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3 years ago

What is the force exerted by a solid surface that opposes gravity?

Physics

2 answers:

jonny [76]3 years ago

6 0

Electromagnetic force between the molecules!

Ann [662]3 years ago

6 0

The force exerted upward that opposes gravity is the Normal Force.

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Which material is the LEAST dense?

givi [52]

Brick has least dense

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4 years ago

Read 2 more answers

What is lever means,types of lever.. <br>will give brainles

serious [3.7K]

Explanation:

A lever is a rigid bar which moves freely about a fixed point called fulcrum....

The types of lever are :

First class lever
Second class lever
Third class lever....

3 0

2 years ago

The earth rotates once per day about an axis passing through the north and south poles, an axis that is perpendicular to the pla

mina [271]

Answer:

(a) v = 463.97 m/s, $a_{c} = 0.0337\ m/s^{2}$

(b) v' = 196.02 m/s, $a'_{c} = 0.0143\ m/s^{2}$

Solution:

As per the question:

Time period of the rotation of earth, T = 24 h = $24\times 3600 = 86400\ s$

Radius of the earth, R = $6.38\times 10^{6}\ m$

Angle, $\theta = 65.0^{\circ}$

Now,

Angular velocity is given by:

$\omega = \frac{2\pi}{T} = \frac{2\pi}{86400} = 7.27\times 10^{- 5} \rad$

(a) To calculate the speed and the centripetal acceleration at the equator:

Linear velocity or the speed, v = $R\omega$

$v = 6.38\times 10^{6}\times 7.27\times 10^{- 5} = 463.967\ m/s$

Centripetal Acceleration, $a_{c} = \frac{v^{2}}{R}$

$a_{c} = \frac{463.967^{2}}{6.38\times 10^{6}} = 0.0337\ m/s^{2}$

(b) To calculate the speed and acceleration at an altitude of $65.0^{\circ}$ N:

The horizontal component of the radius, R' = $Rcos\theta$

$R' = 6.38\times 10^{6}cos65.0^{\circ} = 2.69\times 10^{6}\ m$

Now,

For the speed, v' = $R'\omega = 2.69\times 10^{6}\times 7.27\times 10^{- 5} = 196.02\ m/s$

For the centripetal acceleration,

$a'_{c} = \frac{v'^{2}}{R'}$

$a'_{c} = \frac{196.02^{2}}{2.69\times 10^{6}} = 0.01428\ m/s^{2}$

8 0

4 years ago

An athlete swings a ball, connected to the end of a chain, in a horizontal circle. The athlete is able to rotate the ball at the

garik1379 [7]

(a) 6.04 rev/s

The speed of the ball is given by:

$v=\omega r$

where

$\omega$ is the angular speed

r is the distance of the ball from the centre of the circle

In situation 1), we have

$\omega=8.13 rev/s \cdot 2\pi = 51.0 rad/s$

r = 0.600 m

So the speed of the ball is

$v=(51.0 rad/s)(0.600 m)=30.6 m/s$

In situation 2), we have

$\omega=6.04 rev/s \cdot 2\pi = 37.9 rad/s$

r = 0.900 m

So the speed of the ball is

$v=(37.9 rad/s)(0.900 m)=34.1 m/s$

So, the ball has greater speed when rotating at 6.04 rev/s.

(b) $1561 m/s^2$

The centripetal acceleration of the ball is given by

$a=\frac{v^2}{r}$

where

v is the speed

r is the distance of the ball from the centre of the trajectory

For situation 1),

v = 30.6 m/s

r = 0.600 m

So the centripetal acceleration is

$a=\frac{(30.6 m/s)^2}{0.600 m}=1561 m/s^2$

(c) $1292 m/s^2$

For situation 2 we have

v = 34.1 m/s

r = 0.900 m

So the centripetal acceleration is

$a=\frac{v^2}{r}=\frac{(34.1 m/s)^2}{0.900 m}=1292 m/s^2$

5 0

3 years ago

A heat engine exhausts 8900 j of heat while performing 2800 j of useful work. part a what is the efficiency of this engine?

Anuta_ua [19.1K]

Given: Heat Qout means useful work = 2800 J

Heat Qin = 8900 J

Required; Efficiency = ?

Formula: Efficiency = Qout/Qin = x 100%

= 2800 J/8900 J = 0.3146 X 100 %

Efficiency = 31.46%

7 0

3 years ago