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Scilla [17]
3 years ago
6

What is the potential energy of the bowling ball as it sits on top of a building ?​

Physics
1 answer:
tangare [24]3 years ago
8 0

Answer:

The potential energy of the bowling ball will be mgh

Explanation:

Let the mass of bowling ball =m

The height of building on which bowling ball sits=h

So,

The potential energy of the bowling ball =P.E.= mgh

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Plsssss asappppppppp
KengaRu [80]

Answer:

20 meters.

Explanation:

In the graph, the x-axis (the horizontal axis) represents the time, while the y-axis (the vertical axis) represents the distance.

If we want to find the distance covered in the first T seconds, you need to find the value T in the horizontal axis.

Once you find it, we draw a vertical line, in the point where this vertical line touches the graph, we now draw a horizontal line. This horizontal line will intersect the y-axis in a given value. That value is the total distance travelled by the time T.

In this case, we want to find the total distance that David ran in the first 4 seconds.

Then we need to find the value 4 seconds in the horizontal axis. Now we perform the above steps, and we will find that the correspondent y-value is 20.

This means that in the first 4 seconds, David ran a distance of 20 meters.

6 0
3 years ago
A car speeds up from 18.54 m/s to<br> 29.52 m/s in 13.84 s.<br> The acceleration of the car is:
valkas [14]

Answer:

.7934m/s^{2}

Explanation:

Acceleration = change in velocity / change in time

A = 10.98m/s / 13.84s

A = .7934m/s^{2}

5 0
3 years ago
Read 2 more answers
Car A starts in Sacramento at 11am. It travels along 400 mile route to Los Angeles at 60 mph. Car B starts from Los Angeles at n
damaskus [11]

Answer:

  • 38.89 miles

Explanation:

from the question we have the following:

distance between Sacramento and los angles = 400 miles

speed of car A = 60 mph

start time of car A = 11 am

speed of car B = 75 mph

start time of car B = 12 pm

distance of Fresno from Los Angeles = 150 miles

  • To start off let's allow car A to travel for one hour (from 11 am to 12 pm), during which it would have covered a distance of 60 miles.
  • Now the time would be 12 pm and the distance between the two cars would be 400 - 60 (distance traveled by car A within 11 am to 12 pm) = 340 miles
  • From 12 pm to the time both cars will meet, the distance covered by car A + distance covered by car B would be equal to 340 miles. Therefore
  • Distance covered by car A = speed x time(t) = 60 x t = 60t
  • Distance covered by car B = speed x time(t) = 75 x t = 75t
  • 60t + 75t = 340 miles
  • 135t = 340
  • t = 2.51 hours
  • Recall that at their meeting point, the distance covered by car B = 75t = 75 x 2.62 = 188.89 miles
  • Since Fresno is 150 miles from Los Angeles, car B which is 188.89 miles from Los Angeles at their meeting point would be 188.89 - 150 = 38.89 miles from Fresno
  • 38.89 miles would also be the distance of car A from Fresno since that is their meeting point.

4 0
3 years ago
A projectile is launched diagonally into the air and has a hang time of 24.5 seconds. Approximately how much time is required fo
Rasek [7]

Answer:

t=12.25\ seconds

Explanation:

<u>Diagonal Launch </u>

It's referred to as a situation where an object is thrown in free air forming an angle with the horizontal. The object then describes a known path called a parabola, where there are x and y components of the speed, displacement, and acceleration.

The object will eventually reach its maximum height (apex) and then it will return to the height from which it was launched. The equation for the height at any time t is

x=v_ocos\theta t

\displaystyle y=y_o+v_osin\theta \ t-\frac{gt^2}{2}

Where vo is the magnitude of the initial velocity, \theta is the angle, t is the time and g is the acceleration of gravity

The maximum height the object can reach can be computed as

\displaystyle t=\frac{v_osin\theta}{g}

There are two times where the value of y is y_o when t=0 (at launching time) and when it goes back to the same level. We need to find that time t by making y=y_o

\displaystyle y_o=y_o+v_osin\theta\ t-\frac{gt^2}{2}

Removing y_o and dividing by t (t different of zero)

\displaystyle 0=v_osin\theta-\frac{gt}{2}

Then we find the total flight as

\displaystyle t=\frac{2v_osin\theta}{g}

We can easily note the total time (hang time) is twice the maximum (apex) time, so the required time is

\boxed{t=24.5/2=12.25\ seconds}

4 0
3 years ago
What is the mass of an atom with six protons, seven neutrons, and eight electrons?
qaws [65]

Answer:

13

Explanation:

The mass of an atom= number of protons + number of neutrons

7 0
3 years ago
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