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3 years ago

How could you figure out how fast you can run

Physics

2 answers:

SashulF [63]3 years ago

8 0

You could use a RADAR gun wich shows the mph of cars but u could use it on humans

fomenos3 years ago

4 0

By timing yourself :)

or are you talking about mph?

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MA of the first class lever may be equal to, greater than 1.Why

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Answer:

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Explanation:

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3 years ago

Define solar system. Mention 3 types of energy.

sveticcg [70]

Answer:

The Solar System is the Sun and all the objects that orbit around it

The Sun is orbited by planets, asteroids, comets and other things. ...

It has strong gravity

There are many types of energies

three types are

kinetic energy

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gravitational energy

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3 years ago

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A ball is thrown up in the air. When it falls back down and reaches the ground, it's final downward velocity is 8.0 m/s. What wa

hram777 [196]

Given : Time taken to reach the maximum height t=3 s a=−g=−10m/s
2

The initial velocity of the ball can be calculated by,
Using v=u+at
∴ 0=u−10×3 ⟹u=30 m/s

Using S=ut+
2
1

at
2

∴ S=30×3+
2
1

×(−10)×3
2
=45m

8 0

3 years ago

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Two small charged spheres are located on the y-axis. One is at y = 1.00 m, the other is at y = −1.00 m, and they both have a cha

yuradex [85]

Answer:

(a) 23.946 kV

(b) -0.077 J

Explanation:

(a) The electric potential is given by the following formula:

$V=k\frac{q_1}{r_1}+k\frac{q_2}{r_2}$ (1)

k: Coulomb's constant = 8.98*10^9 Nm^2/C^2

q1 = q2 = 1.60*10^{-6}C

r1 and r2 are the distance from the charges to the point in which electric potential is evaluated.

Firs, you calculate the distance r1 and r2 by taking into account the position of the charges

$r_1=\sqrt{(1.00)^2+(0.670)^2}m=1.20m\\\\r_2=\sqrt{(-1.00)^2+(0.670)^2}m=1.20m$

Next, you replace the values of the parameters to calculate V:

$V=(8.98*10^9Nm^2/C^2)\frac{1.6*10^{-6}C}{1.20m}+(8.98*10^9Nm^2/C^2)\frac{1.6*10^{-6}C}{1.20m}\\\\V=23946.66\ V=23.946\ kV$

(b) The potential electric energy is given by:

$U_T=U_{1,2}+U_{1,3}+U_{2,3}\\\\U_T=k\frac{q_1q_2}{r_{1,2}}+k\frac{q_1q_3}{r_{1,3}}+k\frac{q_2q_3}{r_{2,3}}\\\\r_{1,2}=2.00m\\\\r_{1,3}=1.20m\\\\r_{2,3}=1.20m\\\\U_T=(8.98*10^9)[\frac{(1.6*10^{-6})^2}{2.00m}+\frac{(1.6*10^{-6})(-3.70*10^{-6})}{1.20}+\frac{(1.6*10^{-6})(-3.70*10^{-6})}{1.20}]J\\\\U_T=-0.077J$

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3 years ago

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PtichkaEL [24]

Who is the right answer

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3 years ago

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