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1 year ago

a mango drops to the ground from the top of a tree which is 5 meter high,how does it take the mango to reach the ground

1 answer:

5 0

Gravity tends to pull the object towards the earth's surface, resulting in a drop or a free fall. Fruits falling from a tree, a stone thrown off a cliff, skydiving, and other free-fall motions are examples.

Mango takes 1.009 seconds to reach the ground.

A situation in which an object moves solely under the influence of gravity is referred to as free fall. An external force acts on the ball, causing it to move faster. This free fall acceleration is also known as gravitational acceleration. The term "free fall" refers to a downward movement with no initial force or velocity.
Gravity tends to pull the object towards the earth's surface, resulting in a drop or a free fall. Fruits falling from a tree, a stone thrown off a cliff, skydiving, and other free-fall motions are examples.

h = $\frac{g t^2}{2}$

where,

h = height = 5m ,

t = time ,

g =gravity = 9.8 m/s.

$t^{2}$ = 2h / g = 10/9.8 = 1.02

t= $\sqrt{1.02}$ = 1.009 sec

To learn more about : Free Fall

Ref : brainly.com/question/12167131

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Plain electromagnetic wave (in air) has a frequency of 1 MHz and its B-field amplitude is 9 nT a. What is the wavelength in air?

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Answer:

Part a)

$\lambda = 300 m$

Part b)

$E = 2.7 N/C$

Part c)

$I = 9.68 \times 10^{-3} W/m^2$

$P = 3.22 \times 10^{-11} N/m^2$

Explanation:

Part a)

As we know that frequency = 1 MHz

speed of electromagnetic wave is same as speed of light

So the wavelength is given as

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$\lambda = \frac{3\times 10^8}{1\times 10^6}$

$\lambda = 300 m$

Part b)

As we know the relation between electric field and magnetic field

$E = Bc$

$E = (9 \times 10^{-9})(3\times 10^8)$

$E = 2.7 N/C$

Part c)

Intensity of wave is given as

$I = \frac{1}{2}\epsilon_0E^2c$

$I = \frac{1}{2}(8.85 \times 10^{-12})(2.7)^2(3\times 10^8)$

$I = 9.68 \times 10^{-3} W/m^2$

Pressure is defined as ratio of intensity and speed

$P = \frac{I}{c} = \frac{9.68\times 10^{-3}}{3\times 10^8}$

$P = 3.22 \times 10^{-11} N/m^2$

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3 years ago

The speed at which an object falls and the acceleration at which it falls are

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No, the speed at which an object falls is not equal to the acceleration at which it falls.

Answer:

Option B

Explanation:

Speed is defined as how fast an object can cover a specific distance and in what time it covers. So it is measured as the ratio of distance covered to the time taken to cover that distance. While acceleration is the rate of change of velocity. Moreover, speed is a scalar quantity and acceleration is a vector quantity. So most of the times, the direction will play an important role in the varying values of speed and acceleration. Also, acceleration of an object will depend upon the force and mass of the object. Thus, speed and acceleration will not attain same value always.

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3 years ago

When you are driving on a rural road, if your right wheels run off the pavement, you should hold the steering wheel firmly and

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Answer:

C). Take your foot off the gas pedal. Then brake lightly until you are moving at low speed.

Explanation:

While driving on roads of rural areas, if our right wheel moves off the pavement, we should always hold the steering wheel firmly and then take our foot off the gas pedal, then apply brake lightly until we are moving at a low speed.

When our wheels drift off the pavement area, we should not panic and yank. And instead of turning the wheel back in the left direction towards the road, it is always safer to take off our foot from the gas pedal and then apply brakes slowly. When our vehicle slows down check the incoming traffic behind us and then we should slowly move back on to the pavement.

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Momentum is defined as the change in velocity of any object along with its mass. So mathematically, momentum can be derived using the product of mass with the change in velocity.

$\text { Momentum change }=m \times \Delta v$

As here mass is given as 82 kg and the initial velocity was 5.6 m/s and final velocity is 2.5 m/s.

Initial Momentum = $m \times \text { initial velocity }=82 \times 5.6=459.2 \mathrm{kg} \mathrm{m} / \mathrm{s}$

Final Momentum = $m \times \text { final velocity }=82 \times(-2.5)=-205 \mathrm{kg} \mathrm{m} / \mathrm{s}$

Momentum change = Final Momentum - Initial Momentum

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a mango drops to the ground from the top of a tree which is 5 meter high,how does it take the mango to reach the ground​

a mango drops to the ground from the top of a tree which is 5 meter high,how does it take the mango to reach the ground