The lightbulb is an example of

The earth rotates once per day about an axis passing through the north and south poles. True or False

alexandr402 [8]

Answer: False.

Explanation: The Earth rotates once per day in direction from East to West. The rotation of the Earth daily is responsible for day and night experienced, the areas of the Earth facing the Sun experiences day time while the areas of the Earth away from the Sun in the Earth's shadow experiences night time.

While the revolution of the Earth around the gives rise to a year, as it takes 365 days for the Earth to go round the sun.

5 0

3 years ago

Which FBD would represent a car moving right with a motor force of 250 N, and force of friction of 750N, a weight of 8500N and a

babymother [125]

Answer:

Option C

Explanation:

Given that

Motor force is 250 N

Force of friction is 750 N

Weight is 8500 N

And, the normal force is 8500 N

Now based on the above information

Here length of the rector shows the relative magnitude forward force i.e. 250 N i..e lower than the frictional force i.e. backward and weight i.e. 8500 would be equivalent to the normal force

8 0

2 years ago

What is the distance covered by a Freely falling object 5 seconds after being dropped ? After 6 seconds?

mario62 [17]

This year is 60 years since I learned this stuff, and one of the things I always remembered is the formula for the distance a dropped object falls:

D = 1/2 A T²

Distance = (1/2) (acceleration) (time²)

The reason I never forgot it is because it's SO useful SO often. You really should memorize it. And don't bury it too deep in your toolbox ... you'll be needing it again very soon. (In fact, if you had learned it the first time you saw it, you could have solved this problem on your own today.)

The problem doesn't tell us what planet this is happening on, so let's make it easy and just assume it's on Earth. Then the 'acceleration' is Earth gravity, and that's 9.8 m/s² .

In 5 seconds:

D = 1/2 A T²

D = (1/2) (9.8 m/s²) (5 sec)²

D = (4.9 m/s²) (25 sec²)

D = 122.5 meters

In 6 seconds:

D = 1/2 A T²

D = (1/2) (9.8 m/s²) (6 sec)²

D = (4.9 m/s²) (36 sec²)

D = 176 meters

5 0

3 years ago

50 points !! I need help asap.......Consider a 2-kg bowling ball sits on top of a building that is 40 meters tall. It falls to t

r-ruslan [8.4K]

1) At the top of the building, the ball has more potential energy

2) When the ball is halfway through the fall, the potential energy and the kinetic energy are equal

3) Before hitting the ground, the ball has more kinetic energy

4) The potential energy at the top of the building is 784 J

5) The potential energy halfway through the fall is 392 J

6) The kinetic energy halfway through the fall is 392 J

7) The kinetic energy just before hitting the ground is 784 J

Explanation:

1)

The potential energy of an object is given by

$PE=mgh$

where

m is the mass

g is the acceleration of gravity

h is the height relative to the ground

While the kinetic energy is given by

$KE=\frac{1}{2}mv^2$

where v is the speed of the object

When the ball is sitting on the top of the building, we have

$h=40 m$ , therefore the potential energy is not zero
$v=0$ , since the ball is at rest, therefore the kinetic energy is zero

This means that the ball has more potential energy than kinetic energy.

2)

When the ball is halfway through the fall, the height is

$h=20 m$

So, half of its initial height. This also means that the potential energy is now half of the potential energy at the top (because potential energy is directly proportional to the height).

The total mechanical energy of the ball, which is conserved, is the sum of potential and kinetic energy:

$E=PE+KE=const.$

At the top of the building,

$E=PE_{top}$

While halfway through the fall,

$PE_{half}=\frac{PE_{top}}{2}=\frac{E}{2}$

And the mechanical energy is

$E=PE_{half} + KE_{half} = \frac{PE_{top}}{2}+KE_{half}=\frac{E}{2}+KE_{half}$

which means

$KE_{half}=\frac{E}{2}$

So, when the ball is halfway through the fall, the potential energy and the kinetic energy are equal, and they are both half of the total energy.

3)

Just before the ball hits the ground, the situation is the following:

The height of the ball relative to the ground is now zero: $h=0$ . This means that the potential energy of the ball is zero: $PE=0$

The kinetic energy, instead, is not zero: in fact, the ball has gained speed during the fall, so $v\neq 0$ , and therefore the kinetic energy is not zero