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3 years ago

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What is newton's gravitational constant (G)

1 answer:

Setler [38]3 years ago

5 0

The gravitational constant, denoted by the letter G, is an empirical physical constant involved in the calculation of gravitational effects in Sir Isaac Newton's law of universal gravitation and in Albert Einstein's general theory of relativity

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Marcus is on a train that travels 20 and takes 5 seconds to slow to 10 .

Vera_Pavlovna [14]

Answer:

-2

Explanation:

v0=20

v=10

t=5

a=(v-v0)/t=(10-20)/5=-2

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3 years ago

A 215-kg load is hung on a wire of length of 3.60 m, cross-sectional area 2.00 10-5 m2, and Young's modulus 8.00 1010 N/m2. What

Y_Kistochka [10]

Answer:

4.74 * 10^-3

Explanation:

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3 years ago

A massless, hollow sphere of radius R is entirely filled with a fluid such that its density is p. This same hollow sphere is now

Vesnalui [34]

Answer:

a. 8p

Explanation:

We are given that

Radius of hollow sphere , R1=R

Density of hollow sphere= $\rho$

After compress

Radius of hollow sphere, R2=R/2

We have to find density of the compressed sphere.

We know that

$Density=\frac{mass}{volume}$

$Mass=Density\times volume=Constant$

Therefore, $\rho_1 V_1=\rho_2V_2$

Volume of sphere= $\frac{4}{3}\pi r^3$

Using the formula

$\rho\times \frac{4}{3}\pi R^3=\rho_2\times \frac{4}{3}\pi (R/2)^3$

$\rho R^3=\rho_2\times \frac{R^3}{8}$

$\rho_2=8\rho$

Hence, the density of the compressed sphere= $8\rho$

Option a is correct.

7 0

3 years ago

A 100 kg cart is moving at 3 m/s. Calculate the cart’s kinetic energy.

STALIN [3.7K]

Answer:

450

Explanation:

Given,

Mass= 100kg

Velocity= 3 m/s

Kinetic Energy= ?

Kinetic Energy= 1/2 mv^2

= 1/2× 100× 3^2

= 1/2× 900

= 450.

<em>HOPE</em><em> </em><em>IT</em><em> </em><em>HELPED</em><em> </em><em>:</em><em>)</em>

8 0

3 years ago

Read 2 more answers

What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3 = 55.0 nC placed between q1

notka56 [123]

Complete Question

Part of the question is shown on the first uploaded image

The rest of the question

What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3 = 55.0 nC placed between q1 and q2 at x3 = -1.220 m ? Your answer may be positive or negative, depending on the direction of the force. Express your answer numerically in newtons to three significant figures.

Answer:

The net force exerted on the third charge is $F_{net}= 3.22*10^{-5} \ J$

Explanation:

From the question we are told that

The third charge is $q_3 = 55 nC = 55 *10^{-9} C$

The position of the third charge is $x = -1.220 \ m$

The first charge is $q_1 = -16 nC = -16 *10^{-9} \ C$

The position of the first charge is $x_1 = -1.650m$

The second charge is $q_2 = 32 nC = 32 *10^{-9} C$

The position of the second charge is $x_2 = 0 \ m$

The distance between the first and the third charge is

$d_{1-3} = -1.650 -(-1.220)$

$d_{1-3} = -0.43 \ m$

The force exerted on the third charge by the first is

$F_{1-3} = \frac{k q_1 q_3}{d_{1-3}^2}$

Where k is the coulomb's constant with a value $9*10^{9} \ kg\cdot m^3\cdot s^{-4}\cdot A^2.$

substituting values

$F_{1-3} = \frac{9*10^{9}* 16 *10^{-9} * (55*10^{-9})}{(-0.43)^2}$

$F_{1-3} = 4.28 *10^{-5} \ N$

The distance between the second and the third charge is

$d_{2-3} = 0- (-1.22)$

$d_{2-3} =1.220 \ m$

The force exerted on the third charge by the first is mathematically evaluated as

$F_{2-3} = \frac{k q_2 q_3}{d_{2-3}^2}$

substituting values

$F_{2-3} = \frac{9*10^{9} * (32*10^{-9}) *(55*10^{-9})}{(1.220)^2}$

$F_{2-3} = 1.06*10^{-5} N$

The net force is

$F_{net} = F_{1-3} -F_{2-3}$

substituting values

$F_{net} = 4.28 *10^{-5} - 1.06*10^{-5}$

$F_{net}= 3.22*10^{-5} \ J$

6 0

3 years ago