<span>An object is located 51 millimeters from a diverging lens the object has a height og 13 millimeters and the image height is 3.5 millimeters how far in front of the lens is image located?</span>

3 0

3 years ago

A car is traveling at 108 km/h, stuck behind a slower car. Finally the road is clear and the car pulls over to make a pass. The

mezya [45]

Answer:

The average acceleration of the car is 2.143 meters per square second.

Explanation:

Let assume that car accelerates uniformly, in that case, we can obtain the value of acceleration by using the following equation of motion:

$v = v_{o}+a\cdot t$

Where:

$v_{o}$ - Initial velocity, measured in meters per second.

$v$ - Final velocity, measured in meters per second.

$a$ - Acceleration, measured in meters per square second.

$t$ - Time, measured in seconds.

Now, we clear acceleration within expression:

$a = \frac{v-v_{o}}{t}$

Initial and final velocities are now converted from kilometers per hour into meters per second:

$v_{o} = \left(108\,\frac{km}{h} \right)\cdot \left(1000\,\frac{m}{km} \right)\cdot \left(\frac{1}{3600}\,\frac{h}{s} \right)$

$v_{o} = 30\,\frac{m}{s}$

$v = \left(135\,\frac{km}{h} \right)\cdot \left(1000\,\frac{m}{km} \right)\cdot \left(\frac{1}{3600}\,\frac{h}{s} \right)$

$v = 37.5\,\frac{m}{s}$

If we know that $t = 3.5\,s$ , then, the average acceleration of the car is:

$a = \frac{37.5\,\frac{m}{s}-30\,\frac{m}{s} }{3.5\,s}$

$a = 2.143\,\frac{m}{s^{2}}$

The average acceleration of the car is 2.143 meters per square second.

8 0

3 years ago

I need help... anyone know this?

julsineya [31]

Answer:mp

Explanation:

7 0

3 years ago