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3 years ago

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The two identical trucks are traveling in opposite directions, and truck B is carrying a heavy load. (a) Will the magnitude of t

he force exerted on truck A by truck B be greater than, less than, or equal to the magnitude of the force exerted on truck B by truck A

1 answer:

Dmitry [639]3 years ago

6 0

Answer:

The force exerted on truck A by truck B will definitely be greater than the magnitude of the force exerted on truck B by truck A

Explanation:

According to the concept of momentum

Momentum is the product of a body's mass and its velocity.

Yes both trucks are the same, but truck B is carrying a heavy load which gives it more momentum that A, actually momentum is a function of mass

Momentum M=mass(m) x velocity (v)

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Calculate the ratio of the resistance of 12.0 m of aluminum wire 2.5 mm in diameter, to 30.0 m of copper wire 1.6 mm in diameter

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Answer: 0.258

Explanation:

The resistance $R$ of a wire is calculated by the following formula:

$R=\rho\frac{l}{s}$ (1)

Where:

$\rho$ is the resistivity of the material the wire is made of. For aluminium is $\rho_{Al}=2.65(10)^{-8}m\Omega$ and for copper is $\rho_{Cu}=1.68(10)^{-8}m\Omega$

$l$ is the length of the wire, which in the case of aluminium is $l_{Al}=12m$ , and in the case of copper is $l_{Cu}=30m$

$s$ is the transversal area of the wire. In this case is a circumference for both wires, so we will use the formula of the area of the circumference:

$s=\pi{(\frac{d}{2})}^{2}$ (2) Where $d$ is the diameter of the circumference.

For aluminium wire the diameter is $d_{Al}=2.5mm=0.0025m$ and for copper is $d_{Cu}=1.6mm=0.0016m$

So, in this problem we have two transversal areas:

<u>For aluminium:</u>

$s_{Al}=\pi{(\frac{d_{AL}}{2})}^{2}=\pi{(\frac{0.0025m}{2})}^{2}$

$s_{Al}=0.000004908m^{2}$ (3)

<u>For copper:</u>

$s_{Cu}=\pi{\frac{(d_{Cu}}{2})}^{2}=\pi{(\frac{0.0016m}{2})}^{2}$

$s_{Cu}=0.00000201m^{2}$ (4)

Now we have to calculate the resistance for each wire:

<u>Aluminium wire:</u>

$R_{Al}=2.65(10)^{-8}m\Omega\frac{12m}{0.000004908m^{2}}$ (5)

$R_{Al}=0.0647\Omega$ (6) Resistance of aluminium wire

<u>Copper wire:</u>

$R_{Cu}=1.68(10)^{-8}m\Omega\frac{30m}{0.00000201m^{2}}$ (6)

$R_{Cu}=0.250\Omega$ (7) Resistance of copper wire

At this point we are able to calculate the ratio of the resistance of both wires:

$Ratio=\frac{R_{Al}}{R_{Cu}}$ (8)

$\frac{R_{Al}}{R_{Cu}}=\frac{0.0647\Omega}{0.250\Omega}$ (9)

Finally:

$\frac{R_{Al}}{R_{Cu}}=0.258$ This is the ratio

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A hollow conducting spherical shell has radii of 0.80 m and 1.20 m, The radial component of the electric field at a point that i

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Complete Question

The complete question is shown on the first uploaded image

Answer:

The electric field at that point is $E = 7500 \ N/C$

Explanation:

From the question we are told that

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The radius of the outer circle is $r_o = 1.20 \ m$

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substituting values

$E = \frac{k * q_c }{x^2}$

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