Different masses are hung on a spring scale calibrated in newtons. The force exerted by gravity on 1.0 kg is shown in the image
2 answers:
Answer:
F = 64 N
Explanation:
When 1 kg of weight attached with the spring, the scale reads 9.8 N. Weight of an object is the product of mass and acceleration due to gravity. Let F is the force exerted by gravity on 6.5 kg. It can be calculated as :
F = 63.7 N
or
F = 64 N
So, the force exerted by gravity on 6.5 kg is 64 N.
You might be interested in
Answer:
The coefficient of friction present between the roadway and the wheels of the truck is <u>0.833</u>.
Explanation:
Given:
Radius of the curve (R) = 150 m
Maximum speed of truck (v) = 35.5 m/s
Let the coefficient of friction between the roadway and the wheels of the truck be "μ".
As the truck is moving around a circular curve. So, the force acting on it is centripetal force which acts in the radial inward direction towards the center of the circular curve.
The centripetal force acting on the truck is given as:
Now, the friction between the roadway and the wheels of the truck is responsible for providing the necessary centripetal force. So, frictional force is equal to the centripetal force necessary for circular motion.
Frictional force is given as:
Where, 'N' is the normal force. Since there is no vertical motion, the normal force is equal to weight of truck. So,
Therefore, frictional force,
Now, frictional force = centripetal force
Plug in the given values and solve for 'μ'. This gives,
Therefore, the coefficient of friction present between the roadway and the wheels of the truck is 0.833