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Savatey [412]
2 years ago
14

A 15-g bullet moving at 300 m/s passes through a 2.0 cm thick sheet of foam plastic and emerges with a speed of 90 m/s. Let's as

sume that the speed of the bullet changes takes place uniformly (uniform deceleration). Find the change in momentum of the bullet. Find the time the bullet is in contact with the plastic, Delta t. Find the average force impeded by the bullet's motion through the plastic.
Physics
1 answer:
Shkiper50 [21]2 years ago
8 0

Answer:

Explanation:

a) Change in momentum, Δp = mΔv = m(v - u) = (15 * 10-3) * (90 - 300) = -3.15 kg-m2

b) Acceleration of the bullet, a = (v2 - u2) / 2s = (902 - 3002) / (2 * 0.02) = -2047500 m/s2

So, the bullet is in contact with the plastic for the time,  \bigtriangleupt = \frac{(v - u)}{a} =\frac{(90 - 300)}{(-2047500)} = 1.03 \times 10^{-4} s

c) Average force, F_{avg} =\frac{\bigtriangleup p}{\bigtriangleup t} =\frac{(-3.15)}{(1.03 \times 10^{-4})} = 30.6 kN

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Determine the sign (+ or −) of the torque about the elbow caused by the biceps, τbiceps, the sign of the weight of the forearm,
Alex Ar [27]
Ans: 
1.  τbiceps = +(Positive)
2.  τforearm = -(Negative)
3.  τball = -(Negative)

Explanation:

The figure is attached down below.

1. T<span>orque about the elbow caused by the biceps, τbiceps:
Since Torque = r x F (where r and F are the vectors)
</span>Where r is the vector from elbow to the biceps.
<span>
We can see in the figure that F(biceps) is in upward direction, and by applying the right hand rule from r to F, we get the counterclockwise direction. The torque in counterclockwise direction is positive(+). Therefore, the sign would be +.

2. </span>Torque about the the weight of the forearm, τforearm:
Since Torque = r x F (where r and F are the vectors)
Where r is the vector from elbow to the forearm.

Also weight is the special kind of Force caused by the gravity.

We can see in the figure that W(forearm) is in downward direction, and by applying the right hand rule from r to F, we get the clockwise direction. The torque in clockwise direction is negative(-). Therefore, the sign would be -.

3. Torque about the the weight of the ball, τball:
Since Torque = r x F (where r and F are the vectors)
Where r is the vector from elbow to the ball.

Also weight is the special kind of Force caused by the gravity.

We can see in the figure that W(ball) is in downward direction, and by applying the right hand rule from r to F, we get the clockwise direction. The torque in clockwise direction is negative(-). Therefore, the sign would be -.

8 0
2 years ago
If it takes an airplane 15 minutes to go from 30 mph to 330 mph, what is its acceleration?
zzz [600]
Using the a=vf-vi divided by tf-ti:
A is acceleration
Vf is final velocity- 330
Vi is intial velocity-30
Tf is final time-15
Ti is initial time-0
A = 330-30 divided by 15-0
A = 300 divided by 15
A= 20 m/s^2 
Hope this helps
3 0
2 years ago
Your town is installing a fountain in the main square. If the water is to rise 26.0 m (85.3 feet) above the fountain, how much p
Brums [2.3K]

Answer:

P = 3.55 \times 10^5 Pa

Explanation:

As we know that water from the fountain will raise to maximum height

H = 26.0 m

now by energy conservation we can say that initial speed of the water just after it moves out will be

\frac{1}{2}mv^2 = mgH

v = \sqrt{2gH}

v = \sqrt{2(9.81)(26)}

v = 22.6 m/s

Now we can use Bernuolli's theorem to find the initial pressure inside the pipe

P = P_0 + \frac{1}{2}\rho v^2

P = 10^5 + \frac{1}{2}(1000)(22.6^2)

P = 3.55 \times 10^5 Pa

6 0
2 years ago
A 6.70 −μC particle moves through a region of space where an electric field of magnitude 1500 N/C points in the positive x direc
Alborosie

The given question is incomplete. The complete question is as follows.

A 6.70 −μC particle moves through a region of space where an electric field of magnitude 1500 N/C points in the positive x direction, and a magnetic field of magnitude 1.25 T points in the positive z direction.

A) If the net force acting on the particle is 6.21 \times 10^{-3} N in the positive x direction, find the components of the particle's velocity. Assume the particle's velocity is in the x-y plane.

Enter your answers numerically separated by commas.

Explanation:

The given data is as follows.

           Q = 6.50 \times 10^{-6} C

           E = 1300 N/C in the +x direction

           B = 1.02 T in the +z direction

and,    F_{net} = 6.25 \times 10^{-3} N in the +x direction

Also,       F_{net} = F_{E} - F_{b}

                         = qE - qvB

Now, we will calculate the value of v as follows.

             v = (\frac{1}{B}) \times (E - \frac{F_{net}}{q})

                 = (\frac{1}{1.02 T}) \times (1300 - \frac{6.25 \times 10^{-3}}{6.50 \times 10^{-6}})

                v = 458.507 m/s

Using the value for velocity, we need to know which direction it's going.

You know +x direction for E, +z direction for B and +x for F_{net}.

Using the right hand rule where:

your right thumb goes toward the F_{net}, then your index finger points to B (z direction) Then curl your middle, ring, and pink 90 angle. This shows where v is going which is -y direction.

Thus, we can conclude that v_{x}, v_{y}, v_{z} = 0, -(458.507), 0.

8 0
2 years ago
A car travels forward with constant velocity. It goes over a small stone, which gets stuck in the groove of a tire. The initial
puteri [66]

Answer:

A) vertically upward

Explanation:

Since the tyre is rotating with uniform angular speed and moving with constant linear speed

So as soon as a small stone is stuck into the groove of the tyre the speed of the stone is same as that of the tyre

so now we can say that stone will start revolving with the tyre of the car at constant angular speed and moving with uniform speed also

so here just after that the tangential acceleration of the stone must be zero while radial acceleration must be towards the center of the tyre given as

a_c = \omega^2 R

so we will have direction of net acceleration is towards its center so correct answer will be

A) vertically upward

7 0
2 years ago
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