Answer:
<em>The force that would be applied on the rope just to start moving the wagon is 122 N</em>
Explanation:
Frictional force opposes motion between two surfaces in contact. It is the force that must be applied before a body starts to move. Static friction opposes the motion of two bodies that are in contact but are not moving. The magnitude of static friction to overcome for the body to move can be calculated using equation 1.
F = μ x mg .............................. 1
where F is the frictional force;
μ is the coefficient of friction ( μs, in this case, static friction);
m is mass of the object and;
g is the acceleration due to gravity( a constant equal to 9.81 m/
)
from the equation we are provide with;
μs = 0.25
m = 50 kg
g = 9.81 m/
F =?
Using equation 1
F = 0.25 x 50 kg x 9.81 m/
F = 122.63 N
<em>Therefore a force of 122 N must be applied to the rope just to start the wagon.</em>
Information about Earthquakes would normally be shown on a map called "geologic hazards"
The answers are -2.4 for the distance. 7.2 for the height. Concave for the type
Answer:
The capacitance is cut in half.
Explanation:
The capacitance of a plate capacitor is directly proportional to the area A of the plates and inversely proportional to the distance between the plates d. So if the distance was doubled we should expect that the capacitance would be cut in half. That can be verified by the following equation that is used to compute the capacitance in such cases:
C = (\epsilon)*(A/d)
Where \epsilon is a constant that represents the characteristics for the insulator between the plates. A is the area of the plates and d is the distance between them. When we double d we have a new capacitance, given by:
C_new = (\epsilon)*(A/2d)
C_new = (1/2)*[(\epsilon)*(A/d)]
Since C = (\epsilon)*(A/d)] we have:
C_new = (1/2)*C
(a) Determine the circumference of the Earth through the equation,
C = 2πr
Substituting the known values,
C = 2π(1.50 x 10¹¹ m)
C = 9.424 x 10¹¹ m
Then, divide the answer by time which is given to a year which is equal to 31536000 s.
orbital speed = (9.424 x 10¹¹ m)/31536000 s
orbital speed = 29883.307 m/s
Hence, the orbital speed of the Earth is ~29883.307 m/s.
(b) The mass of the sun is ~1.9891 x 10³⁰ kg.
