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2 years ago

A man weighs 90 N on the surface of earth of radius R at what height above the surface of earth is weight will be 40 N.

Physics

1 answer:

lapo4ka [179]2 years ago

4 0

Answer:

3/2 R

Explanation:

Let the Mass of man be m, Mass of earth be M.

By Newtons, law of gravitation,

F = GMm/R^2, Where G is gravitational constant

90 = GMm/R^2

When man weighs 40N,

GMm/(New R)^2 = 40

= 4/9 (90)

GMm/(New R)^2 = 4/9 (GMm/R^2)

(New R)^2 = 9/4 R^2

New R = 3/2 R

Therefore, height is 3/2 R

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Three positive charges A, B, and C, and a negative charge D are placed in a line as shown in the diagram. All four charges are o

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Answer:

a. charge C experiences the greatest net force, and charge B receives the smallest net force

b. ratio=9

Explanation:

Electrostatic Force

Two point-charges $q_1$ and $q_2$ separated a distance d will exert a force on each other of a magnitude given by the Coulomb's formula

$\displaystyle F=\frac{k\ q_1\ q_2}{r^2}$

Where k is the proportional constant of value

$k=9*10^9\ N.m^2/c^2$

The diagram provided in the question shows four identical charges (let's assume their value is Q) separated by identical distance (of value d). The force between the charges next to others is

$\displaystyle F_1=\frac{k\ Q\ Q}{d^2}$

$\displaystyle F_1=\frac{k\ Q^2}{d^2}$

The force between charges separated 2d is

$\displaystyle F_2=\frac{k\ Q^2}{(2d)^2}$

$\displaystyle F_2=\frac{k\ Q^2}{4d^2}$

And the force between the charges A and D is

$\displaystyle F_3=\frac{k\ Q^2}{(3d)^2}$

$\displaystyle F_3=\frac{k\ Q^2}{9d^2}$

Now, let's analyze each charge and the force applied to them by the others

Let's recall equally signed charges repel each other and differently signed charges attrach each other

Charge A. It receives force to the left from B and C and to the right from D

$\displaystyle F_A=-F_1-F_2+F_3=-\frac{k\ Q^2}{d^2}-\frac{k\ Q^2}{4d^2}+\frac{k\ Q^2}{9d^2}$

$\displaystyle F_A=\frac{k\ Q^2}{d^2}(-1-\frac{1}{4}+\frac{1}{9})$

$\displaystyle F_A=-\frac{41}{36}F_1$

Charge B. It receives force to the right from A and D and to the left from C

$\displaystyle F_B=F_1-F_1+F_2=\frac{k\ Q^2}{d^2}-\frac{k\ Q^2}{d^2}+\frac{k\ Q^2}{4d^2}$

$\displaystyle F_B=\frac{1}{4}F_1$

Charge C. It receives forces to the right from all charges.

$\displaystyle F_C=F_2+F_1+F_1=\frac{k\ Q^2}{4d^2}+\frac{k\ Q^2}{d^2}+\frac{k\ Q^2}{d^2}$

$\displaystyle F_C=\frac{9}{4}F_1$

Charge D. It receives forces to the left from all charges

$\displaystyle F_D=-F_3-F_2-F_1=-\frac{k\ Q^2}{9d^2}-\frac{k\ Q^2}{4d^2}-\frac{k\ Q^2}{d^2}$

$\displaystyle F_D=-\frac{49}{36}F_1$

Comparing the magnitudes of each force is just a matter of computing the fractions

$\displaystyle \frac{41}{36}=1.13,\ \frac{1}{4}=0.25,\ \frac{9}{4}=2.25,\ \frac{49}{36}=1.36$

We can see the charge C experiences the greatest net force, and charge B receives the smallest net force

The ratio of the greatest to the smallest net force is

$\displaystyle \frac{\frac{9}{4}}{\frac{1}{4}}=9$

The greatest force is 9 times the smallest net force

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2 years ago

The figure below shows a shooting competition, where air rifles fire soft metal at distant targets

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Answer:

where is the figure?

Explanation:

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2 years ago

A student was traveling to see his grandmother who lives 15 miles north oh his home he started from rest and maintained a pace o

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Explanation:

Given parameters:

Distance = 15miles north = 24140.2m

Initial velocity = 0m/s

Final velocity = 4m/s

Unknown:

Speed, velocity and acceleration = ?

Solution:

The speed is the distance divide by time. It is a scalar quantity and has no directional attribute.

Speed = $\frac{distance }{time}$

The speed of the student is 4m/s

Velocity is the displacement divided by time. It is a vector quantity which specifies the direction and magnitude;

Velocity = $\frac{displacement}{time}$

The velocity of the student is 4m/s due north

Acceleration is the change in velocity with time;

To find the acceleration, we use

v² = u² + 2as

v is the final velocity

u is the initial velocity

a is the acceleration

s is the distance

4² = 0² + 2x a x 24140.2

a = $\frac{16}{2 x 24140.2}$ = 0.00033m/s²

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Answer:

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Explanation:

We must use conservation of linear momentum before and after the collision, $p_i=p_f$

Before the collision we have:

$p_i=p_1+p_2=m_1v_1+m_2v_2$

where these are the masses are initial velocities of both players.

After the collision we have:

$p_f=(m_1+m_2)v_f$

since they clong together, acting as one body.

This means we have:

$m_1v_1+m_2v_2=(m_1+m_2)v_f$

Or:

$v_f=\frac{m_1v_1+m_2v_2}{m_1+m_2}$

Which for our values is:

$v_f=\frac{(98.5kg)(6.05m/s)+(119kg)(-3.5m/s)}{(98.5kg)+(119kg)}=0.825m/s$

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2 years ago

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2 years ago