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drek231 [11]
3 years ago
9

What is the distance fallen for a freely falling object 1 s after being dropped from a rest position? What is the distance for a

4-s drop?
Physics
1 answer:
kondor19780726 [428]3 years ago
5 0
<h2>Distance traveled in 1 second after drop is 4.9 m</h2><h2>Distance traveled in 4 seconds after drop is 78.4 m</h2>

Explanation:

We have s = ut + 0.5at²

For a free falling object initial velocity u = 0 m/s and acceleration due to gravity, g = 9.8 m/s²

Substituting

                 s = 0 x t + 0.5 x 9.8 x t²

                 s = 4.9t²

We need to find distance traveled in 1 s and 4 s

Distance traveled in 1 second

                   s = 4.9 x 1² = 4.9 m

Distance traveled in 4 seconds

                   s = 4.9 x 4² = 78.4 m

Distance traveled in 1 second after drop = 4.9 m

Distance traveled in 4 seconds after drop = 78.4 m

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How much work is done on a satellite in a circular orbit about earth?
alexgriva [62]

Answer:

0 J

Explanation:

W = Work done on the satellite in circular orbit about earth by earth

F = Force on gravity on satellite by earth

d = displacement of the satellite

\theta = Angle between the force on gravity and displacement = 90

We know that, Work done is given as

W = F d Cos\theta\\W = F d Cos90\\W = F d (0)\\W = 0 J

3 0
3 years ago
A coin mass 0.005 kg is dropped from a height of 3 m. How much mechanical energy does it have right before it hits the ground ?
lesya692 [45]
Mechanical energy equals the sum of potential and kinetic energy. During the process, all PE converts into KE, assuming air resistance is neglected. So, the mechanical energy does not change and is equal to the initial potential energy.
ME
=mgh
=0.005 x 9.81 x 3
=0.147J
6 0
3 years ago
A steel ball bearing with a radius of 1.5 cm forms an image of an object that has been placed 1.1 cm away from the bearing’s sur
Nonamiya [84]

Answer:

Check the explanation

Explanation:

given

R = 1.5 cm

object distance, u = 1.1 cm

focal length of the ball, f = -R/2

= -1.5/2

= -0.75 cm

let v is the image distance

use, 1/u + 1/v = 1/f

1/v = 1/f - 1/u

1/v = 1/(-0.75) - 1/(1.1)

v = -0.446 cm <<<<<---------------Answer

magnification, m = -v/u

= -(-0.446)/1.1

= 0.405 <<<<<<<<<---------------Answer

The image is virtual

The image is upright

given

R = 1.5 cm

object distance, u = 1.1 cm

focal length of the ball, f = -R/2

= -1.5/2

= -0.75 cm

let v is the image distance

use, 1/u + 1/v = 1/f

1/v = 1/f - 1/u

1/v = 1/(-0.75) - 1/(1.1)

v = -0.446 cm <<<<<---------------Answer

magnification, m = -v/u

= -(-0.446)/1.1

= 0.405 <<<<<<<<<---------------Answer

Kindly check the diagram in the attached image below.

5 0
3 years ago
A horizontal 953 N merry-go-round of radius 1.68 m is started from rest by a constant horizontal force of 73.9 N applied tangent
solniwko [45]

Answer:

K.E=365.2 J

Explanation:

Given data

Weight w =953 N

radius r=1.68 m

F=73.9 N

t=2.55 s

g=9.8 m/s²

To find

Kinetic Energy K.E

Solution

From the moment of inertia

I=(1/2)MR^{2}\\ as \\W=mg\\So\\I=(1/2)(W/g)R^{2}\\I=(1/2)(953/9.8)(1.68)^{2}\\I=137.232kg.m^{2}

The angular acceleration is given as

a=T/I\\a=\frac{FR}{I}\\ a=\frac{(73.9)(1.68)}{137.232}\\a=0.905rad/s^{2}

The angular velocity is given as

w=at\\w=(0.905)(2.55)\\w=2.31rad/s

So the Kinetic Energy is given as

K.E=(1/2)Iw^{2}\\ K.E=(1/2)(137.232)(2.31)^{2}\\ K.E=365.2J

3 0
3 years ago
What happens to reactants when a chemical reaction takes place?
yanalaym [24]

Answer:

they collide with one another

5 0
3 years ago
Read 2 more answers
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