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3 years ago

Can someone help me?!!

Physics

1 answer:

Fantom [35]3 years ago

5 0

<h2>Hello!</h2>

The answer is:

The first option, the force tending to lift Rover is equal to 14.5 N.

To calculate the force that is tending to lift Rover vertically, we need to calculate the vertical component force.

Since we know that the angle between the force and the ground is 29°, we can calculate the vertical component of the force using the following formula:

$F_y=Force*Sin(29\°)$

We are given that the force is equal to 30.0 N, so, calculating we have:

$F_y=Force*Sin(29\°)$

$F_y=30N*Sin(29\°)=14.5N$

Also, we can calculate the horizontal component of the force using the following formula:

$F_x=Force*Cos(29\°)$

$F_x=30N*Cos(29\°)=26.24N$

Hence, we have that the correct option is the first option, the force tending to lift Rover is equal to 14.5 N.

Have a nice day!

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Consider the power dissipated in a series R–L–C circuit with R=280Ω, L=100mH, C=0.800μF, V=50V, and ω=10500rad/s. The current an

ki77a [65]

Answer:

0.28802

2.57162 W

14.28 W

53.55 W

6.07142 W

Explanation:

R = 280Ω

L = 100 mH

C = 0.800 μF

V = 50 V

ω = 10500rad/s

For RLC circuit impedance is given by

$Z=\sqrt{R^2+(X_L-X_C)^2}\\\Rightarrow Z=\sqrt{R^2+(\omega L-\dfrac{1}{\omega C})^2}\\\Rightarrow Z=\sqrt{280^2+(10500\times 100\times 10^{-3}-\dfrac{1}{10500\times 0.8\times 10^{-6}})^2}\\\Rightarrow Z=972.1483\ \Omega$

Power factor is given by

$F=\dfrac{R}{Z}\\\Rightarrow F=\dfrac{280}{972.1483}\\\Rightarrow F=0.28802$

The power factor is 0.28802

The average power to the circuit is given by

$P=\dfrac{V^2}{Z}\\\Rightarrow P=\dfrac{50^2}{972.1483}\\\Rightarrow P=2.57162\ W$

The average power to the circuit is 2.57162 W

Power to resistor

$P_R=IR\\\Rightarrow P_R=5.1\times 10^{-2}\times 280\\\Rightarrow P_R=14.28\ W$

Power to resistor is 14.28 W

Power to inductor

$P_L=IX_L\\\Rightarrow P_L=5.1\times 10^{-2}\times 10500\times 100\times 10^{-3}\\\Rightarrow P_L=53.55\ W$

Power to the inductor is 53.55 W

Power to the capacitor

$P_C=IX_C\\\Rightarrow P_C=5.1\times 10^{-2}\times \dfrac{1}{10500\times 0.8\times 10^{-6}}\\\Rightarrow P_C=6.07142\ W$

The power to the capacitor is 6.07142 W

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2 years ago

Determine the kinetic energy of a 10 Kg roller coaster car that is moving with a speed of 2m/s.

skelet666 [1.2K]

In this question all required information's are already provided. Based on these details the answer to the question can be easily determined. Let us now write down all the information's that are already given.

Mass of the roller coaster = 1000 kg

Velocity of the roller coaster = 20.0 m/s

We know the formula for finding the kinetic energy is

Kinetic energy = 0.5 * mass * (velocity) ^2

= 0.5 * 1000 * (20)^2

= 0.5 * 1000 * 400

= 200000 Joules

So the Kinetic energy of the roller coaster is 200000 joules.

i hope this helps you friend good luck on your quiz or lesson

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1 year ago

A small airplane flies 780 miles with an average speed of 260 miles per hour. 1.5 hours after the plane leaves, a Boeing 747 lea

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Answer:

520 miles per hour

Explanation:

Let the speed of the Boeing 747 be x miles per hour.

The small airplane covers distance of 780 miles with speed 260 miles per hour.

Also,

After 1.5 hours the Boeing 747 leave the same place and reaches at same time. Both covered distance of 780 miles.

So,

<u>Time taken by Boeing 747 + 1.5 hours = Time taken by small plane.</u>

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So,

780 / x + 1.5 = 780/ 260

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3 years ago

The answer and how to do it?? Thanks

denis-greek [22]

Answer:

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Explanation:

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(4N + 10 x 1kg)=(1kg)a

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