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Gala2k [10]
3 years ago
9

What is newton's gravitational constant (G)​

Physics
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

6 0
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:

6 0
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
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