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slavikrds [6]
3 years ago
14

According to Newton, there were two things needed for an object to fall around or orbit the earth. Label the diagram below with

these two things
Physics
1 answer:
iVinArrow [24]3 years ago
4 0

Centripetal force and centrifugal force


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a bowling ball has a mass of 6 kilograms. a tennis ball has a mass of 0.06 kilogram. how much inertia does the bowling ball have
Andreyy89

Answer:

100 times

Explanation:

Since inertia is directly proportional to the mass of an object, the higher the mass the higher the inertia. In this case, 6 Kg is 100 times heavier than 0.06 Kg to imply The bowling ball  has 100 times more inertia than the tennis  ball because it has 100 times more mass

5 0
3 years ago
When an 81 kg adult uses a spiral staircase to climb to the second floor of his house, his gravitational potential energy increa
professor190 [17]

Answer:

option (e) is correct.

Explanation:

mass of adult, M = 81 kg

Gravitational Potential energy of adult, U = 2000 J

Let the height raised is h.

U = mass of adult x g x h

2000 = 81 x 9.8 x h

h = 2.52 m

mass of adult, m = 18 kg

height raised, h = 2.52 m

Potential energy of child, U' = m x g x h

U' = 18 x 2.52 x 9.8 = 444.44 J

Thus, option (e) is correct.

7 0
3 years ago
Did Kepler propose the modern model of the solar system?
rewona [7]
Yes Kepler did propose the modern model of the solar system.
8 0
3 years ago
Deprecion como la ejerzo
Mrac [35]

Solo dígales cómo se siente, dígales que no está contento porque está triste y no sabe qué hacer al respecto, que está deprimido, y que solo quiere que las cosas mejoren.

8 0
3 years ago
A 1.8-kg object is attached to a spring and placed on frictionless, horizontal surface. A force of 40 N stretches a spring 20 cm
Sergio [31]

Answer:

a) k = 200 N/m

b) E = 4 J

c) Δx = 6.3 cm

Explanation:

a)

  • In order to find force constant of the spring, k, we can use the the Hooke's Law, which reads as follows:

       F = - k * \Delta x (1)

  • where F = 40 N and Δx =- 0.2 m (since the force opposes to the displacement from the equilibrium position, we say that it's a restoring force).
  • Solving for k:

       k =- \frac{F}{\Delta x} =-\frac{40 N}{-0.2m} = 200 N/m (2)

b)

  • Assuming no friction present, total mechanical energy mus keep constant.
  • When the spring is stretched, all the energy is elastic potential, and can be expressed as follows:

        U = \frac{1}{2}* k* (\Delta x)^{2} (3)

  • Replacing k and Δx by their values, we get:

       U = \frac{1}{2}* k* (\Delta x)^{2} = \frac{1}{2}* 200 N/m* (0.2m)^{2} = 4 J (4)

c)

  • When the object is oscillating, at any time, its energy will be part elastic potential, and part kinetic energy.
  • We know that due to the conservation of energy, this sum will be equal to the total energy that we found in b).
  • So, we can write the following expression:

        \frac{1}{2}* k* \Delta x_{1} ^{2} + \frac{1}{2} * m* v^{2}  = \frac{1}{2}*k*\Delta x^{2}   (5)

  • Replacing the right side of (5) with (4), k, m, and v by the givens, and simplifying, we can solve for Δx₁, as follows:

        \frac{1}{2}* 200N/m* \Delta x_{1} ^{2} + \frac{1}{2} * 1.8kg* (-2.0m/s)^{2}  = 4J   (6)

⇒      \frac{1}{2}* 200N/m* \Delta x_{1} ^{2}   = 4J  - 3.6 J = 0.4 J (7)

⇒     \Delta x_{1}   = \sqrt{\frac{0.8J}{200N/m} } = 6.3 cm (8)

6 0
3 years ago
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