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3 years ago

What is the force between two proton’s that are 0.005m apart?

Physics

1 answer:

Jet001 [13]3 years ago

7 0

Answer:

Force, $F=9.21\times 10^{-24}\ N$

Explanation:

Distance between proton is 0.005 m

It is required to find the force between two protons. Protons have of positive charge of $1.6\times 10^{-19}\ C$ . Two protons will have a force of repulsion between them. The force is given by :

$F=\dfrac{kq^2}{r^2}\\\\F=\dfrac{9\times 10^9\times (1.6\times 10^{-19})^2}{(0.005)^2}\\\\F=9.21\times 10^{-24}\ N$

So, the force between two protons is $9.21\times 10^{-24}\ N$ .

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If I drop a watermelon from the top of one of the tower dorms at CSU, and it takes 3.34 seconds to hit the ground, calculate how

WINSTONCH [101]

T= 3.34

Vi= 0

A= 9.81

D= ?

d=Vit+1/2at^2

d= 1/2(9.81)(3.34)2

d= 54.7 or 55 meters tall

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3 years ago

An athlete in a gym applies a constant force of 50 N to the pedals of a bicycle to keep the rotation rate of the wheel at 10 rev

marishachu [46]

Answer:

Explanation:

Given that,

Force applied to pedal F = 50N

Angular velocity ω = 10rev/s

We know that, 1rev = 2πrad

Then, ω = 10rev/s = 10×2π rad/s

ω = 20π rad/s

Length of pedal r = 30cm = 0.3m

Power?

Power is given as

P = τ×ω

We need to find the torque τ

τ = r × F

Since r is perpendicular to F

Then, τ = 0.3 × 50

τ = 15 Nm

Then,

P = τ×ω

P = 15 × 20π

P = 942.48 Watts

power delivered to the bicycle by the athlete is 942.48 W

6 0

3 years ago

Why are the trends for electronegativity and ionization energy similar

Goshia [24]

Answer:

Explanation:

Ionization Energy Trends

Ionization energy is the energy required to remove an electron from a neutral atom in its gaseous phase. Conceptually, ionization energy is the opposite of electronegativity. ... As a result, it is easier for valence shell electrons to ionize, and thus the ionization energy decreases down a group

3 0

3 years ago

A 91.5 kg football player running east at 3.73 m/s tackles a 63.5 kg player running east at 3.09 m/s. what is their velocity aft

PIT_PIT [208]

Their velocity afterward is v=3.467 m/s

Explanation:

Given:

Mass of the first football player= 91.5 kg

Initial velocity of the football player 3.73 m/s

Mass of second football player=63.56 kg

Initial velocity of the second football player=3.09 m/s

To find:

Final velocity of both players=?

Solution:

According to the law of conservation of momentum,

Initial momentum =final momentum

mathematically represented as

$m_1u_1+m_2u_2=m_1v_1+m_2v_2$ ...........................(1)

where

$u_1$ =intial velocity of the football player

$u_2$ = inital velocity of second football player

$v_1$ =finall velocity of the first football player

$v_2$ =final velocity of second football player

after tackling , both the football players moves with the same velocity,

so $v_1=v_2=v$

Hence equation (1) becomes

$m_1u_1+m_2u_2=(m_1+m_2)v$

$v=\frac{m_1u_1+m_2u_2}{(m_1+m_2)}$

now substituting the values,

$v=\frac{(91.5\times+3.73)+(63.5\times3.09)}{(91.5+63.5)}$ $v=\frac{(341.29+196.210)}{155}$

$v=\frac{537.5}{155}$

v=3.467 m/s

7 0

3 years ago

A skydiver jumps out of a plane and immediately begins falling toward the Earth.

worty [1.4K]

Answer:

5.2m/s^2

Explanation:

A body that moves with constant acceleration means that it moves in "a uniformly accelerated motion", which means that if the velocity is plotted with respect to time we will find a line and its slope will be the value of the acceleration, it determines how much it changes the speed with respect to time.

When performing a mathematical demonstration, it is found that the equations that define this movement are as follows.

$Vf=Vo+a.t (1)\\{Vf^{2}-Vo^2}/{2.a} =X(2)\\X= VoT+0.5at^{2} (3)\\X=(Vf+Vo)T/2 (4)$

Where

Vf = final speed

Vo = Initial speed

T = time

A = acceleration

X = displacement

In conclusion to solve any problem related to a body that moves with constant acceleration we use the 4 above equations and use algebra to solve

for this case we can use the ecuation number 3

x=100m

t=6.2s

Vo=0m/s

$X= VoT+0.5at^{2}\\X= 0.5at^{2}\\a=\frac{X}{0.5t^2} \\a=\frac{100}{0.5(6.2)^2}=5.2m/s^2$

7 0

3 years ago