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DerKrebs [107]
3 years ago
11

A ball encounters no air resistance when thrown into the air with 100 J of kinetic energy, which is transformed to gravitational

potential energy at the top of its trajectory. What is its kinetic energy when it returns to its original height
Physics
2 answers:
nikdorinn [45]3 years ago
4 0

Answer:

The kinetic energy when it returns to its original height is 100 J

Explanation:

The ball is thrown up with a Kinetic Energy K. E. = 0.5×m×v² = 100 J

Therefore the final height is given by

u² = v² -2·g·s

Where:

u = final velocity = 0

v = initial velocity

s = final height

Therefore v² = 2·g·s = 19.62·s

P.E = Potential Energy = m·g·s

Since v² = 2·g·s

Substituting the value of v² in the kinetic energy formula, we obtain

K. E. = 0.5×m×2·g·s = m·g·s = P.E. = 100 J

When the ball returns to the original height, we have

v² = u² + 2·g·s

Since u = 0 = initial velocity in this case we have

v² = 2·g·s and the Kinetic energy = 0.5·m·v²

Since m and s are the same then 0.5·m·v² = 100 J.

finlep [7]3 years ago
4 0

Answer:

100 J

Explanation:

If there is no air resistance, then the kinetic energy is conserved. That at first the ball has 100 J of kinetic energy, when it reaches the top of its trajectory the ball has 100 J of gravitational potential energy (the kinetic energy is zero and all of them at the initial point has transformed into gravitational potential energy) , and finally when it returns to its original height it has again 100 J of kinetic energy (inverse effect than before).

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Moustapha jones drives east at 100km/hr for 3 hours then back west at 80km/hr for 1.5 hours. which pair of answers gives his ave
olya-2409 [2.1K]

The average speed of the car is 93.33 km/hr and the average velocity of the car is 40 km/hr.

The total distance cover in east direction is=100*3=300 km

The total distance cover in the west direction=80*1.5=120 km

The total distance covered is =300+120=420 km

And Total displacement of the car is =300-120=180 km

As we know that the average speed is given as

Avg Speed =Total Distance / Total time

=420/4.5=93.33 km/hr

As we know that the average velocity is given as

Avg Speed =Total Displacement/ Total time

=180/4.5=40 km/hr

Therefore, The average speed of the car is 93.33 km/hr and the average velocity of the car is 40 km/hr.

7 0
3 years ago
1) On the way to the moon, the Apollo astro-
kramer
(1) You must find the point of equilibrium between the two forces,

<span>G * <span><span><span>MT</span><span>ms / </span></span><span>(R−x)^2 </span></span>= G * <span><span><span>ML</span><span>ms / </span></span><span>x^2
MT / (R-x)^2 = ML / x^2

So,

x = R * sqrt(ML * MT) - ML / (MT - ML)
R = is the distance between Earth and Moon.

</span></span></span>The result should be,
x = 3.83 * 10^7m
from the center of the Moon, and 

R - x = 3.46*10^8 m
from the center of the Earth.


(2) As the distance from the center of the Earth is the number we found before,
d = R - x = 3.46*10^8m
The acceleration at this point is
g = G * MT / d^2
g = 3.33*10^-3 m/s^2
6 0
3 years ago
To test the performance of its tires, a car
Rom4ik [11]

The coefficient of static friction is 0.222

Explanation:

In order for the car to remain in circular motion, the frictional force must be able to provide the necessary centripetal force. Therefore, the car will start skidding when the two forces are equal:

\mu mg=m\frac{v^2}{r}

where the term on the left is the frictional force, while the term on the right is the centripetal force, and where

\mu is the coefficient of static friction

m is the mass of the car

g is the acceleration of gravity

v is the speed of the car

r is the radius of the track

In this problem, we have:

r = 564 m

v = 35 m/s

g=9.8 m/s^2

And re-arranging the equation for \mu, we can find the coefficient of static friction:

\mu = \frac{v^2}{gr}=\frac{35^2}{(9.8)(564)}=0.222

Learn more about friction:

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5 0
3 years ago
A light wave travels through air at a speed of 3.0x108 m/s. Green light has a wavelength of about 5.76x1014Hz. What is the wavel
solong [7]

Answer:

521 nm

Explanation:

Given the values and units we are given, I'm assuming  5.76*10^14 Hz is frequency.

The formula to use here is λ * υ = c, where λ is wavelength, υ is frequency, and c is the speed of light.

λ = \frac{3*10^8\frac{m}{s} }{5.76*10^{14}Hz} = {5.20833*10^{-7} m}\approx{521 *10^{-9}m}={521 nm}

4 0
3 years ago
A convex spherical mirror having a radius of curvature 18 cm (focal length = 1/2 radius of curvature for a spherical mirror) pro
tekilochka [14]

Answer:

distance between object and image =  18.9 cm

Explanation:

given data

radius of curvature = 18 cm

focal length = 1/2 radius of curvature

magnification = 40%

to find out

distance between object and image

solution

we know lens formula that is

1/f = 1/v + 1/u     ....................1

here f = 18 /2 and v and u is object and image distance

and we know m = 40% = 0.40

so 0.40 = -v / u

so here v = - 0.40 u

so from equation 1

1/f = 1/v + 1/u

2/18 = - 1/0.40u + 1/u

u = -13.5 cm   ..................2

and

v = -0.40 (- 13.5)

v = 5.4 cm     ......................3

so from equation 2 and 3

distance between object and image =  5.4 + 13.5

distance between object and image =  18.9 cm

6 0
3 years ago
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