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adelina 88 [10]
3 years ago
12

A person of mass m is standing on the surface of the Earth, of mass M E . What is the acceleration that the Earth experiences du

e to the person's gravitational pull?
Physics
1 answer:
Lana71 [14]3 years ago
7 0

Answer:

a_E=\dfrac{Gm}{r^2}

Explanation:

M_E = Mass of the Earth =  5.972 × 10²⁴ kg

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

r = Radius of Earth = 6371000 m

m = Mass of person

The force on the person will balance the gravitational force

M_Ea_E=\dfrac{GmM_E}{r^2}\\\Rightarrow a_E=\dfrac{Gm}{r^2}

The acceleration that the Earth will feel is a_E=\dfrac{Gm}{r^2}

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Hans Langseth's beard measured 5.33 m in 1927. Consider two charges, q1 = 2.42 nC and an unspecified charge, q2, are separated 5
schepotkina [342]

Answer:

-7.89 * 10^(-9) C

Explanation:

Parameters given:

q1 = 2.42 nC = 2.42 * 10^(-9) C

Distance between q1 and q2 = 5.33 m

q3 = 1.0 nC = 1 * 10^(-9) C

Distance between q1 and q3 = 1.9 m

Distance between q2 and q3 = 5.33 - 1.9 = 3.43 m

The net force acting on q3 is:

F = F(q1, q3) + F(q2, q3)

F = (k*q1*q3)/1.9² + (k*q2*q3)/3.43²

F = (9 * 10^(9) * 2.42 * 10^(-9) * 1 * 10^(-9))/3.61 + (9 * 10^(9) * q2 * 1 * 10^(-9))/11.7649

F = 6.033 * 10^(-9) + 0.765*q2

If the net force is zero:

0 = 6.033 * 10^(-9) + 0.765*q2

-0.765*q2 = 6.033 * 10^(-9)

=> q2 = -[6.033 * 10^(-9)]/0.765

q2 = -7.89 * 10^(-9) C

3 0
3 years ago
1. In Newton’s ring experiment, the diameter of the 5th ring is 0.30 cm and diameter of 15th the ring is 0.62 cm. Find the diame
IgorC [24]

Answer:

Diameter of Newton’s 5th ring = 0.30 cm

Diameter of Newton’s 15th ring = 0.62 cm

Diameter of Newton’s 25th ring = ?

From Newton’s rings experiment we infer that

D2n+m − D2n = 4λmR

For the 5th and 15th rings we have

D215 − D25 = 4λ * 10 * R _______ (1) (m = 10)

For 15th and 25th rings

D225 − D215 = 4λ * 10 * R _______ (2) (m = 10)

We equate the two derivatives

Equation (2) = Equation (1)

D225 − D215 = D215 − D25

D225 = 2D215 – D25

Substituting the values into the equation

D225 = 2 * 0.62 * 0.62 – 0.3 * 0.3 =0.6788 cm2

D25 = 0.8239 cm

4 0
3 years ago
The​ time, t, required to drive a fixed distance varies inversely as the​ speed, r. It takes 3 hr at a speed of 14 ​km/h to driv
kenny6666 [7]

Answer:

time taken with speed 23 km/h will be 1.8 hours or 1 hour 48 minutes

Explanation:

Given:

Time is inversely proportional to the speed

mathematically,

t ∝ (1/r)

let the proportionality constant be 'k'

thus,

t = k/r

therefore, for case 1

time = 3 hr

speed = 14 km/hr

3 = k/14

also,

for case 2

let the time be = t

r = 23 km/h

thus,

we have

t = k/23

on dividing equation 2 by 1

we get

\frac{t}{3}=\frac{k/23}{k/14}

or

t=\frac{14\times3}{23}

or

t = 1.8 hr = or 1 hour 48 minutes ( 0.8 hours × 60 minutes/hour = 48 minutes)

4 0
3 years ago
Please help me find the equations guys
Alexeev081 [22]

The line at the bottom of the picture ... probably the first line on a list of choices  .. is the correct equation.

4 0
3 years ago
3 cases where kinetic energy become potential energy
finlep [7]
--  Toss a rock straight up.  The kinetic energy you give it
with your hand becomes potential energy as it rises. 
Eventually, when its kinetic energy is completely changed
to potential energy, it stops rising.

--  When you're riding your bike and going really fast, you come
to the bottom of a hill.  You stop pedaling, and coast up the hill.
As your kinetic energy changes to potential energy, you coast
slower and slower.  Eventually, your energy is all potential, and
you stop coasting.

--  A little kid on a swing at the park.  The swing is going really fast
at the bottom of the arc, and then it starts rising.  As it rises, the
kinetic energy changes into potential energy, more and more as it
swings higher and higher.  Eventually it reaches a point where its
energy is all potential; then it stops rising, and begins falling again.
8 0
3 years ago
Read 2 more answers
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