The electrons equal or copy the number of ____

a roller coaster is at the top of a hill and rolls to the top of a lower hill.If mechanical energy is conserved,on the top of wh

pychu [463]

Hello!!

Here we have a simple matter of conservation of energy. ME=PE+KE.

At point A we have PE=mgh and KE=1/2mv^2. At point A all we have is PE since the coaster isn’t rolling yet. But by conservation of energy, we know that it will have enough energy to roll down and get to and equal height on another hill. Providing we are neglecting friction and drag and resistance forces which we are in this case. So we can conclude that the KE will be greater at Point B since ME=PE+KE and for ME to remain the same and we know the PE is less on lower hill, so we can conclude that KE on lower hill is greater to keep ME the same and have conservation of energy.

Hope this helps you understand the concept!! Any questions please just ask!! Thank you so much!!

7 0

4 years ago

When passing from a low index of refraction medium to a high index of refraction medium which way does the refracted ray bend: t

Vaselesa [24]

Answer:

toward the normal

Explanation:

Light travels at different speed in different mediums.

Refractive index is equal to velocity of the light 'c' in empty space divided by the velocity 'v' in the substance.

Or ,

n = c/v.

Light travels at a slower speed in water as compared to air because there are more number of interfering molecules in the path of the light in case of water as compared to liquid.

When a light travels from lower denser medium say water to higher denser medium say water, it bends towards the perpendicular (normal) as its speed reduces in that medium.

6 0

4 years ago

The parachute on a drag racing car deploys at the end of a run. If the car has a mass of 820 kg and the car is moving 36 m/s, wh

Lelechka [254]

In order to determine the required force to stop the car, proceed as follow:

Calculate the deceleration of the car, by using the following formula:

$v^2=v^2_o-2ax$

where,

v: final speed = 0m/s (the car stops)

vo: initial speed = 36m/s

x: distance traveled = 980m

a: deceleration of the car= ?

Solve the equation above for a, replace the values of the other parameters and simplify:

$\begin{gathered} a=\frac{v^2_o-v^2}{2x} \\ a=\frac{(36\frac{m}{s})^2-(0\frac{m}{s})^2}{2(980m)}=0.66\frac{m}{s^2} \end{gathered}$

Next, consider that the formula for the force is:

$F=ma$

where,

m: mass of the car = 820 kg

a: deceleration of the car = 0.66m/s^2

Replace the previous values and simplify:

$F=(820kg)(0.66\frac{m}{s^2})=542.20N$

Hence, the required force to stop the car is 542.20N

4 0

1 year ago

A cord of mass 0.65 kg is stretched between two supports 28 m apart. if the tension in the cord is 150 n, 2 how long will it tak

MaRussiya [10]

Ans: Time <span>taken by a pulse to travel from one support to the other = 0.348s
</span>
Explanation:
First you need to find out the speed of the wave.

Since
Speed = v = $\sqrt{ \frac{T}{\mu} }$

Where
T = Tension in the cord = 150N
μ = Mass per unit length = mass/Length = 0.65/28 = 0.0232 kg/m

So
v = $\sqrt{ \frac{150}{0.0232} }$ = 80.41 m/s

Now the time-taken by the wave = t = Length/speed = 28/80.41=0.348s

5 0

3 years ago

Andy Petite pitches a 0.8 kg baseball with a velocity of 67 m/s. Josh Hamilton

kodGreya [7K]

The Impulse delivered to the baseball is 89 kgm/s.

To solve the problem above, we use the formula of impulse.

⇒ Formula:

I = m(v-u)................. Equation 1

Where:

I = Impulse delivered to the baseball
m = mass of the baseball
v = Final velocity of the baseball
u = initial speed of the baseball

From the question,

⇒ Given:

m = 0.8 kg
u = 67 m/s
v = -44 m/s

⇒ Substitute these values into equation 1

I = 0.8(-44-67)
I = 0.8(-111)
I = -88.8
I ≈ -89 kgm/s

Note: The negative tells that the impulse is in the same direction as the final velocity and therefore can be ignored.

Hence, The Impulse delivered to the baseball is 89 kgm/s.

Learn more about impulse here: brainly.com/question/7973509

7 0

2 years ago

The electrons equal or copy the number of _____