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2 years ago

5

What force must be exerted on the master cylinder of a hydraulic lift to support the weight of a 2000-kg car (a large car) resti

ng on the slave cylinder

1 answer:

Tema [17]2 years ago

7 0

Explanation:

Below is an attachment containing the solution.

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A man pushes a heavy cart with a force exerted of 250 Newtons to keep it moving at a constant velocity. What is the kinetic fric

miv72 [106K]

It can't be less than 250 N or the cart wouldn't move at all. That means there is only 1 answer. It's between not enough info or 250 N. The answer is 250 N. If it was any more, there would be acceleration.

3 0

3 years ago

La luz roja visible tiene una longitud de onda de 680 nanómetros (6,8 x 10-7 m). La velocidad de la luz es de 3.0 x108 m / s. ¿C

Lina20 [59]

Answer:

Frequency, $f=4.41\times 10^{14}\ Hz$

Explanation:

Visible red light has a wavelength of 680 nanometers (6.8 x 10⁻⁷ m). The speed of light is 3.0 x 10 ⁸ m / s. What is the frequency of visible red light?

It is given that,

Wavelength of a visible red light is, $\lambda=6.8\times 10^{-7}\ m$

Speed of light is, $c=3\times 10^8\ m/s$

We need to find the frequency of visible red light. It can be calculated using below relation.

$c=f\lambda\\\\f=\dfrac{c}{\lambda}\\\\f=\dfrac{3\times 10^8}{6.8\times 10^{-7}}\\\\f=4.41\times 10^{14}\ Hz$

So, the frequency of visible red light is $4.41\times 10^{14}\ Hz$ .

3 0

2 years ago

A 125 g pendulum bob hung on a string of length 35 cm has the same period as when the bob is hung from a spring and caused to os

Bingel [31]

Answer:

k = 3.5 N/m

Explanation:

It is given that the time period the bob in pendulum is the same as its time period in spring mass system:

$Time\ Period\ of\ Pendulum = Time\ Period\ of\ Spring-Mass\ System\\2\pi \sqrt{\frac{l}{g}} = 2\pi \sqrt{\frac{m}{k}$

$\frac{l}{g} = \frac{m}{k}\\\\ k = g\frac{m}{l}$

where,

k = spring constant = ?

g = acceleration due to gravity = 9.81 m/s²

m = mass of bob = 125 g = 0.125 kg

l = length of pendulum = 35 cm = 0.35 m

Therefore,

$k = (9.81\ m/s^2)(\frac{0.125\ kg}{0.35\ m})\\\\$

<u>k = 3.5 N/m</u>

4 0

2 years ago

When hiking at altitude, the air pressure drops to 825,000Pa. Solve the force that site exerts against the side of a rectangular

OLga [1]

Answer:

The force is $1237500N$

Explanation:

We can apply the following expression

$pressure=\frac{Force}{Area}$

To find the area of rectangular tent wall:

$Area=length *width\\Area=2.00m*0.750m\\Area=1.5m^{2}$

Now to find the force:

$pressure=825,000pa$

$pressure=\frac{Force}{Area}$

$Force=pressure*Area\\Force=825000pa*1.5m^{2} \\Force=1237500N$

7 0

2 years ago

A reservoir located in the mountain 250 m above sea level flows through a pipe to a hydroelectric plant in a town at sea level.

Pavlova-9 [17]

Answer:

v₂ = 70 m / s

Explanation:

For this exercise let's use Bernoulli's equation

where subscript 1 is for the top of the mountain and subscript 2 is for Tuesday's level

P₁ + ½ ρ v₁² + ρ g y₁ = P₂ +1/2 ρ v₂² + ρ g y₂

indicate that the pressure in the two points is the same, y₁ = 250 m, y₂ = 0 m, the water in the upper part, because it is a reservoir, is very large for which the velocity is very small, we will approximate it to 0 (v₁ = 0), we substitute

ρ g y₁ = ½ ρ v₂²

v₂ = $\sqrt {2g \ y_1}$

let's calculate

v₂ = √( 2 9.8 250)

v₂ = 70 m / s

6 0

3 years ago