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2 years ago

When he drove home his car broke down it wasnt due to a problem with the wheels it was the?

Physics

2 answers:

Goshia [24]2 years ago

3 0

It would have to be the engine

Nikitich [7]2 years ago

3 0

Engine. As that is what powers the car.

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The ideal mechanical advantage of an inclined plane is 3.5, and its efficiency is 0.6. What is the mechanical advantage of the i

Nadya [2.5K]

Mechanical advantage = ideal mechanical advantage x efficiency = 3.5 x 0.6 = 2.1
The mechanical advantage of the inclined plane is 2.1

7 0

3 years ago

Que es electromagnetico?

adoni [48]

Se llaman cuerpos eléctricos.

3 0

3 years ago

How much must a woman weigh ( force) if the pressure she exerts while standing on one foot has an area of 0.6m squared exerts a

lions [1.4K]

Answer:

W = 9.6 N

Explanation:

Given that,

Area on 1 foot, A = 0.6 m²

Pressure, P = 16 Pa

The pressure is given by force acting per unit area. So,

$P=\dfrac{F}{A}\\\\P=\dfrac{W}{A}\\\\W=16\times 0.6\\\\W=9.6\ N$

So, the required weight is 9.6 N.

4 0

3 years ago

A 22kg Accelerates at a rate of 2.3 m/s. What is the magnitude of the net force acting on the bike?

Tcecarenko [31]

magnitude of the net force = mass x acceleraton

= 22 x 2.3

=50.6 N

7 0

3 years ago

A stone is dropped into a river from a bridge at a height h above the water. Another stone is thrown vertically down at a time t

Mumz [18]

Answer:

$v_{y_0} = \frac{\frac{g}{2}t(t - 2\sqrt{\frac{2h}{g}})}{\sqrt{\frac{2h}{g}} - t}$

Explanation:

We will apply the equations of kinematics to both stones separately.

First stone:

Let us denote the time spent after the second stone is thrown as 'T'.

$y - y_0 = v_{y_0}(t+T) + \frac{1}{2}a(t+T)^2\\0 - h = 0 + \frac{1}{2}(-g)(t+T)^2\\(t+T)^2 = \frac{2h}{g}\\T = \sqrt{\frac{2h}{g}}-t$

Second stone:

$y - y_0 = v_{y_0}T + \frac{1}{2}aT^2\\0 - h = v_{y_0}T -\frac{1}{2}gT^2\\-h = v_{y_0}(\sqrt{\frac{2h}{g}} - t) - \frac{g}{2}(\sqrt{\frac{2h}{g}} - t)^2\\-h = v_{y_0}(\sqrt{\frac{2h}{g}} - t) - \frac{g}{2}(\frac{2h}{g} + t^2 - 2t\sqrt{\frac{2h}{g}})\\-h = v_{y_0}\sqrt{\frac{2h}{g}} - v_{y_0}t - h -\frac{g}{2}t^2 + gt\sqrt{\frac{2h}{g}}\\v_{y_0}(\sqrt{\frac{2h}{g}} - t) = \frac{g}{2}t^2 - gt\sqrt{\frac{2h}{g}}\\v_{y_0} = \frac{\frac{g}{2}t(t - 2\sqrt{\frac{2h}{g}})}{\sqrt{\frac{2h}{g}} - t}$

6 0

3 years ago

Read 2 more answers