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Alja [10]
3 years ago
12

A hole is drilled to the center of the earth and a ball is dropped into it. When the ball is at the earth's center, compared wit

h their respective values at the earth's surface,
Physics
1 answer:
adell [148]3 years ago
8 0

Answer:

At the center of the earth if the ball is dropped in the hole then it's mass will be the same but the weight of the ball will be zero. As the gravity at the center of earth will be zero. As,

W = mg

However at the surface of earth the ball will have the maximum weight.

Explanation:

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ohaa [14]

Answer:

by a seprating funnel or let it stand it will settle on its own

Explanation:

5 0
3 years ago
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Find the moment of inertia Ihoop of a hoop of radius r and mass m with respect to an axis perpendicular to the hoop and passing
Juliette [100K]

Answer: MR²

is the the moment of inertia  of a hoop of radius R and mass M with respect to an axis perpendicular to the hoop and passing through its center

Explanation:

Since in the hoop , all mass elements  are situated at the same distance from the centre , the following expression for the moment of inertia can be written as follows.

I = ∫ r² dm

= R²∫ dm

MR²

where M is total mass and R is radius of the hoop .

3 0
3 years ago
A small sphere with mass m is attached to a massless rod of length L that is pivoted at the top, forming a simple pendulum. The
USPshnik [31]

Answer:

a) see attached, a = g sin θ

b)

c)   v = √(2gL (1-cos θ))

Explanation:

In the attached we can see the forces on the sphere, which are the attention of the bar that is perpendicular to the movement and the weight of the sphere that is vertical at all times. To solve this problem, a reference system is created with one axis parallel to the bar and the other perpendicular to the rod, the weight of decomposing in this reference system and the linear acceleration is given by

          Wₓ = m a

          W sin θ = m a

          a = g sin θ

b) The diagram is the same, the only thing that changes is the angle that is less

                θ' = 9/2  θ

             

c) At this point the weight and the force of the bar are in the same line of action, so that at linear acceleration it is zero, even when the pendulum has velocity v, so it follows its path.

The easiest way to find linear speed is to use conservation of energy

Highest point

            Em₀ = mg h = mg L (1-cos tea)

Lowest point

          Emf = K = ½ m v²

          Em₀ = Emf

          g L (1-cos θ) = v² / 2

              v = √(2gL (1-cos θ))

4 0
3 years ago
A 40.0 kg wagon is towed up a hill inclined at 18.5 degrees with respect to the horizontal. The tow rope is parallel to the incl
sp2606 [1]

Answer:

7.9m/s

Explanation:

We are given that

Mass of wagon=40 kg

\theta=18.5^{\circ}

Tension=140 N

Initial velocity of wagon=u=0

Displacement=s=80 m

Net force applied  on wagon=F=T-mgsin\theta=140-40(9.8)sin18.5=15.7 N

By using g=9.8m/s^2

a=\frac{F}{a}=\frac{15.7}{40}=0.39m/s^2

We know that

v^2-u^2=2as

Using the formula

v^2=2\times 0.39\times 80

v=\sqrt{2\times 0.39\times 80}=7.9m/s

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