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marysya [2.9K]
3 years ago
13

Your car and the trailer it is towing are initially at rest. you apply an acceleration of 1.75 m/s2 to get the vehicles moving.

the mass of your car is 1200 kg and the mass of the trailer is 540 kg. (a) determine the net force acting on the car.
Physics
1 answer:
kirza4 [7]3 years ago
4 0
The net force is defined as the vector sum of all the forces that are present in an object. The formula for net force is F=ma. Everything else cancels out like the force of gravity on the car, and the reactive force of the road surface, etc. It can be concluded that the force exerted by the trailer on the car is the same as that exerted by the car on the trailer. Given that the acceleration is 1.75 m/s^2, the car is 1200 kilograms, and the trailer is 540 kilograms. Thus, the net force acting on the car is equal to 21qkm/hr^2 
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A fruit bat falls from the roof of a cave. We know that her potential energy was
bulgar [2K]

Answer:

v = 15.65 m/s

Explanation:

We use conservation of mechanical energy between initial (i) and final (f) states:

Pi + KEi = Pf + KEf

At the top of the cave at the instant the bat starts to fall, there is only potential energy since the bat's velocity is zero.

Pi = m g h = 600 J

and the KEi = 0 J (no velocity)

Knowing the height of the cave's roof (12.8 m) , we can find the mass of the bat:

m = 600 J / (g 12.5) = 4.9 kg

Using conservation of mechanical energy, the final state is:

Pf + KEf = 600 J

with Pf = 0 (just touching the ground)

KEf= 1/2  4.9 (v^2)

and we solve for the velocity:

600 J = 0 + 1/2  4.9 (v^2)

v^2 = 600 * 2 / 4.9 = 244.9

v = 15.65 m/s

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2 years ago
A uniformly charged ball of radius a and charge –Q is at the center of a hollowmetal shell with inner radius b and outer radius
vlabodo [156]

Answer:

<u>r < a:</u>

E = \frac{1}{4\pi \epsilon_0}\frac{Qr}{a^3}

<u>r = a:</u>

E = \frac{1}{4\pi a^2}\frac{Q}{\epsilon_0}

<u>a < r < b:</u>

E = \frac{1}{4\pi \epsilon_0}\frac{Q}{r^2}

<u>r = b:</u>

E = \frac{1}{4\pi b^2}\frac{Q}{\epsilon_0}

<u>b < r < c:</u>

E = 0

<u>r = c:</u>

E = \frac{1}{4\pi \epsilon_0}\frac{Q}{c^2}

<u>r < c:</u>

E = \frac{1}{4\pi \epsilon_0}\frac{Q}{r^2}

Explanation:

Gauss' Law will be applied to each region to find the E-field.

\int \vec{E}d\vec{a} = \frac{Q_{encl}}{\epsilon_0}

An imaginary sphere is drawn with radius r, which is equal to the point where the E-field is asked. The area of this imaginary sphere is multiplied by E, and this is equal to the charge enclosed by this imaginary surface divided by ε0.

<u>r<a:</u>

Since the ball is uniformly charged and not hollow, then the enclosed charge can be found by the following method: If the total ball has a charge -Q and volume V, then the enclosed part of the ball has a charge Q_enc and volume V_enc. Then;

\frac{Q}{V} = \frac{Q_{encl}}{V_{encl}}\\\frac{Q}{\frac{4}{3}\pi a^3} = \frac{Q_{encl}}{\frac{4}{3}\pi r^3}\\Q_{encl} = \frac{Qr^3}{a^3}

Applying Gauss' Law:

E4\pi r^2 = \frac{-Qr^3}{\epsilon_0 a^3}\\E = -\frac{1}{4\pi \epsilon_0}\frac{Qr}{a^3}\\E = \frac{r}{4\pi a^3}\frac{Q}{\epsilon_0}

The minus sign determines the direction of the field, which is towards the center.

<u>At r = a: </u>

E = \frac{1}{4\pi a^2}\frac{Q}{\epsilon_0}

<u>At a < r < b:</u>

The imaginary surface is drawn between the inner surface of the metal sphere and the smaller ball. In this case the enclosed charge is equal to the total charge of the ball, -Q.

<u />E4\pi r^2 = \frac{-Q}{\epsilon_0}\\E = -\frac{1}{4\pi \epsilon_0}\frac{Q}{r^2}<u />

<u>At r = b:</u>

<u />E = -\frac{1}{4\pi b^2}\frac{Q}{\epsilon_0}<u />

Again, the minus sign indicates the direction of the field towards the center.

<u>At b < r < c:</u>

The hollow metal sphere has a net charge of +2Q. Since the sphere is a conductor, all of its charges are distributed across its surface. No charge is present within the sphere. The smaller ball has a net charge of -Q, so the inner surface of the metal sphere must possess a net charge of +Q. Since the net charge of the metal sphere is +2Q, then the outer surface of the metal should possess +Q.

Now, the imaginary surface is drawn inside the metal sphere. The total enclosed charge in this region is zero, since the total charge of the inner surface (+Q) and the smaller ball (-Q) is zero. Therefore, the Electric region in this region is zero.

E = 0.

<u>At r < c:</u>

The imaginary surface is drawn outside of the metal sphere. In this case, the enclosed charge is +Q (The metal (+2Q) plus the smaller ball (-Q)).

E4\pi r^2 = \frac{Q}{\epsilon_0}\\E = \frac{1}{4\pi \epsilon_0}\frac{Q}{r^2}

<u>At r = c:</u>

E = \frac{1}{4\pi \epsilon_0}\frac{Q}{c^2}

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