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3 years ago

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Help questions 5&6 about force

1 answer:

Iteru [2.4K]3 years ago

8 0

Force ehhehehe force hehehehe force anyway yeah that’s right

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Andy pushes a box to the right with 5 N of force. Kristen pushes the box to the left with 3 N of force. What will be the resulti

REY [17]

The forces are applied in opposite direction.
So the net force will be the difference of both forces.
net force =5-3=2N

This force will be in direction of 5N(bigger force) means in direction which Andy is pushing.

6 0

3 years ago

A circular wire loop has a radius of 7.50 cm. a sinusoidal electromagnetic plane wave traveling in air passes throughthe loop, w

ExtremeBDS [4]

Intensity of electromagnetic wave is given as

$I = 2[\frac{B_{rms}^2}{2\mu_0}\times c]$

given that

$I = 0.0275 W/m^2$

$0.0275 = \frac{B_{rms}^2}{\mu_0} \times c$

here we know that

$\mu_0 = 4\pi \times 10^{-7}$

$c = 3 \times 10^8 m/s$

now we have

$0.0275 = \frac{B_{rms}^2}{4 \pi \times 10^{-7}}(3 \times 10^8)$

$B_{rms} = 1.07 \times 10^{-8} T$

now we will have

$B_0 = \sqrt2 B_{rms}$

$B_0 = 1.52 \times 10^{-8} T$

frequency of wave is given as

$f = \frac{c}{\lambda}$

$f = \frac{3 \times 10^8}{6.90} = 4.35 \times 10^7 Hz$

now the induced EMF is given as

$EMF = B_0A2\pi f$

$EMF = 1.52 \times 10^{-8} \times \pi(0.075)^2 \times (2\pi \times 4.25 \times 10^7)$

$EMF = 0.0733 Volts$

7 0

4 years ago

Read 2 more answers

The answer is not B.

mezya [45]

Answer:

D

Explanation:

15m/ s

i hope this helps

4 0

2 years ago

What is the best piece of advice you have ever received or given?

Afina-wow [57]

Answer:Your life is your responsibility.

Explanation:

4 0

3 years ago

Read 2 more answers

To test the quality of a tennis ball, you drop it onto the floor from a height of 4.00 m. It rebounds to a height of 2.00 m. If

arlik [135]

Answer:

Part a)

$a = 1260.3 m/s^2$

Part b)

Direction = upwards

Explanation:

When ball is dropped from height h = 4.0 m

then the speed of the ball just before it will strike the ground is given as

$v_f^2 - v_i^2 = 2 a d$

$v_1^2 - 0^2 = 2(9.81)(4.0)$

$v_1 = 8.86 m/s$

Now ball will rebound to height h = 2.00 m

so the velocity of ball just after it will rebound is given as

$v_f^2 - v_i^2 = 2 a d$

$0 - v_2^2 = 2(-9.81)(2.00)$

$v_2 = 6.26 m/s$

Part a)

Average acceleration is given as

$a = \frac{v_f - v_i}{\Delta t}$

$a = \frac{6.26 - (-8.86)}{12.0 \times 10^{-3}}$

$a = 1260.35 m/s^2$

Part B)

As we know that ball rebounds upwards after collision while before collision it is moving downwards

So the direction of the acceleration is vertically upwards

7 0

3 years ago