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3 years ago

How do you measure the wavelength of a transverse wave?

Physics

1 answer:

DENIUS [597]3 years ago

8 0

By looking at how wiggily the bar is lol

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If a 70-kg swimmer pushed off a pool wall with a force of 250N at what rate will the swimmer accelerate from the wall

Vlad [161]

F = ma
250 = 70 x a
a = 250/70
a = 3.57

5 0

3 years ago

Which of the following means that an image is real?

DochEvi [55]

To declare an image is real you would call it “authentic”

8 0

3 years ago

Read 2 more answers

A worker is thinking about two ways to get a box up 1.2 m onto a loading dock. He can use a force of 250 N to lift it straight u

harina [27]

Answer:

The second option has a lower power output. P=30 W

Explanation:

Mechanical Power

It is a physical magnitude that measures the rate a work W is done over time t.

$\displaystyle P=\frac{W}{t}$

Since W=F.d

$\displaystyle P=\frac{F.d}{t}$

The first option means the worker will lift the box by a distance of 1.2 meters in 3 seconds by applying 250 N of force. That produces a power of

$\displaystyle P=\frac{(250). (1.2)}{3}=100\ Watt$

The second option requires the worker applies 75 N of force and travel a distance of 4 meters for 10 seconds, thus the power is

$\displaystyle P=\frac{(75). (4)}{10}=30\ Watt$

The second option has a lower power output

7 0

3 years ago

What kind of charge is produced when a silk shirt is rubbed with human body

jeka94

Answer:

Frictional charges

Explanation:

6 0

3 years ago

A skater extends her arms, holding a 2 kg mass in each hand. She is rotating about a vertical axis at a given rate. She brings h

Usimov [2.4K]

Explanation:

It is known that relation between torque and angular acceleration is as follows.

$\tau = I \times \alpha$

and, I = $\sum mr^{2}$

So, $I_{1} = 2 kg \times (1 m)^{2} + 2 kg \times (1 m)^{2}$

= 4 $kg m^{2}$

$\tau_{1} = 4 kg m^{2} \times \alpha_{1}$

$\tau_{2} = I_{2} \alpha_{2}$

So, $I_{2} = 2 kg \times (0.5 m)^{2} + 2 kg \times (0.5 m)^{2}$

= 1 $kg m^{2}$

as $\tau_{2} = I_{2} \alpha_{2}$

= $1 kg m^{2} \times \alpha_{2}$

Hence, $\tau_{1} = \tau_{2}$

$4 \alpha_{1} = \alpha_{2}$

$\alpha_{1} = \frac{1}{4} \alpha_{2}$

Thus, we can conclude that the new rotation is $\frac{1}{4}$ times that of the first rotation rate.

8 0

3 years ago