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2 years ago

What do we need to know in order to make motion safe for all?

Physics

1 answer:

Ghella [55]2 years ago

7 0

Answer:

motion changes with time of the position or orientation of a body. Motion that changes the orientation of a body is called rotation.

Explanation:

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Two vehicles approach an intersection: a truck moving eastbound at 16.0 m/s and an SUV moving southbound at 20.0 m/s. Suppose th

Alex Ar [27]

Answer:

The value of magnitude of the velocity vector of the truck relative to the SUV $V_{xy}= 25.62 \frac{m}{sec}$

Explanation:

Velocity of the truck $V_{x}$ = 16 $\frac{m}{s}$

Velocity of the SUV $V_{y}$ = - 20 $\frac{m}{s}$

The magnitude of the velocity vector of the truck relative to the SUV is

$V_{xy} =\sqrt{ V_{x} ^{2} + V_{y}^{2}}$

Put the value of $V_{x}$ & $V_{y}$ in the above equation we get

$V_{xy} = \sqrt{(16)^{2} + (-20)^{2}}$

$V_{xy} = \sqrt{656}$

$V_{xy}= 25.62 \frac{m}{sec}$

This is the value of magnitude of the velocity vector of the truck relative to the SUV.

8 0

3 years ago

Determine What is the efficiency of a machine whose work output is 4 joules, and whose work input is 10 joules?

anastassius [24]

Answer:

The efficiency of the machine is 40 %.

Explanation:

We have,

Output work by the machine is 4 joules and the input work is 10 joules.

It is required to find the efficiency of a machine. The efficiency of a machine is given by the ratio of output work to the input work. It is given as :

$\eta=\dfrac{4}{10}\times 100\\\\\eta=40\%$

So, the efficiency of the machine is 40 %.

8 0

3 years ago

A 50 pF capacitor and a 200 pF capacitor are both charged to 4.2 kV. They are then disconnected from the voltage source and are

8_murik_8 [283]

Answer:

The energy lost is zero.

Explanation:

Given that,

First capacitor = 50 pF

Second capacitor = 200 pF

Potential = 4.2 kV

We need to calculate the energy lost

Using formula of energy lost

$E = E_{initial}-E_{final}$

Put the value into the formula

$E=\dfrac{1}{2}C_{1}V^2+\dfrac{1}{2}C_{2}V^2-\dfrac{}{}(C_{1}+C_{2})V^2$

$E=\dfrac{1}{2}\times50\times10^{-12}\times(4.2\times10^{3})^2+\dfrac{1}{2}\times200\times10^{-12}\times(4.2\times10^{3})^2-\dfrac{1}{2}\times(50\times10^{-12}+200\times10^{-12})\times(4.2\times10^{3})^2$