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vodomira [7]
3 years ago
6

A student swings a container of water in a vertical circle of radius 1.0 m. Calculate the minimum speed of the container so that

the water remains in the container at the top of the circle. ANS. 3.1 m/s E-4, P-12?​
Physics
1 answer:
12345 [234]3 years ago
6 0

Answer:

Explanation:

The centripetal acceleration requirement must equal gravity at the top of the circle

mg = mv²/R

  v = √Rg

  v = √(1.0(9.8))

  v = 3.1304951...

  v = 3.1 m/s

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6 0
2 years ago
The escape speed from an object is v2 = 2GM/R, where M is the mass of the object, R is the object's starting radius, and G is th
Rom4ik [11]

Answer:

Approximate escape speed = 45.3 km/s

Explanation:

Escape speed

        v=\sqrt{\frac{2GM}{R}}

Here we have

   Gravitational constant = G = 6.67 × 10⁻¹¹ m³ kg⁻¹ s⁻²

   R = 1 AU = 1.496 × 10¹¹ m

   M = 2.3 × 10³⁰ kg

Substituting

    v=\sqrt{\frac{2\times 6.67\times 10^{-11}\times 2.3\times 10^{30}}{1.496\times 10^{11}}}=4.53\times 10^4m/s=45.3km/s

Approximate escape speed = 45.3 km/s

6 0
3 years ago
Pick an activity you enjoy, such as running or riding a scooter, and describe how newton's laws apply to that activity.
deff fn [24]
I like playing basketball. So I'm the object in motion. Until an unbalanced force comes and hits me I fall and stay at rest. 
5 0
4 years ago
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1. Ramon puts two magnetic toy trains very close to each other on a track. What will happen next, and
FrozenT [24]

Answer:

If one side of the train is positive and the other is negative they will attract if they are the same then they will repel.

Explanation:

If both are positive they will repel if both are negative they will repel and if they are opposites they will attract.

5 0
3 years ago
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A ball is launched with initial speed v from ground level up a frictionless slope (This means the ball slides up the slope witho
amid [387]

Answer:

hmax = 1/2 · v²/g

Explanation:

Hi there!

Due to the conservation of energy and since there is no dissipative force (like friction) all the kinetic energy (KE) of the ball has to be converted into gravitational potential energy (PE) when the ball comes to stop.

KE = PE

Where KE is the initial kinetic energy and PE is the final potential energy.

The kinetic energy of the ball is calculated as follows:

KE = 1/2 · m · v²

Where:

m = mass of the ball

v = velocity.

The potential energy is calculated as follows:

PE = m · g · h

Where:

m = mass of the ball.

g = acceleration due to gravity (known value: 9.81 m/s²).

h = height.

At  the maximum height, the potential energy is equal to the initial kinetic energy because the energy is conserved, i.e, all the kinetic energy was converted into potential energy (there was no energy dissipation as heat because there was no friction). Then:

PE = KE

m · g · hmax = 1/2 · m · v²

Solving  for hmax:

hmax = 1/2 · v² / g

4 0
4 years ago
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