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densk [106]
3 years ago
9

Consider an electron and a proton separated by a distance of 4.5 nm. (a) what is the magnitude of the gravitational force betwee

n them? 5.1128e-20 incorrect: your answer is incorrect. n (b) what is the magnitude of the electric force between them?
Physics
1 answer:
Dmitry_Shevchenko [17]3 years ago
6 0
A.) For letter a, we use the law of universal gravitation using the constant G = 6.674×10−<span>11 m3</span>⋅kg−1⋅s−<span>2

Grav. F = G*m1*m2*(1/d^2)

m1 is mass of electron = </span>9.11 × 10-31<span> kg
m2 is mass of proton = </span>1.67 × 10<span>-27 kg
d = 4.5 nm = 4.5 x 10^-9 m

Grav F = 5.01 x 10^-51 N

b.) </span>For letter b, we use the Coulomb's using the constant k = 9×10^9 N

Electric force = k*Q1*Q2*(1/d^2)
Q1 is charge of electron = -1.6 × 10-19 C
Q2 is charge of proton = +1.6 × 10-19 C

Electric force = 1.14 x 10^-11 N
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The phenomenon of vehicle "tripping" is investigated here. The sport-utility vehicle is sliding sideways with speed v1 and no an
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Answer:

v_1  = 3.5 \ m/s

Explanation:

Given that :

mass of the SUV is  = 2140 kg

moment of inertia about G , i.e I_G = 875 kg.m²

We know from the conservation of angular momentum that:

H_1= H_2

mv_1 *0.765 = [I+m(0.765^2+0.895^2)] \omega_2

2140v_1*0.765 = [875+2140(0.765^2+0.895^2)] \omega_2

1637.1 v_1 = 3841.575 \omega_2

\omega_2 = \frac{1637.1 v_1}{3841.575}

\omega _2 = 0.4626 \ v_1

From the conservation of energy as well;we have :

T_2 +V_{2  \to 3} = T_3 \\ \\ \\  \frac{1}{2} I_A \omega_2^2 - mgh =0

[\frac{1}{2} [875+2140(0.765^2+0.895^2)](0.4262 \ v_1)^2 -2140(9.81)[\sqrt{0.76^2+0.895^2} -0.765]] =0

706.93 \ v_1^2 - 8657.49 =0

706.93 \ v_1^2  = 8657.49

v_1^2  =  \frac{8657.49}{706.93 }

v_1 ^2 =  12.25

v_1  = \sqrt{ 12.25

v_1  = 3.5 \ m/s

6 0
3 years ago
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