A diverging lens has a focal length of -30.0 cm. An object is placed 18.0 cm in front of this lens.

How much does a 59 kg woman weigh on earth?

leonid [27]

9.8........................

5 0

3 years ago

What is the centripetal force that would be required to keep a 4.0 kg mass moving in a horizontal circle with a radius of 0.80 m

KIM [24]

Answer:

D. 1.8 × 102 newtons radially inward

Explanation:

The magnitude of the centripetal force is given by:

$F=m\frac{v^2}{r}$

where

m is the mass of the object

v is the tangential speed

r is the radius of the circular trajector

In this problem, we have m = 4.0 kg, v = 6.0 m/s and r = 0.80 m, therefore substituting into the equation we get

$F=(4.0 kg)\frac{(6.0 m/s)^2}{0.80 m}=180 N$

The centripetal force is the force that keeps the object in a circular trajectory, so it is a force that is always directed inward (towards the centre of the circular path) and radially. Therefore, the correct answer is

D. 1.8 × 102 newtons radially inward

4 0

3 years ago

Read 2 more answers

An object in uniform circular motion must be changing its speed in order move in a circular path

mafiozo [28]

Answer:

False

Explanation:

An object in uniform circular motion must be changing its velocity in order to move in a circular path.

In fact, remind that velocity is a vector which consists of a magnitude (the speed) and a direction.

When an object is moving in uniform circular motion, the direction of the motion is constantly changing (since the trajectory is a circle): so, this means that the velocity is also changing. However, this does not imply that the speed of the object is changing. In fact, in a uniform circular motion, the speed of the object remains constant.

8 0

3 years ago

A particle vibrates in Simple Harmonic Motion with amplitude. What will be its displacement in one time-period? A 2A 4A 0

yan [13]

'Displacement' is the distance and direction between the starting point and
ending point, regardless of the path followed to get there.

A particle that's executing simple harmonic motion is always in the same place
where it was one time period ago, and where it will be later after another time
period has passed.

So its displacement during exactly one time period is exactly zero.

3 0

3 years ago

A sled of mass 15 kg slides across the ground with a friction force of 80.85 N. What is k between the sled and the ground?

Dvinal [7]

The expression for the frictional force between the sled and the ground is:
$F=k m g$
where $k$ is the coefficient of friction, m is the mass of the object and $g=9.81 m/s^2$ is the gravitational acceleration.

The friction force in our problem is F=80.85 N. The mass of the object is m=15 kg. Re-arranging the formula, we can find the value of k:
$k= \frac{F}{mg}= \frac{80.85 N}{(15 kg)(9.81 m/s^2)}=0.55$

8 0

3 years ago

Read 2 more answers