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4 years ago

I really need help this was due Friday! I will mark brainiest!

Physics

1 answer:

Svetllana [295]4 years ago

6 0

Answer:

Average speed = (total distance covered) / (time to cover the distance)

You only need the first and last points from the table.

First point: Position=0, Time=0

Last point: Position= 16.34, Time = 2.34 hours

Total distance covered = 16.34 (miles ? km ?)

Time to cover the distance = 2.34 hours

Average speed = (16.34 miles) / (2.34 hours)

Average speed = 6.983 mile/hour

Explanation:

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Find the kinetic energy of a 0.1 kg toy truck moving at a speed of 1.1 m/s

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0.0605J is your answer. Use the formula KE=1/2mv^2

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4 years ago

If a player through a basketball to the target with an initial velocity of 17 m/s making an angle of 30 degrees with the horizon

Svetllana [295]

Answer:

The final position made with the vertical is 2.77 m.

Explanation:

Given;

initial velocity of the ball, V = 17 m/s

angle of projection, θ = 30⁰

time of motion, t = 1.3 s

The vertical component of the velocity is calculated as;

$V_y = Vsin \theta\\\\V_y = 17 \times sin(30)\\\\V_y = 8.5 \ m/s$

The final position made with the vertical (Yf) after 1.3 seconds is calculated as;

$Y_f = V_yt - \frac{1}{2}g t^2\\\\Y_f = (8.5 \times 1.3 ) - (\frac{1}{2} \times 9.8 \times 1.3^2)\\\\Y_f = 11.05 \ - \ 8.281\\\\Y_f = 2.77 \ m$

Therefore, the final position made with the vertical is 2.77 m.

3 0

3 years ago

n electromagnetic wave in vacuum has an electric field amplitude of 611 V/m. Calculate the amplitude of the corresponding magnet

enot [183]

Answer:

The corresponding magnetic field is

Explanation:

From the question we are told that

The electric field amplitude is $E_o = 611\ V/m$

Generally the magnetic field amplitude is mathematically represented as

$B_o = \frac{E_o }{c }$

Where c is the speed of light with a constant value

$c = 3.0 *0^{8} \ m/s$

$B_o = \frac{611 }{3.0*10^{8}}$

$B_o = 2.0 4 *10^{-6} \ Vm^{-2} s$

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$B_o = 2.0 4 *10^{-6} \ T$

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ra1l [238]

Answer:

The third particle should be at 0.0743 m from the origin on the negative x-axis.

Explanation:

Let's assume that the third charge is on the negative x-axis. So we have:

$E_{1}+E_{3}-E_{2}=0$

We know that the electric field is:

$E=k\frac{q}{r^{2}}$

Where:

k is the Coulomb constant
q is the charge
r is the distance from the charge to the point

So, we have:

$k\frac{q_{1}}{r_{1}^{2}}+k\frac{q_{3}}{r_{3}^{2}}-k\frac{q_{2}}{r_{2}^{2}}=0$

Let's solve it for r(3).

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Therefore, the third particle should be at 0.0743 m from the origin on the negative x-axis.

I hope it helps you!

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I think the right answer is c

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