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Fed [463]
2 years ago
15

Einstein and Lorentz, being avid tennis players, play a fast-paced game on a court where they stand 20.0 m from each other. Bein

g very skilled players, they play without a net. The tennis ball has mass 0.0580 kg. You can ignore gravity and assume that the ball travels parallel to the ground as it travels between the two players. Unless otherwise specified, all measurements are made by the two men.
(a) Lorentz serves the ball at 80.0 m/s. What is the ball's kinetic energy?
(b) Einstein slams a return at 1.80 x 10^8 m/s. What is the ball's kinetic energy?
Physics
1 answer:
kobusy [5.1K]2 years ago
4 0

Answer:

Explanation:

a ) Kinetic energy of ball = 1/2 m v² , m is mass of the ball and v is its velocity

= 1/2 x .058 x 80²

= 185.6 J .

b ) The velocity of ball is very high so we shall find out the relativistic kinetic energy , the formula is given below

kinetic energy = mc² (    \frac{1}{\sqrt{1-\frac{v^2}{c^2} } } -1  )

= mc² x \frac{1}{\sqrt{1-\frac{v^2}{c^2} } }   - mc²

= .058 x 9 x 10¹⁶ x \frac{1}{\sqrt{1-\frac{1.8^2}{3^2} } }  - .058 x 9 x 10¹⁶

= .058 x 9 x 10¹⁶ x 1 / .8 - .058 x 9 x 10¹⁶

.058 x 9 x 10¹⁶ ( 1.25 - 1 )

= .058 x 9 x 10¹⁶ x .25

= .1305 x 10¹⁶

= 13.05 x 10¹⁴ J .

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When Earth’s Northern Hemisphere is tilted toward the Sun during June, some would argue that the cause of our seasons is that th
Ludmilka [50]

Answer:

Distance of Earth from the Sun has nothing to do with the seasons only the tilt is responsible for the change in seasons.

Explanation:

The Earth's tilt does cause the seasons but the distance from the sun and has nothing to do with the change in seasons. In June, when the Northern Hemisphere is tilted in the direction of the Sun during the Northern Hemisphere summer the Earth is actually farthest from the Sun. In January, when the Southern Hemisphere is tilted in the direction of the Sun during the Northern Hemisphere winter the Earth is actually closest to the Sun. This is caused due to the elliptical orbit of the Earth. So, distance of Earth from the Sun has nothing to do with the seasons.

4 0
2 years ago
A 225 kg red bumper car is moving at 3.0 m/s. It hits a stationary 180 kg blue bumper car. The red and blue bumper cars combine
Alex Ar [27]

Given


m1(mass of red bumper): 225 Kg


m2 (mass of blue bumper): 180 Kg


m3(mass of green bumper):150 Kg


v1 (velocity of red bumper): 3.0 m/s


v2 (final velocity of the combined bumpers): ?




The law of conservation of momentum states that when two bodies collide with each other, the momentum of the two bodies before the collision is equal to the momentum after the collision. This can be mathemetaically represented as below:


Pa= Pb


Where Pa is the momentum before collision and Pb is the momentum after collision.


Now applying this law for the above problem we get


Momentum before collision= momentum after collision.


Momentum before collision = (m1+m2) x v1 =(225+180)x 3 = 1215 Kgm/s


Momentum after collision = (m1+m2+m3) x v2 =(225+180+150)x v2

=555v2

Now we know that Momentum before collision= momentum after collision.


Hence we get


1215 = 555 v2


v2 = 2.188 m/s


Hence the velocity of the combined bumper cars is 2.188 m/s

4 0
2 years ago
Read 2 more answers
A sphere of radius 5.00 cm carries charge 3.00 nC. Calculate the electric-field magnitude at a distance 4.00 cm from the center
OlgaM077 [116]

Answer:

a)   E = 8.63 10³ N /C,  E = 7.49 10³ N/C

b)   E= 0 N/C,  E = 7.49 10³ N/C  

Explanation:

a)  For this exercise we can use Gauss's law

         Ф = ∫ E. dA = q_{int} /ε₀

We must take a Gaussian surface in a spherical shape. In this way the line of the electric field and the radi of the sphere are parallel by which the scalar product is reduced to the algebraic product

The area of ​​a sphere is

        A = 4π r²

 

if we use the concept of density

        ρ = q_{int} / V

        q_{int} = ρ V

the volume of the sphere is

      V = 4/3 π r³

         

we substitute

         E 4π r² = ρ (4/3 π r³) /ε₀

         E = ρ r / 3ε₀

the density is

         ρ = Q / V

         V = 4/3 π a³

         E = Q 3 / (4π a³) r / 3ε₀

         k = 1 / 4π ε₀

         E = k Q r / a³

 

let's calculate

for r = 4.00cm = 0.04m

        E = 8.99 10⁹ 3.00 10⁻⁹ 0.04 / 0.05³

        E = 8.63 10³ N / c

for r = 6.00 cm

in this case the gaussine surface is outside the sphere, so all the charge is inside

         E (4π r²) = Q /ε₀

         E = k q / r²

let's calculate

         E = 8.99 10⁹ 3 10⁻⁹ / 0.06²

          E = 7.49 10³ N/C

b) We repeat in calculation for a conducting sphere.

For r = 4 cm

In this case, all the charge eta on the surface of the sphere, due to the mutual repulsion between the mobile charges, so since there is no charge inside the Gaussian surface, therefore the field is zero.

         E = 0

In the case of r = 0.06 m, in this case, all the load is inside the Gaussian surface, therefore the field is

        E = k q / r²

      E = 7.49 10³ N / C

6 0
2 years ago
Select the correct answer from each drop-down menu. A baking tray is made of metal because it’s of heat. An oven mitt is used to
Margarita [4]

An oven mitt is used to take the tray out of the oven because it’s an insulator.

5 0
2 years ago
Read 2 more answers
Nerve impulses are carried along axons, the elongated fibers that transmit neural signals. We can model an axon as a tube with a
jeka94

Answer:

The resistance of the axon is 1.27\times 10^7\ \Omega.

Explanation:

Given that,

Inner diameter of the model of an axon, d=10\ \mu m

Radius of the model, r=5\ \mu m=5\times 10^{-6}\ m

Resistivity of the fluid inside the tube wall, \rho=0.5\ \Omega -m

Length of the axon, l = 2 mm = 0.002 m

We know that the resistance in terms of resistivity of an object is given by :

R=\rho\dfrac{l}{A}\\\\R=0.5\times \dfrac{0.002}{\pi (5\times 10^{-6})^2}\\\\R=1.27\times 10^7\ \Omega

So, the resistance of the axon is 1.27\times 10^7\ \Omega. Hence, this is the required solution.

8 0
2 years ago
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