of 992 m/s.
What is the change in momentum of the bullet?
Physics
1 answer:
artcher [175]2 years ago
6 0

The change in momentum of the bullet : 59.52 kg m/s

<h3>Further explanation</h3>

Given

m=0.06 kg

Δt=0.025 s

vo=0(from rest)

vt= 992 m/s

Required

The change in momentum

Solution

The change in momentum  = ΔP

ΔP  =m(vt-vo)

ΔP =0.06(992-0)

ΔP =59.52 kg m/s

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Isaac drops a rubber ball drom height of 2.0m and it bounces to a height of 1.5m. a) What fraction of it's initial energy is los
True [87]

Answer:

a)  ΔE = 25 %

b) v = 8,85 m/s

c) The energy was used against air resistance

Explanation:

In any situation total energy of a body is equal to potential energy plus

kinetic energy, then, just at the moment when Isaac dop the ball the situation is:

Ei = Ep + Ek         where     Ep = m*g*h    and  Ek = 1/2*m*v²

As v = 0  (Isaac drops the ball)

Ei = Ep  =  m*g*h    = 2*m*g

At the end (when the ball bounced to 1,5 m

E₂ = Ep₂  + Ek₂         again at that point  v =0 and

E₂ = 1,5*m*g*

Ei =  E₂ + E(lost)

E(lost) = Ei - E₂

E(lost) = 2*m*g* - 1,5*m*g      and the fraction of energy lost is

E(lost)/Ei

ΔE = (2*m*g* - 1,5*m*g )/ 2*m*g

ΔE = 0,5*m*g / 2*m*g

ΔE = 0,5/2

ΔE = 0,25      or     ΔE = 25 %

b) The speed of the ball is

Potential energy is converted in kinetic energy just when the ball is touching the ground, then

m*g*h = 1/2*m*v²

2*h*g = 1/2 *v²

v² = 4*g*h

v² = 4*2*9,8

v² = 78,4

v = 8,85 m/s

If the impact is an elastic collision, then Ek before and after the impact is the same.

7 0
3 years ago
AN airplane travels 4000 m in 20s on a heading 0f 35 degrees north west. Calculate average velocity .
alexandr402 [8]
Isn't velocity Distance over time? if the degree isn't adding resistance it should be 4000 ÷ 20 which gives you 200mps ("per second") which is the velocity without resistance.
6 0
3 years ago
Sam is pulling a box up to the second story of his apartment via a string. The box weighs 16.5 kg and starts from rest on the gr
Katyanochek1 [597]

Answer:

Weight (mass) = 16.5 kg

velocity = 0 m/a

acceleration =2.6 m/s^2

displacement = 13.2m

now,

acceleration = velocity/ time

2.6 = 0 / t

t = o / 2.6

t = o

8 0
3 years ago
Grindstone: Rotational Dynamics and Kinematics You have a grindstone (a disk) that is 133.0 kg, has a 0.635 m radius, and is tur
Colt1911 [192]

Answer:

Ff = 839.05 N

Explanation:

We can use the equation:

Ff = μ*N

where <em>N</em> can be obtained as follows:

∑ Fc = m*ac   ⇒   N - F = m*ac = m*ω²*R    ⇒  N = F + m*ω²*R

then if

F = 32 N

m = 133 Kg

R = 0.635 m

ω = 95 rev /min = (95 rev / min)(2π rad / 1 rev)(1 min / 60 s) = 9.9484 rad /s

we get

N = 32 N + (133 Kg)*(9.9484 rad /s)²*(0.635 m) = 8390.53 N

Finally

Ff = μ*N = 0.10*(8390.53 N) = 839.05 N

3 0
3 years ago
what is the force if a block of wood with mass 20 kg slides along a frictionless surface at 2 m/s squared
CaHeK987 [17]

Answer:

40 N

Explanation:

F=ma where F is the applied force, m is the mass of object and a is the acceleration.

Since there is no friction, substituting 20 Kg for m and 2 m/s squared for a then we obtain

F=20*2=40 N

5 0
3 years ago
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