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Eduardwww [97]
4 years ago
7

Forces contribute to the net force on a car rolling down a ramp.

Physics
1 answer:
Jlenok [28]4 years ago
8 0
The contact force or support force which is perpendicular to the objects path
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The magnitude E of an electric field depends on the radial distance r according to E = A/r4, where A is a constant with unit vol
Lesechka [4]

Answer:

\Delta V = 0.053 A

Explanation:

Electric field in a given region is given by equation

E = \frac{A}{r^4}

as we know the relation between electric field and potential difference is given as

\Delta V = -\int E. dr

so here we have

\Delta V = - \int (\frac{A}{r^4}) .dr

\Delta V = \frac{A}{3r_1^3} - \frac{A}{3r_2^3}

here we know that

r_1 = 1.71 m  and r_2 = 2.89 m

so we will have

\Delta V = \frac{A}{3}(\frac{1}{1.71^3} - \frac{1}{2.89^3})

so we will have

\Delta V = 0.053 A

8 0
3 years ago
What is the internal resistor of the cell in closed circuit?
Drupady [299]

Generally, the internal resistance of the new battery is small, about 0.2 euros, while the old battery is large, close to 1 euro ,

5 0
4 years ago
How does kinetic energy affect the stopping distance of a small vehicle compared to a large vehicle?
Zarrin [17]
<span>The kinetic energy is the work done by the object due to its motion. It is represented by the formula of the half the velocity squared multiply by the mass of the object. In this problem, you have two  vehicles, the other one is large and the other one is small. Let us assume that they travel with the same velocity. Note that the kinetic energy is proportional to the mass of the object. So when you increase the mass of the other, it also increases the kinetic energy of that object. The same holds true for the two vehicles. The larger the vehicle, its kinetic energy is also large and therefore its stopping distance will be longer than that of the smaller vehicle.</span>
8 0
3 years ago
A 15.0 kg crate, initially at rest, slides down a ramp 2.0 m long and inclined at an angle of 20.0° with the horizontal. Using t
Elis [28]

The component of the crate's weight that is parallel to the ramp is the only force that acts in the direction of the crate's displacement. This component has a magnitude of

<em>F</em> = <em>mg</em> sin(20.0°) = (15.0 kg) (9.81 m/s^2) sin(20.0°) ≈ 50.3 N

Then the work done by this force on the crate as it slides down the ramp is

<em>W</em> = <em>F d</em> = (50.3 N) (2.0 m) ≈ 101 J

The work-energy theorem says that the total work done on the crate is equal to the change in its kinetic energy. Since it starts at rest, its initial kinetic energy is 0, so

<em>W</em> = <em>K</em> = 1/2 <em>mv</em> ^2

Solve for <em>v</em> :

<em>v</em> = √(2<em>W</em>/<em>m</em>) = √(2 (101 J) / (2.0 m)) ≈ 10.0 m/s

3 0
3 years ago
What is the difference between a license and a certificate in the healthcare system?
Tems11 [23]

Answer:

A

Explanation:

7 0
3 years ago
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