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inysia [295]
3 years ago
13

Swinging a tennis racket against a ball is an example of a third class lever. Please select the best answer from the choices pro

vided. T F
Physics
2 answers:
kari74 [83]3 years ago
7 0
that statement is true

a Third class lever applied when the effort place between the load and the fulcrum.

For example, in a forearm serve
Fulcrum : The elbow
Effort : The effort that putted by the biceps muscle
Load : The arm

Kobotan [32]3 years ago
4 0

Answer:

The given statement is true.

Explanation:

One of the examples of a simple machine is Levers. It reduces the human effort to do work and make work easier. There are three classes of the lever.

The given case is "swinging a tennis racket against a ball". It is an example of the third class lever. In this type of class, the effort is placed in between the load and the fulcrum.

Hence, the swinging a tennis racket against a ball is an example of a third class lever. This statement is true.

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What is the phase of the moon when the moon is positioned between the sun and earth?
vladimir2022 [97]

Answer:

Solar eclipse

Explanation:

A solar eclipse is when the moon passes between the sun and Earth, causing it to go dark and to give the moon a halo effect :D hope this helped!

8 0
2 years ago
MathPhys Help pls Tysm
HACTEHA [7]

Answer:

8.75

Explanation:

First, find the force of friction.

Kinetic energy = work done by friction

½ mv² = Fd

½ (3.9 kg) (2.9 m/s)² = F (1.4 m)

F = 11.7 N

Next, find the distance at the new velocity.

Kinetic energy = work done by friction

½ mv² = Fd

½ (3.9 kg) (2.5 × 2.9 m/s)² = (11.7 N) d

d = 8.75 m

3 0
3 years ago
Read 2 more answers
Data that shows consistent, regular, repetitive form displays a pattern.<br><br> True<br><br> False
anyanavicka [17]
I think the answer is True
6 0
3 years ago
Read 2 more answers
An electron in the beam of a cathod-ray tube is accelerated by a potential difference of 2.12 kV . Then it passes through a regi
son4ous [18]

Answer:

B=9.1397*10^-4 Tesla

Explanation:

To find the velocity first we put kinetic energy og electron is equal to potential energy of electron

K.E=P.E

\frac{1}{2}*m*v^{2}  =e*V

where :

m is the mass of electron

v is the velocity

V is the potential difference

v=\sqrt{\frac{2*e*V}{m} }    eq 1

Radius of electron moving in magnetic field is given by:

R=\frac{m*v}{q*B}       eq 2

where:

m is the mass of electron

v is the velocity

q=e=charge of electron

B is the magnitude of magnetic field

Put v from eq 1 into eq 2

R=\frac{m*\sqrt{\frac{2*e*V}{m} } }{e B}

B=\sqrt{\frac{2*m*V}{e*R^{2} } }

B=\sqrt{\frac{2*(9.31*10^{-31})*(2.12*10^{3})  }{(1.60*10^{-19})*(0.170)^{2}  } }

B=9.1397*10^-4 Tesla

3 0
3 years ago
Toy car in a science experiment covers 1.6 meters in half a second. If a the car travels at a steady speed, how far will it go i
Tanzania [10]
The answer is D. 32 m.

The simple equation that connects speed (v), time (t), and distance (d) can be expressed as:
v= \frac{d}{t}         ⇒ d=v*t

It is given:
v =  \frac{1.6m}{0.5s} = \frac{1.6m*2}{0.5s*2}= \frac{3.2m}{1s}  = 3.2 m/s
t = 10 s
d = ?

So:
d= v*t=3.2m/s*10s = 32m
3 0
3 years ago
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