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Georgia [21]
2 years ago
9

The gravitational force between two volleyball players is 3.9 × 10^-7 N . If the kasses of the players are 63 kg and 78 kg, what

is their separation?​
Physics
1 answer:
N76 [4]2 years ago
3 0

Answer:

I honestly have no idea bro.

Explanation:

Hope that helps have a great night.

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Explain what the star looks like to Jane
S_A_V [24]
The star looks like a desirable hunk of masculinity to Jane. But to John, the star looks like a wimpy momma's boy who might compete with him for Jane's attention. Jane and John have different impressions of the star because of their gender-specific instincts that have evolved during thousands of millenia of human evolution.
7 0
3 years ago
Read 2 more answers
An object is propelled along a straight-line path by a force. If the new force were doubled, is acceleration would
Aneli [31]
Still go straight but would obviously go up in speed!!





Hope this helps plz mark as brainlist and 5 star
6 0
3 years ago
Read 2 more answers
Suppose that the height of the incline is h = 14.7 m. Find the speed at the bottom for each of the following objects. 1.solid sp
tensa zangetsu [6.8K]

Answer:

1. 14.4 m/s  2. 13.2 m/s 3. 12.0 m/s 4. 13.9 m/s

Explanation:  

Assuming no friction present, the different objects roll without slipping, so there is a constant relationship between linear and angular velocity, as follows:

ω= v/r

If no friction exists, the change in total kinetic energy must be equal in magnitude to the change in the gravitational potential energy:

∆K = -∆U

 ½ *m*v² + ½* I* ω²  = m*g*h

Simplifying and replacing the value of the angular velocity:

½ * v² + ½ I *(v/r)² = g*h (1)

In order to answer the question, we just need to replace h by the value given, and I (moment of inertia) for the value for each different object, as follows:

  •  Solid Sphere I = 2/5* m *r²

                Replacing in (1):

                ½ * v² + ½ (2/5 *m*r²) *(v/r)² = g*h

                Replacing by the value given for h, and solving for v:

                v = √(10/7*9.8 m/s2*14.7 m)  = 14. 4 m/s

  • Spherical shell I=2/3*m*r²

                Replacing in (1):

                ½ * v² + ½ (2/3 *m*r²) *(v/r)² = g*h

                Replacing by the value given for h, and solving for v:

                v = √(6/5*9.8 m/s2*14.7 m)  = 13.2 m/s

  • Hoop   I= m*r²

                Replacing in (1):

                ½ * v² + ½ (m*r²) *(v/r)² = g*h

               Replacing by the value given for h, and solving for v:

               v = √(9.8 m/s2*14.7 m)  = 12.0 m/s

  • Cylinder I = 1/2 * m* r²

                 Replacing in (1):

                ½ * v² + ½ (1/2 *m*r²) *(v/r)²= g*h

                 Replacing by the value given for h, and solving for v:

                v = 2*√(1/3*9.8 m/s2*14.7 m)  = 13.9 m/s

5 0
3 years ago
A pneumatic system consists of several identical 1.5 inch diameter cylinders to lift 450-pound pallets in a warehouse. the gauge
DerKrebs [107]
The applicable equation:

P = F/A

P = pressure
F = Force or weight
A = surface area

Pressure on each cylinder = (W/n)/A
Where n = number of cylinders. Additionally, pressure in the reservoir is equivalent to the pressure in each cylinder.
Net pressure =  75 - 14.7 = 60.3 psi

Therefore,
60.3 = (W/n)/A = (450/n)/(πD^2/4) = (450/n)/(π*1.5^2/4) = (450/n)/(1.7671)
60.3*1.7671 = 450/n
106.03 = 450/n
n = 450/106.3 = 4.244 ≈ 5

The number of cylinders is 5.
3 0
3 years ago
A proton traveling along the x-axis enters a region at x = 0 where the x-component of the electric field is given by E = Ao/x1/2
storchak [24]

.Answer:

The value of the work done is \bf{ 5.29 qA_{0}}.

Explanation:

When a charged particle having charge q is moving through an electric field E, the net force (F) on the charge is

F = qE~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(1)

and the work done (W) by the particle is

W = \int\limits^x_0 {F} \, dx ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(2)

Given, E = \dfrac{A_{0}}{x^{1/2}}.

Substitute the value of electric field in equation (1) and then substitute the result in equation (2).

W &=& \int\limits^7_0 {q\dfrac{A_{0}}{x^{1/2}}} \, dx \\&=& qA_{0} \int\limits^7_0 {x^{-1/2}} \, dx \\&=& 2qA_{0}[x^{1/2}]_{0}^{7}\\&=& 5.29 qA_{0}

7 0
3 years ago
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