Consider a student drops a block of ice. which of the following forces are acting on the block of ice as it falls

Can someone answer this please I’ll give brainliest

Alinara [238K]

Answer:

The last one is false

Explanation:

Energy can be neither created or destroyed. It can only move from one type of energy to another.

6 0

2 years ago

A small metal ball is suspended from the ceiling by a thread of negligible mass. The ball is then set in motion in a horizontal

Gemiola [76]

Answer:

Time taken, $T=2\pi \sqrt{\dfrac{l\ cos\theta}{g}}$

Explanation:

It is given that, a small metal ball is suspended from the ceiling by a thread of negligible mass. The ball is then set in motion in a horizontal circle so that the thread’s trajectory describes a cone as shown in attached figure.

From the figure,

The sum of forces in y direction is :

$T\ cos\theta-mg=0$

$T=\dfrac{mg}{cos\theta}$

Sum of forces in x direction,

$T\ sin\theta=\dfrac{mv^2}{r}$

$mg\ tan\theta=\dfrac{mv^2}{r}$ .............(1)

Also, $r=l\ sin\theta$

Equation (1) becomes :

$mg\ tan\theta=\dfrac{mv^2}{l\ sin\theta}$

$v=\sqrt{gl\ tan\theta.sin\theta}$ ...............(2)

Let t is the time taken for the ball to rotate once around the axis. It is given by :

$T=\dfrac{2\pi r}{v}$

Put the value of T from equation (2) to the above expression:

$T=\dfrac{2\pi r}{\sqrt{gl\ tan\theta.sin\theta}}$

$T=\dfrac{2\pi l\ sin\theta}{\sqrt{gl\ tan\theta.sin\theta}}$

On solving above equation :

$T=2\pi \sqrt{\dfrac{l\ cos\theta}{g}}$

Hence, this is the required solution.

4 0

2 years ago

Difference between pulling and pushing force

Mrac [35]

Answer:

Push and pull both are forces , but the difference is in their direction at which it is applied . If the force applied in the direction of motion of the particle then we call it as push . If that force applied in the direction OPPOSITE to the motion of particle then it is termed as pull

7 0

2 years ago

A car with a mass of 833 kg rounds an unbanked curve in the road at a speed of 28.0 m/s. If the

SSSSS [86.1K]

I had the same question

6 0

2 years ago

A 56.6-kg crate rests on a level floor at a shipping dock. The coefficients of static and kinetic friction are 0.517 and 0.260,

kifflom [539]

Answer:

Explanation:

Magnitude of frictional force = μ mg

μ is either static or kinetic friction.

To start the crate moving , static friction is calculated .

a ) To start crate moving , force required = μ mg where μ is coefficient of static friction .

force required =.517 x 56.6 x 9.8 = 286.76 N .

b ) to slide the crate across the dock at a constant speed , force required

= μ mg where μ is coefficient of kinetic friction , where μ is kinetic friction

= .26 x 56.6 x 9.8 = 144.21 N .

3 0

2 years ago