If the earth were twice the distance from the sun that it is now, the gravitational force exerted on it by the sun would be: a)
1 answer:
Answer:
a) 1/4 what it is now
Explanation:
As we know that force of gravitation between two planets at some distance "r" from each other is given as
now since we know that if the distance between Earth and Sun is changed
So the force of gravity will be given as
now we know that the distance between sun and earth is changed to twice the initial distance between them
so we have
so new gravitational force between sun and earth is given as
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Answer:
1) Please find attached, created with Microsoft Visio
2) The acceleration of the masses connected by the light string is 0.00735 m/s²
3) The tension in the cord is 0.147 N
4) The time it would take the block to go 1.2 m to the edge of the table is approximately 18.07 s
5) The velocity of the cart as soon as it gets to the edge of the table is 0.042 m/s
Explanation:
1) Please find attached, the required free body diagram, showing the tension, weight and frictional (zero friction) forces acting on the cart and the mass created with Microsoft Visio
2) The acceleration of the masses connected by the light string is given as follows;
F = Mass, m × Acceleration, a
The mass of the truck, M = 20.0 kg
The mass attached to the string, hanging rom the pulley, m = 0.0150 kg
The force, F acting on the system = The pulling force on the cart = The tension on the cable = The weight of the hanging mass = 0.0150 × 9.8 = 0.147 N
The pulling force acting on the cart, F = M × a
∴ F = 0.147 N = 20.0 kg × a
a = 0.147 N/(20.0 kg) = 0.00735 m/s²
The acceleration of the truck = a = 0.00735 m/s²
3) The tension in the cord = F = 0.147 N
4) The time, t, it would take the block to go 1.2 m to the edge of the table is given by the kinematic equation, s = u·t + 1/2·a·t²
Where;
s = The distance to the edge of the table = 1.2 m
u = The initial velocity = 0 m/s (The cart is assumed to be initially at rest)
a = The acceleration of the cart = 0.00735 m/s²
t = The time taken
Substituting the known values, gives;
s = u·t + 1/2·a·t²
1.2 = 0 × t + 1/2 ×0.00735 × t²
1.2 = 1/2 ×0.00735 × t²
t² = 1.2/(1/2 ×0.00735) ≈ 326.5306
t = √(1.2/(1/2 ×0.00735)) ≈ 18.07
The time it would take the block to go 1.2 m to the edge of the table = t ≈ 18.07 s
5) The velocity, v, of the cart as soon as it gets to the edge of the table is given by the kinematic equation, v² = u² + 2·a·s as follows;
v² = u² + 2·a·s
u = 0 m/s
v² = 0² + 2 × 0.00735 × 1.2 = 0.001764
v = √(0.001764) = 0.042
The velocity of the cart as soon as it gets to the edge of the table = v = 0.042 m/s.
Answer:
The work done on the gas = 2PV
Explanation:
In this question, we need to apply the basic gas laws to determine the word done.
For this question, we need to draw a PV diagram (a pressure-volume diagram), which I have made and attached in the attachment. So please refer to that attachment. I will be using this diagram to solve for the work done on the gas.
So, Please refer to the attachment number 1. where x -axis is of volume and y-axis is of pressure.
As we know that, the work done on the gas is equal to the area under the curve.
W = Area of the triangle
W = 0.5 x ( base) x ( height)
W = 0.5 x (BC) x (AC)
W = 0.5 x (3V-V) x (3P-P)
W = 2PV
Hence, the work done on the gas = 2PV
