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Drupady [299]
3 years ago
7

A car of mass m, traveling at constant speed, rides over the top of a round hill. How do the normal force of the road on the car

and the gravitational force of the Earth on the car compare at this point?A. N > mg B. N < mg C. N = mg
Physics
1 answer:
dybincka [34]3 years ago
4 0

Answer:

The normal force will be lower than the gravitational force acting on the car. Therefore the answer is N < mg, which is <em>option B</em>.

Explanation:

Over a round hill, the centripetal force acting toward the the radius of the hill supports the gravitational force (mg) of the car. This notion can be expressed mathematically as follows:

At the top of a round hill

Normal force = Gravitational force - centripetal force

At the foot of a round hill

Normal Force = centripetal force + Gravitational force

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STatiana [176]

Answer:

Answer: 2nd Law - The soccer ball weighs less than the bowling ball, so you need less force to get it to have the same acceleration as the bowling ball.

Explanation:

7 0
3 years ago
you walk from the park to your friend's house, then back to your house. (a) What is the distance traveled? (b) What is your disp
777dan777 [17]

Explanation:

The total distance in a path is called distance.

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a) Here, the distance travelled between the park to your friend's house and back is

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b) Displacement would be the shortest distance between the park and your house.

a) Distance walked between your house to library and back is

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3 0
3 years ago
When a basketball player dribbles a ball, it falls to the floor and bounces up. Is a force required to make it bounce? Why? If a
Marizza181 [45]
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5 0
3 years ago
What is the first step in the formation of a protostar?
Fittoniya [83]

Star formation begins in relatively small molecular clouds called dense cores.[7] Each dense core is initially in balance between self-gravity, which tends to compress the object, and both gas pressure and magnetic pressure, which tend to inflate it. As the dense core accrues mass from its larger, surrounding cloud, self-gravity begins to overwhelm pressure, and collapse begins. Theoretical modeling of an idealized spherical cloud initially supported only by gas pressure indicates that the collapse process spreads from the inside toward the outside.[8] Spectroscopic observations of dense cores that do not yet contain stars indicate that contraction indeed occurs. So far, however, the predicted outward spread of the collapse region has not been observed.[9]

The gas that collapses toward the center of the dense core first builds up a low-mass protostar, and then a protoplanetary disk orbiting the object. As the collapse continues, an increasing amount of gas impacts the disk rather than the star, a consequence of angular momentum conservation. Exactly how material in the disk spirals inward onto the protostar is not yet understood, despite a great deal of theoretical effort. This problem is illustrative of the larger issue of accretion disk theory, which plays a role in much of astrophysics.

Regardless of the details, the outer surface of a protostar consists at least partially of shocked gas that has fallen from the inner edge of the disk. The surface is thus very different from the relatively quiescent photosphere of a pre-main sequence or main-sequence star. Within its deep interior, the protostar has lower temperature than an ordinary star. At its center, hydrogen is not yet undergoing nuclear fusion. Theory predicts, however, that the hydrogen isotope deuterium is undergoing fusion, creating helium-3. The heat from this fusion reaction tends to inflate the protostar, and thereby helps determine the size of the youngest observed pre-main-sequence stars.[11]

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The actual radiation emanating from a protostar is predicted to be in the infrared and millimeter regimes. Point-like sources of such long-wavelength radiation are commonly seen in regions that are obscured by molecular clouds. It is commonly believed that those conventionally labeled as Class 0 or Class I sources are protostars.[12][13] However, there is still no definitive evidence for this identification.

4 0
3 years ago
A positive test charge of 5.00 E-5 C is places in an electric field. The force on it is 0.751 N. The magnitude of the electric f
zalisa [80]

Explanation:

Charge=5.00 E-5

Force=0.751N

F=qE

0.751=5.00 E-5*E

E=1.502*10⁴

option a

8 0
3 years ago
Read 2 more answers
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