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2 years ago

How can understanding Development help you understand yourself?

1 answer:

7 0

Developing a better understanding of yourself may also improve your capacity to better understand the thoughts and feelings of other people.

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An electron is released from rest at a distance of 0.300 m from a large insulating sheet of charge that has uniform surface char

jarptica [38.1K]

Yes that is the correct answer

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3 years ago

Runaway truck ramps are common on mountainous highways in case the brakes fail on large trucks. If a

dusya [7]

Answer:

$W=-21,870,000\ J$

Explanation:

<u>Work and Kinetic Energy </u>

The work an object does due to its motion is equal to the change of its kinetic energy. Being ko and k1 the initial and final kinetic energy respectively and m the mass of the object, then

$W=\Delta k=k_1-k_0$

Since

$\displaystyle k=\frac{mv^2}{2}$

We have

$\displaystyle W=\frac{mv_1^2}{2}-\frac{mv_0^2}{2}$

The truck has a mass of 60,000 kg and is moving at 27 m/s. The runaway truck ramp must stop the truck, so the final speed is 0. Thus

$\displaystyle W=\frac{(60,000)0^2}{2}-\frac{(60,000)(27)^2}{2}$

$W=0-21870000\ J$

$\boxed{W=-21,870,000\ J}$

3 0

3 years ago

Two small space probes have been slowed to 10m/s as they approach the moon from the same direction. Probe 1 has a mass of 86kg a

bulgar [2K]

Answer:

Explanation:

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2 years ago

a rectangular tank measures 12.5 metres long 10.0 wide and 2.0 metre high calculate the mass of the water in the tank when it is

Tom [10]

Answer:

250000kg

Explanation:

Mass =density * volume

Density=1000kg/m³

Volume=l*w*h

12.5*10.0*2.0=250m³

Mass=1000 *250

=250000kg

7 0

2 years ago

A lady bug is sitting on the bottom of a can while you twirl it overhead on a string that is 65.0

MA_775_DIABLO [31]

The linear speed of the ladybug is 4.1 m/s

Explanation:

First of all, we need to find the angular speed of the lady bug. This is given by:

$\omega=\frac{2\pi}{T}$

where

T is the period of revolution

The period of revolution is the time taken by the ladybug to complete one revolution: in this case, since it does 1 revolution every second, the period is 1 second:

T = 1 s

Therefore, the angular speed is

$\omega=\frac{2\pi}{1 s}=6.28 rad/s$