A tool for studying motion is called?

Calculate the recoil speed of a 1.4 kg rifle shooting 0.006 kg bullets with muzzle speed of 800 m/a.3.43 m/s

Angelina_Jolie [31]

Answer:

a) 3.43 m/s

Explanation:

Due to the law of conservation of momentum, the total momentum of the bullet - rifle system must be conserved.

The total momentum before the bullet is shot is zero, because they are both at rest, so:

$p_i = 0$

Instead the total momentum of the system after the shot is:

$p_f = mv+MV$

where:

m = 0.006 kg is the mass of the bullet

M = 1.4 kg is the mass of the rifle

v = 800 m/s is the velocity of the bullet

V is the recoil velocity of the rifle

The total momentum is conserved, therefore we can write:

$p_i = p_f$

Which means:

$0=mv+MV$

Solving for V, we can find the recoil velocity of the rifle:

$V=-\frac{mv}{M}=-\frac{(0.006)(800)}{1.4}=-3.43 m/s$

where the negative sign indicates that the velocity is opposite to direction of the bullet: so the recoil speed is

a) 3.43 m/s

5 0

3 years ago

What is this question formula a=V²-U² /2S?

erik [133]

Answer:

sorry I don't really know about that question.

Explanation:

7 0

3 years ago

Three crates with various contents are pulled by a force Fpull=3615 N across a horizontal, frictionless roller‑conveyor system.

SIZIF [17.4K]

The question is incomplete. Here is the complete question.

Three crtaes with various contents are pulled by a force Fpull=3615N across a horizontal, frictionless roller-conveyor system.The group pf boxes accelerates at 1.516m/s2 to the right. Between each adjacent pair of boxes is a force meter that measures the magnitude of the tension in the connecting rope. Between the box of mass m1 and the box of mass m2, the force meter reads F12=1387N. Between the box of mass m2 and box of mass m3, the force meter reads F23=2304N. Assume that the ropes and force meters are massless.

(a) What is the total mass of the three boxes?

(b) What is the mass of each box?

Answer: (a) Total mass = 2384.5kg;

(b) m1 = 915kg;

m2 = 605kg;

m3 = 864.5kg;

Explanation: The image of the boxes is described in the picture below.

(a) The system is moving at a constant acceleration and with a force Fpull. Using Newton's 2nd Law:

$F_{pull}=m_{T}.a$

$m_{T}=\frac{F_{pull}}{a}$

$m_{T}=\frac{3615}{1.516}$

$m_{T}=2384.5$

Total mass of the system of boxes is 2384.5kg.

(b) For each mass, analyse each box and make them each a free-body diagram.

For $m_{1}$ :

The only force acting On the $m_{1}$ box is force of tension between 1 and 2 and as all the system is moving at a same acceleration.

$m_{1} = \frac{F_{12}}{a}$

$m_{1} = \frac{1387}{1.516}$

$m_{1}$ = 915kg

For $m_{2}$ :

There are two forces acting on $m_{2}$ : tension caused by box 1 and tension caused by box 3. Positive referential is to the right (because it's the movement's direction), so force caused by 1 is opposing force caused by 3:

$m_{2} = \frac{F_{23}-F_{12}}{a}$

$m_{2} = \frac{2304-1387}{1.516}$

$m_{2}$ = 605kg

For $m_{3}$ :

$m_{3} = m_{T} - (m_{1}+m_{2})$

$m_{3} = 2384.5-1520.0$

$m_{3}$ = 864.5kg

8 0

3 years ago

Sam is conducting an experiment in a closed room. He places two sound-generating instruments, A and B, in different places in th

diamong [38]

Answer: The correct answer is "Instrument A is placed closer to Sam than instrument B".

Explanation:

The sound can be soft or loud. Loudness depends on the amplitude of the sound wave. Higher the amplitude, more will be loudness. Lower the amplitude, lesser will be loudness.

Pitch depends on the frequency.

In the given problem, the instruments A and B generate sound waves of the same amplitude and at the same time.

Loudness depends on the sound energy produced as the energy of the sound is directly proportional to the square of the amplitude. It also depends on the distance between the source and the receiver.

Sam records a louder sound from instrument A than from instrument B. It means that there is mismatch in loudness. It can happen due to the placement of the instrument A closer to Sam than instrument B.

Therefore, the correct option is "Instrument A is placed closer to Sam than instrument B".

4 0

4 years ago

A plane wall with constant properties is initially at a uniform temperature To. Suddenly, the surface at x = L is exposed to a c

Rzqust [24]

Answer:

The distribution is as depicted in the attached figure.

Explanation:

From the given data

The plane wall is initially with constant properties is initially at a uniform temperature, To.
Suddenly the surface x=L is exposed to convection process such that T∞>To.
The other surface x=0 is maintained at To

Uniform volumetric heating q' such that the steady state temperature exceeds T∞.

Assumptions which are valid are

There is only conduction in 1-D.
The system bears constant properties.
The volumetric heat generation is uniform

From the given data, the condition are as follows

Initial Condition

At t≤0

$T(x,0)=T_o$

This indicates that initially the temperature distribution was independent of x and is indicated as a straight line.

Boundary Conditions

At x=0

$T(0,t)=T_o$

This indicates that the temperature on the x=0 plane will be equal to To which will rise further due to the volumetric heat generation.

At x=L

$-k\frac{\partial T}{\partial x}]_{x=L}=h[T(L,t)-T_{\infty}]$

This indicates that at the time t, the rate of conduction and the rate of convection will be equal at x=L.

The temperature distribution along with the schematics are given in the attached figure.

Further the heat flux is inferred from the temperature distribution using the Fourier law and is also as in the attached figure.

It is important to note that as T(x,∞)>T∞ and T∞>To thus the heat on both the boundaries will flow away from the wall.

3 0

3 years ago