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Vladimir [108]
2 years ago
6

When the mass of the cylinder increased by a factor of 3, from 1.0 kg to 3.0 kg, what happened to the cylinder’s gravitational p

otential energy? It decreased by a factor of 3. It decreased by a factor of 2. It increased by a factor of 2. It increased by a factor of 3.
Physics
1 answer:
Gnesinka [82]2 years ago
6 0

Answer:

It increased by a factor of 3.

Explanation:

The gravitational potential energy of an object is given by

U=mgh

where

m is the mass

g is the gravitational acceleration

h is the heigth of the object relative to some reference point (for instance, the ground)

As we see from the formula, the gravitational potential energy is directly proportional to the mass, m: therefore, if the mass of the cylinder is increased by a factor 3, then the gravitational potential energy will also increase by a factor 3.

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Calculate the force of gravity between two objects of masses 1300 kg and 7800 kg, which are 0.23 m apart.
uranmaximum [27]

Answer:

F = Gm1m2/r^2 where G = 6.67x10^-11, m1 =1300, m2 = 7800, r = 0.23m

F = 6.67x10^-11 *1300*7800/(0.23)^2 = 0.0127852N

Explanation:

6 0
2 years ago
Which of the following best describes a capacitor?
galben [10]

Answer:

B

Explanation:

The capacitor is a component which has the ability to store energy in the form of an electrical charge  making a potential difference on those two metal plates

A capacitor consists of two or more parallel conductive (metal) plates. They are electrically seperated by an insulating material (ex: air, mica,ceramic etc.) which is called as Dielectric Layer

Due to this insulating layer, DC current can not flow through the capacitor.But it allows a voltage to be present across the plates in the form of an electrical charge.

4 0
2 years ago
Read 2 more answers
A section of highway has the following flowdensity relationship q = 50k − 0.156k2 [with q in veh/h and k in veh/mi]. What is the
lions [1.4K]

Answer:

a) capacity of the highway section = 4006.4 veh/h

b) The speed at capacity = 25 mph

c) The density when the highway is at one-quarter of its capacity = k = 21.5 veh/mi or 299 veh/mi

Explanation:

q = 50k - 0.156k²

with q in veh/h and k in veh/mi

a) capacity of the highway section

To obtain the capacity of the highway section, we first find the k thay corresponds to the maximum q.

q = 50k - 0.156k²

At maximum flow density, (dq/dk) = 0

(dq/dt) = 50 - 0.312k = 0

k = (50/0.312) = 160.3 ≈ 160 veh/mi

q = 50k - 0.156k²

q = 50(160.3) - 0.156(160.3)²

q = 4006.4 veh/h

b) The speed at the capacity

U = (q/k) = (4006.4/160.3) = 25 mph

c) the density when the highway is at one-quarter of its capacity?

Capacity = 4006.4

One-quarter of the capacity = 1001.6 veh/h

1001.6 = 50k - 0.156k²

0.156k² - 50k + 1001.6 = 0

Solving the quadratic equation

k = 21.5 veh/mi or 299 veh/mi

Hope this Helps!!!

3 0
2 years ago
A 5.0-kg crate is resting on a horizontal plank. The coefficient of static friction is 0.50 and the coefficient of kinetic frict
Harlamova29_29 [7]

Answer:

The mass of the crate is 5kg.

We know that the force of friction can be obtained by:

F = N*k

where k is the coefficient of friction, where we use the static one if the object is at rest, and the kinetic one if the object os moving. N is the normal force

If we tilt the base making an angle of 30° with the horizontal, now the normal force against the plank will be equal to the fraction of the weight in the direction normal to the surface of the plank.

Knowing that the angle is 30°, then the fraction of the weight that pushes against the normal is Cos(30°)*W = cos(30°)*5kg*9.8m/s^2 = 42.4N

The fraction of the force in the parallel direction to the plank (the force that would accelerate the crate downwards) is:

F = sin(30°)*5k*9,8m/s = 24.5N

now, the statical friction force is:

Fs = 42.4N*0.5 = 21.2N

The statical force is less than the 24.5N, so the crate will move downwards, then the force that acts on the crate is the kinetic force of friction:

Fk = 42.4N*0.4 = 16.96N

Then, the total force that acts on the crate is:

total force = F - Fk = 24.5N - 16.69N = 7.54N and the direction of this force points downside along the parallel direction of the plank.

3 0
2 years ago
For which of the following problems would a scientist most likely use carbon-14?
spayn [35]

Answer:

To calculate the age of a piece of bone

Explanation:

Carbon 14 is an isotope of carbon that is unstable and decays into Nitrogen 14 by emitting an electron. The decay rate of radioactive material is  normally expressed in terms of its "half-life" (the time required by half the radioactive nuclei of a sample to undergo radioactive decay). The nice thing about carbon 14 is that its "half-life" is about 5730 years, which gives a nice reference to measure the age of fossils that are some thousand years old.

Carbon 14 dating is used to determine the age of objects that have been living organisms long ago. They measure how much carbon 14 is left in the object after years of decaying without having exchange with the ambient via respiration, ingestion, absorption, etc. and therefore having renewed the normal amount of carbon 14 that is in the ambient.

A rock is not a living organism, so its age cannot be determined by carbon 14 dating.

3 0
2 years ago
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