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2 years ago

20.What factor is more important in determining electrical force

Physics

1 answer:

Agata [3.3K]2 years ago

6 0

Answer:

20.The first factor is the amount of charge on each object. The greater the charge, the greater the electric force. The second factor is the distance between the charges. The closer together the charges are, the greater the electric force is

Explanation:

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A ball is thrown vertically upward (assumed to be the positive direction) with a speed of 24.0 m/s from a height of 3.0 m. (a) H

SashulF [63]

Answer:

a) 29.36 m

b) 2.44 s

c) 2.57 s

d) 25.117 m/s

Explanation:

t = Time taken

u = Initial velocity = 24 m/s

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²

$v=u+at\\\Rightarrow 0=24-9.81\times t\\\Rightarrow \frac{-24}{-9.81}=t\\\Rightarrow t=2.44 \s$

Time taken by the ball to reach the highest point is 2.44 seconds

$s=ut+\frac{1}{2}at^2\\\Rightarrow s=24\times 2.44+\frac{1}{2}\times -9.81\times 2.44^2\\\Rightarrow s=29.35\ m$

The highest point reached by the ball above its release point is 29.36 m

c) Total height is 3+29.35 = 32.35 m

$s=ut+\frac{1}{2}at^2\\\Rightarrow 32.35=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{32.35\times 2}{9.81}}\\\Rightarrow t=2.57\ s$

The ball reaches the ground 2.57 seconds after reaching the highest point

$v=u+at\\\Rightarrow v=0+9.81\times 2.57\\\Rightarrow v=25.2117\ m/s$

The ball will hit the ground at 25.2117 m/s

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2 years ago

[1] The assembly starts from rest and reaches an angular speed of 150 rev/min under the action of a 20-N force T applied to the

ExtremeBDS [4]

Answer:

t = 5.89 s

Explanation:

To calculate the time, we need the radius of the pulley and the radius of the sphere which was not given in the question.

Let us assume that the radius of the pulley ( $r_p$ ) = 0.4 m

Let the radius of the sphere (r) = 0.5 m

w = angular speed = 150 rev/min = (150 × 2π / 60) rad/s = 15.708 rad/s

Tension (T) = 20 N

mass (m) = 3 kg each

$\int\limits^0_t {Tr_p} \, dt=H_2-H_1\\( Tr_p)t=4rm(rw)\\( Tr_p)t=4r^2mw$

$t = \frac{4r^2mw}{Tr_P}$

Substituting values:

$t = \frac{4r^2mw}{Tr_P}= \frac{4*(0.5)^2*3*15.708}{20*0.4}=5.89s$

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2 years ago

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Answer:

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