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gladu [14]
3 years ago
15

a rifle is used during the biathlon event in the Olympics the rifle has a mass of 2.5 kg the bullet has amass of 0.02 kg after t

hat shot the bullet travels at a speed of 525 m/s what is the total momentum of the rifle bullet system before the bullet is shot?
Physics
2 answers:
lilavasa [31]3 years ago
8 0

Answer:

the momentum of the system must be ZERO just before the bullet is shot

Explanation:

As we know that initially the bullet is inside the gun with stationary condition

So here we can say that gun and bullet was both at rest initially

so we have

P = m_1v_1 + m_2v_2

since both are at rest so we have

v_1 = v_2 = 0

so total momentum of the system will be

P = 0

So the momentum of the system must be ZERO just before the bullet is shot

Nataliya [291]3 years ago
6 0
The momentum before is zero because nothing is moving. Note that because momentum is a conserved vector, the total momentum after the shot is also zero. You just have to add the products of mass and velocity in opposite directions. So the answer is zero
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Sever21 [200]

Every force has an equal but opposite reaction, the forces are balanced out on the brick.

Weight (mg) acts downwards from the brick but the normal contact force from the table is acting upwards therefore resultant force is 0N equilibrium

You know the direction of the force is based upon the resultant force, so if you were to push the brick horizontally across the table it would travel horizontally

6 0
3 years ago
A 15kg beam that is 10m long is placed on a fulcrum that is 3m from the end an 80kg person sits at the end closer to the fulcrum
Andrei [34K]

Answer:

 m₃ = 30 kg

Explanation:

This is a problem of rotational equilibrium, let's write Newton's law for rotational equilibrium.

Let's fix our reference system in the support, the positive torques are those that create an anti-clockwise turn

Let's look for the distances to the point of support

The distance of man

      x₁ = -3 m

The distance of the bar is

      x₂ = L / 2 -3

     x₂ = 10/2 -3

     x₂ = 2 m

Remote object at the end

      x₃ = L-3

     x₃ = 10-3

     x₃ = 7 m

They give us the mass of man (m1) and the mass of the bar (m2)

Let's write the torques

      W₁ x₁ - W₂ x₂ - w₃ x₃ = 0

       m₁ g 3 - m₂ g 2 - m₃ g 7 = 0

       m₃ = (3m₁ -2m₂) / 7

Let's calculate

      m₃ = (3 80 -2 15) / 7

     m₃ = 30 kg

7 0
3 years ago
A proton beam in an accelerator carries a current of 130 μa. if the beam is incident on a target, how many protons strike the ta
Ivanshal [37]
The current intensity is the product between the total charge that flows through a certain point (in our case, the target) in a time interval \delta t:
I= \frac{Q}{\Delta t}
We know the current, I=130 \mu A=130 \cdot 10^{-6} A, and the time interval, \Delta t=17 s, so we can find the total charge:
Q=I \Delta t= 2.21 \cdot 10^{-3}C&#10;

The total charge Q is the product between the number of protons N and the charge of each protons, e, which is e=1.6 \cdot 10^{-19}C:
Q=Ne
we can  re-write the equation solving for N, so we can find the number of protons striking the target in 17 s:
N= \frac{Q}{e}= \frac{2.21 \cdot 10^{-3}C}{1.6 \cdot 10^{-19}C} =1.38 \cdot 10^{16}
4 0
3 years ago
HELP QUICK: A pilot performs a vertical maneuver around a circle with a radius R. When the
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Answer:

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Explanation:

5 0
3 years ago
The tub of a washer goes into its spin-dry cycle, starting from rest and reaching an angular speed of 4.0 rev/s in 9.0 s. At thi
iris [78.8K]

Answer:

48 rev

Explanation:

a) we can calculate the distance covered by the tube using the formula:

θ = (ω + ωo)t/2

where ω is the final angular speed, θ is the distance covered, t is the time taken, ωo is the initial angular speed.

Firstly, we calculate the distance covered from 0 to 9 s then from 9s to 24 s.

within 9s, the tub runs from rest (0) to 4 rev/s, hence:

t = 9s, wo = 0, w = 4 rev/s = (4 * 2π) rad/s = 8π rad/s. Hence:

θ = (ω + ωo)t/2 = (0 + 8π)9 / 2 = 36π rad

θ = 36π rad = (36π)/2π rev = 18 rev

Also, within 15 s, the tub runs from 4 rev/s to rest, hence:

t = 15 s, wo = 4 rev/s = 8π rad/s, w = 0 rad/s. Hence:

θ = (ω + ωo)t/2 = (8π + 0)15 / 2 = 60π rad

θ = 60π rad = (60π)/2π rev = 30 rev

Therefore the total revolutions by the tube during 24 s interval = 30 rev + 18 rev = 48 rev

4 0
3 years ago
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