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2 years ago

The effort is always less than the load for a second-class lever. true or false

Physics

2 answers:

ddd [48]2 years ago

8 0

False

Have a nice day

Rasek [7]2 years ago

5 0

False is the answer

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A street light is on top of a 8 foot pole. Joe,

zubka84 [21]

8/4 = y/y-x

8y - 8x = 4y

y = 2x

y = 2 x 4

y = 8

Hope this helps

5 0

3 years ago

In the pulley system shown below, a 360 N weight is slowly lifted. Assuming the system is 100% efficient and each pulley is weig

STALIN [3.7K]

Your answer is D. 361 N

8 0

2 years ago

Read 2 more answers

A gas has an initial volume of 2.5 L at a temperature of 275 K and a pressure of 2.1 atm. The pressure of the gas increases to 2

olchik [2.2K]

Answer:

2.10L

Explanation:

Given data

V1= 2.5L

T1= 275K

P1= 2.1atm

P2= 2.7 atm

T2= 298K

V2= ???

Let us apply the gas equation

P1V1/T1= P2V2/T2

substitute into the expression we have

2.1*2.5/275= 2.7*V2/298

5.25/275= 2.7*V2/298

Cross multiply

275*2.7V2= 298*5.25

742.5V2= 1564.5

V2= 1564.5/742.5

V2= 2.10L

Hence the final volume is 2.10L

7 0

2 years ago

A stretched string has a mass per unit length of 5.40 g/cm and a tension of 17.5 N. A sinusoidal wave on this string has an ampl

kondaur [170]

Answer:

Part a)

$y_m = 0.157 mm$

part b)

$k = 101.8 rad/m$

Part c)

$\omega = 579.3 rad/s$

Part d)

here since wave is moving in negative direction so the sign of $\omega$ must be positive

Explanation:

As we know that the speed of wave in string is given by

$v = \sqrt{\frac{T}{m/L}}$

so we have

$T = 17.5 N$

$m/L = 5.4 g/cm = 0.54 kg/m$

now we have

$v = \sqrt{\frac{17.5}{0.54}}$

$v = 5.69 m/s$

now we have

Part a)

$y_m$ = amplitude of wave

$y_m = 0.157 mm$

part b)

$k = \frac{\omega}{v}$

here we know that

$\omega = 2\pi f$

$\omega = 2\pi(92.2) = 579.3 rad/s$

so we have

$k = \frac{579.3}{5.69}$

$k = 101.8 rad/m$

Part c)

$\omega = 579.3 rad/s$

Part d)

here since wave is moving in negative direction so the sign of $\omega$ must be positive

4 0

3 years ago

A plastic box has an initial volume of 2.00 m 3 . It is then submerged below the surface of a liquid and its volume decreases to

nikitadnepr [17]

Answer:

Volume strain is 0.02

Explanation:

Volume strain is defined as the change in volume to the original volume.

It is given that,

Initial volume of the plastic box is 2 m³

It is then submerged below the surface of a liquid and its volume decreases to 1.96 m³

We need to find the volume strain on the box. It is defined as the change in volume divided by the original volume. So,

$\delta V=\dfrac{V_f-V_i}{V_i}\\\\\delta V=\dfrac{1.96-2}{2}\\\\\delta V=0.02$

So, the volume strain on the box is 0.02.

6 0

2 years ago