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3 years ago

12

If the change in velocity increases what happens to the acceleration during the same time period?

1 answer:

viva [34]3 years ago

3 0

Answer:

Acceleration increases

Explanation:

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You are walking towards teh back of the bus that is movind forward with a constant velocity. Descriube your motion relative to t

Arisa [49]

Answer:

When walking towards the back of a bus while it moves forward its apparent velocity seems slower than it actually is with respect to the ground.

Explanation:

When a someone follows a moving bus from behind by walking then the apparent speed of the bus is slightly slower by the same magnitude as the speed of walking because both are moving in the same direction.

This is the concept of relative velocity when the apparent velocity of the same object appears different to different viewers at the same time being on different frame of motion.

When the same bus is observed from a point on the road the speed of the bus will appear greater than the former case discussed above because the bus here is moving with respect to a stationary observer.

5 0

3 years ago

The wavelength of a wave is 4 metre & its time period is 2 second. What is speed of the wave?

Harlamova29_29 [7]

Answer: The speed of wave is 2m/s.

Explanation:

The speed of wave equals to the distance/time. So when we divide 4m and 2s, we get that the speed of wave is 2m/s

5 0

3 years ago

6) The electron volt (eV) is a convenient unit of energy for expressing atomic-scale energies. It is the amount of energy that a

ivolga24 [154]

Answer:

The energy of the photon is $x = 2.86 eV$

Explanation:

From the question we are told that

The first orbit is $n_1 = 5$

The second orbit is $n_2 = 2$

According to Bohr model

The energy of difference of the electron as it moves from on orbital to another is mathematically represented as

$\Delta E = k [\frac{1}{n^2 _1} + \frac{1}{n^2 _2} ]$

Where k is a constant which has a value of $k = -2.179 *10^{-18} J$

So

$\Delta E = - 2.179 * 10^{-18} [\frac{1}{5^2 _1} + \frac{1}{2^2 _2} ]$

$= 4.576 *10^{-19}J$

Now we are told from the question that

$1 eV = 1.602 * 10^{-19} J$

so x eV = $= 4.576 *10^{-19}J$

Therefore

$x = \frac{4.576*10^{-19}}{1.602 *10^{-19}}$

$x = 2.86 eV$

5 0

3 years ago

Which Circut Model is more efficient, why?

vovangra [49]

Answer:

right one is more efficient

Explanation:

7 0

3 years ago

A 2.00x10^6 hertz radio signal is sent a distance of 7.30x10^10 meters from earth to a spaceship orbiting Mars. How much time do

PolarNik [594]

Answer:

It takes $2.43\times 10^2$ for the radio signal to travel from earth to the spaceship.

Explanation:

Given:

The frequency of the radio signal = $2.00 \times 10^6$ hertz

Distance = $7.30\times 10^10$ meters

To Find:

The time taken for the radio signal to travel from the earth to the spaceship= ?

Solution:

we know that the time taken can be found by dividing the distance travelled by the speed at which the signal travels

So

$Time = \frac{Distance}{Speed}$

Here the speed of the radio signal is equal the the speed of light

Now substituting the values,

Time taken is

=> $Time = \frac{7.30\times 10^10}{3 \times 10^8}$

=> $Time = 2.43\times 10^{10-8}$

=> $Time = 2.43\times 10^2$

5 0

3 years ago