Answer:

Explanation:
The torque applied by a force can be calculated as

where
F is the magnitude of the force
d is the length of the arm
is the angle between the direction of the force and the arm
In this problem, we have
F = 15 N
d = 2.0 m

Substituting into the equation, we find

Answer:
There is absolutely No relationship between the weight of an object (which is constant) and the frictional force. If a block is sliding on a surface, that surface will be exerting a force on the block. That force can be resolved into a component parallel to the surface (which we call the frictional component), and a component perpendicular to the surface (called the normal component). For many situations, we find experimentally that the frictional component is approximately proportional to the normal component. The frictional component divided by the normal component is defined to be a quantity called the coefficient of kinetic or sliding friction. The coefficient of kinetic friction obviously depends on the nature of the surfaces involved. The normal component on an object can be decreased if you pull in the direction of the normal component (the weight does not change). However pulling this way on the object not only decreases the normal component, but it also decreases the frictional component since they are proportional. This is why it is easier to slide something if you pull up on it while you push it. If you push down, the normal and frictional components increase so it is harder to slide the object. The weight of an object is the downward force exerted by Earth’s gravity on that object, and it does not change no matter how you push or pull on the object.
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Answer:
centripetal force is calculated by mass(kg) × tangetial velocity(m/s) ÷ radius (m)
Explanation:
so 30000g= 30kg
50km/h = 13.88m/s
600cm= 6m
30×13.88÷6= 69.4N
N= Newton's
hope this helps.
btw I'm 16 and love physics so I tried my best in this hope it went well!!
Answer:
correct answer is 1 and 3
Explanation:
In direct measurement with an instrument, the precision or absolute error of the instrument is given by its appreciation, in this case we see that the measurements have two decimal places, so the appreciation of the instrument must be 0.01 cm
Based on this appreciation, the valid measurements are 5.52 and 5.5.
the other two measurements have errors much higher than the assessment of the instrument, for which there must have been some errors in the measurement.
The correct answer is 1 and 3