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victus00 [196]
2 years ago
7

Einstein and lorentz, being avid tennis players, play a fast-paced game on a court where they stand 19.0 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.
Physics
1 answer:
Dvinal [7]2 years ago
7 0
So sometimes when you actually do you work it’s happens and you get answers idk about that
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1. Although mercury is a metal, it is a liquid at room temperature. Mercury melts at about -39°C. If
grigory [225]

Answer:

I think so because if it starts at a low temperature for that material, it should melt when you bring it up to that temperature.

7 0
2 years ago
The process of achieving fitness in the human body is almost instant and can occur overnight. True or False
777dan777 [17]
This statement is false nothing in the human body can occur overnight
3 0
3 years ago
Coherent light of wavelength 525 nm passes through two thin slits that are 4.15×10^(−2) mm apart and then falls on a screen 7
IRINA_888 [86]

A) 4.7 cm

The formula for the angular spread of the nth-maximum from the central bright fringe for a diffraction from two slits is

sin \theta=\frac{n \lambda}{d}

where

n is the order of the maximum

\lambda is the wavelength

a is the distance between the slits

In this problem,

n = 5

\lambda=525 nm =5.25\cdot 10^{-7} m

a=4.15\cdot 10^{-2} mm=4.15\cdot 10^{-5} m

So we find

\theta=sin^{-1} (\frac{(5)(5.25\cdot 10^{-7} m)}{4.15\cdot 10^{-5} m})=3.62^{\circ}

And given the distance of the screen from the slits,

D=75.0 cm = 0.75 m

The distance of the 5th  bright fringe from the central bright fringe will be given by

y=D tan \theta = (0.75 m)tan 3.62^{\circ}=0.047 m = 4.7 cm

B) 8.1 cm

The formula to find the nth-minimum (dark fringe) in a diffraction pattern from double slit is a bit differente from the previous one:

sin \theta=\frac{(n+\frac{1}{2}) \lambda}{d}

To find the angle corresponding to the 8th dark fringe, we substitute n=8:

\theta=sin^{-1} (\frac{(8+\frac{1}{2})(5.25\cdot 10^{-7} m)}{4.15\cdot 10^{-5} m})=6.17^{\circ}

And the distance of the 8th dark fringe from the central bright fringe will be given by

y=D tan \theta = (0.75 m)tan 6.17^{\circ}=0.081 m = 8.1 cm

5 0
3 years ago
In a lever, the effort arm is two times as long as the load arm. The resultant force will be
jasenka [17]
Answer is B. 

In a lever, the effort arm is 2 times as a long as the load arm. The resultant force will be twice the applied force.

Hope it helped you.

-
Charlie
5 0
3 years ago
Read 2 more answers
A ball is tossed vertically upward. When it reaches its highest point (before falling back downward) Group of answer choices the
nignag [31]

Answer:

the velocity is zero, the acceleration is directed downward, and the force of gravity acting on the ball is directed downward

Explanation:

Is this exercise in kinematics

          v = v₀ - g t

where g is the acceleration of the ball, which is created by the attraction of the ball to the Earth.

At the highest point

velocity must be zero.

The acceleration depends on the Earth therefore it is constant at this point and with a downward direction.

The force of the earth on the ball is towards the center of the Earth, that is, down

all other alternatives are wrong

7 0
2 years ago
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