An astronaut holds a rock 100m above the surface of Planet XX. The rock is then thrown upward with a speed of 15m/s, as shown in
1 answer:
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Answer:
a) 19536 joules of work are done.
b) The work is done by the engine on the structure of the cart.
c) There are three options: (i) Keeping the engine and changing the travelled distance, (ii) Changing the engine and keeping the travelled distance, (iii) Changing the engine and the travelled distance.
d) 24442 joules of work are done.
e) We may change for a bigger engine if it allows a greater acceleration and higher peak speed.
f) The bigger engine uses more gas to go 22 meters.
g) An empty semi truck uses more gas than a car since the first has much more mass than a car and is designed for moves big loads and for being fast.
Explanation:
a) If force applied in the cart is uniform, that is, constant in magnitude and direction and is parallel to distance travelled by the car, the work done on the cart is defined by the following equation:
(1)
Where:
- Force applied by the cart, measured in newtons.
- Distance travelled by the car, measured in meters.
- Work done on the cart, measured in joules.
If we know that and
, then the work done on the cart is:
19536 joules of work are done.
b) The work is done by the engine on the structure of the cart.
c) There are three options: (i) Keeping the engine and changing the travelled distance, (ii) Changing the engine and keeping the travelled distance, (iii) Changing the engine and the travelled distance.
d) If we know that and
, then the work on the cart is:
24442 joules of work are done.
e) We may change for a bigger engine if it allows a greater acceleration and higher peak speed.
f) The gas consumption is directly proportional to the square of velocity and mass of the cart and, hence, to the work done on the cart. In consequence, we conclude that the bigger engine uses more gas to go 22 meters.
g) An empty semi truck uses more gas than a car since the first has much more mass than a car and is designed for moves big loads and for being fast.
Answer:
The centripetal force acting on the skater is <u>48.32 N.</u>
Explanation:
Given:
Radius of circular track is,
Tangential speed of the skater is,
Mass of the skater is,
We are asked to find the centripetal force acting on the skater.
We know that, when an object is under circular motion, the force acting on the object is directly proportional to the mass and square of tangential speed and inversely proportional to the radius of the circular path. This force is called centripetal force.
Centripetal force acting on the skater is given as:
Now, plug in the given values of the known quantities and solve for centripetal force, . This gives,
Therefore, the centripetal force acting on the skater is 48.32 N.