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Drupady [299]
3 years ago
12

Four solid plastic cylinders all have radius 2.55 cm and length 5.70 cm. Find the charge of each cylinder given the following ad

ditional information about each one. Cylinder (a) carries charge with uniform density 15.4 nC/m2 everywhere on its surface.
Physics
1 answer:
vazorg [7]3 years ago
5 0

Answer:3

Explanation:idk

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A load of mass 5kg is raised through a height of 2m. (g=10mls)​
s2008m [1.1K]

The gain in gravitational potential energy of the mass = 100 kg m^2 / s^2.

<u>Explanation:</u>

Gravitational potential energy is an energy in which an object possesses because of its position in a gravitational field. The most common use of gravitational potential energy is for an object near the Earth's surface where the gravitational acceleration can be assumed to be constant at about 9.8 m/s2.

The formula for the gravitational potential energy is the product of mass, gravity, and height.

             GPE = m *g *h

where m represents mass in kg,

           g represents the gravity,

           h represents the height.

          GPE    = 5 * 10 * 2 = 100 kg m^2 / s^2.

7 0
3 years ago
What is numbers 2 and 4? Please help, and many thanks to those who do chose to help.
Elan Coil [88]

Answer:

2. 6 km/h

4. 150 miles/h

5 0
3 years ago
Which of the following best describes the shape of Earth's orbit?
nikklg [1K]

Elliptical, as shown by most projections


4 0
3 years ago
How far would a 6 kg ball move in 10 seconds if it is being pushed with a force of 100 N
Jet001 [13]
The ball would move about, 674 kilometers. I think I’m correct. Somebody correct me if I’m not.
3 0
3 years ago
Read 2 more answers
When a small object is launched from the surface of a fictitious planet with a speed of 52.9 m/s, its final speed when it is ver
NISA [10]

Answer:

The escape speed of the planet is 41.29 m/s.

Explanation:

Given that,

Speed = 52.9 m/s

Final speed = 32.3 m/s

We need to calculate the launched with excess kinetic energy

Using formula of kinetic energy

K.E=\dfrac{1}{2}mv^2

K.E=\dfrac{1}{2}\times m\times(32.3)^2

We need to calculate the escape speed of the planet

Using formula of kinetic energy

\text{escape kinetic energy}=\text{launch kinetic energy}-\text{excess kinetic energy}

\dfrac{1}{2}mv^2=\dfrac{1}{2}mv^2-\dfrac{1}{2}mv^2

\dfrac{1}{2}\times v^2=\dfrac{1}{2}\times(52.9)^2-\dfrac{1}{2}\times(32.3)^2

v=\sqrt{2\times(\dfrac{1}{2}\times(52.9)^2-\dfrac{1}{2}\times(32.3)^2)}

v=41.29\ m/s

Hence, The escape speed of the planet is 41.29 m/s.

4 0
3 years ago
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