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IgorLugansk [536]

3 years ago

5

A 1.5kg cannon is loaded with a 0.52 kg ball . The cannon is ignited and launches the ball forward with speed of 75m/s. Determin

e the post explosion velocity of the cannon.

1 answer:

Mila [183]3 years ago

4 0

Answer:

-26 m/s (backward)

Explanation:

We can solve this problem by using the law of conservation of momentum.

In fact, the total momentum momentum of the cannon + ball system must be conserved before and after the explosion.

Before the explosion, they are both at rest, so the total momentum is zero:

p = 0

After the explosion, the total momentum is:

$p=MV+mv$

where

M = 1.5 kg is the mass of the cannon

m = 0.52 kg is the mass of the ball

v = +75 m/s is the velocity of the ball

V is the velocity of the cannon

Since the momentum is conserved, we can equate the two expressions:

$0=MV+mv$

And solving, we find V:

$V=-\frac{mv}{M}=-\frac{(0.52)(+75)}{1.5}=-26 m/s$

where the negative sign means the direction is opposite to that of the ball.

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A car horn emits a frequency of 400 Hz. A car traveling at 20.0 m/s sounds the horn as it approaches a stationary pedestrian. Wh

Temka [501]

Answer:

The observed frequency by the pedestrian is 424 Hz.

Explanation:

Given;

frequency of the source, Fs = 400 Hz

speed of the car as it approaches the stationary observer, Vs = 20 m/s

Based on Doppler effect, as the car the approaches the stationary observer, the observed frequency will be higher than the transmitted (source) frequency because of decrease in distance between the car and the observer.

The observed frequency is calculated as;

$F_s = F_o [\frac{v}{v_s + v} ] \\\\$

where;

F₀ is the observed frequency

v is the speed of sound in air = 340 m/s

$F_s = F_o [\frac{v}{v_s + v} ] \\\\400 = F_o [\frac{340}{20 + 340} ] \\\\400 = F_o (0.9444) \\\\F_o = \frac{400}{0.9444} \\\\F_o = 423.55 \ Hz \\$

F₀ ≅ 424 Hz.

Therefore, the observed frequency by the pedestrian is 424 Hz.

8 0

3 years ago

Each of the following diagrams shows a spaceship somewhere along the way between Earth and the Moon (not to scale); the midpoint

djyliett [7]

Answer:

$F_5 >F_4>F_1 >F_2>F_3$

Where $F_i$ represent the force for each of the 5 cases $i -1,2,3,4,5$ presented on the figure attached.

Explanation:

For this case the figure attached shows the illustration for the problem

We have an inverse square law with distance for the force, so then the force of gravity between Earth and the spaceship is lower when the spaceship is far away from Earth.

Th formula is given by:

$F = G \frac{m_{Earth} m_{Spaceship}}{r^2}$

Where G is a constant $G = 6.674 x10^{-11} m^2/ (ks s^2)$

$m_{Earth}$ represent the mass for the earth

$m_{spaceship}$ represent the mass for the spaceship

$r$ represent the radius between the earth and the spaceship

For this reason when the distance between the Earth and the Spaceship increases the Force of gravity needs to decrease since are inversely proportional the force and the radius, and for the other case when the Earth and the spaceship are near then the radius decrease and the Force increase.

Based on this case we can create the following rank:

$F_5 >F_4>F_1 >F_2>F_3$

Where $F_i$ represent the force for each of the 5 cases $i -1,2,3,4,5$ presented on the figure attached.

6 0

3 years ago

A solid conducting sphere with radius R that carries positive charge Q is concentric with a very thin insulating shell of radius

ahrayia [7]

Answer:

The specific question is not stated, however the general idea is given in the attached picture. The electric field in each region can be found by Gauss’ Law.

at r < R:

Since the solid sphere is conducting, the total charge Q is distributed over the surface, and the electric field inside the sphere is zero.

E = 0.

at R < r < 2R:

The electric field can be found by Gauss’ Law as in the attachment. The green pencil shows this exact region.

at 2R < r:

The electric field can again be found by Gauss’ Law, the blue pencil shows the calculations for this region.

Explanation:

Gauss’ Law is straightforward when applied to spheres. The area of the sphere is $A = 4\pi r^2$ , and the enclosed charge is given in the question as Q for the inner sphere, and 2Q for the whole system.

3 0

3 years ago

A heat source could be considered a source of pollution Please select the best answer from the choices provided

jeyben [28]

The answer is true, is called Thermal pollotion:).

5 0

2 years ago

Three people pull simultaneously on a stubborn donkey. Jack pulls directly ahead of the donkey with a force of 77.3 N, Jill pull

DiKsa [7]

Answer:

$F_{net} = 232.8 N$

towards right so it is -15 degree

Explanation:

Net force in forward direction due to all three is given as

$F_x = F_1 + F_2cos45 + F_3cos45$

here we know that

$F_1 = 77.3 N$

$F_2 = 61.7 N$

$F_3 = 147 N$

$F_x = 77.3 + 61.7 cos45 + 147 cos45$

$F_x = 224.9 N$

Similarly in Y direction we will have

$F_y = F_3 sin45 - F_2 sin45$

$F_y = (147 - 61.7)sin45$

$F_y = 60.3 N$

Now the net force on the donkey is given as

$F_{net} = \sqrt{F_x^2 + F_y^2}$

$F_{net} = \sqrt{224.9^2 + 60.3^2}$

$F_{net} = 232.8 N$

Now direction of force is given as

$tan\theta = \frac{F_y}{F_x}$

$tan\theta = \frac{60.3}{224.9}$

$\theta = 15^o$ towards right so it is -15 degree

3 0

3 years ago