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tensa zangetsu [6.8K]
3 years ago
14

Emily holds a banana of mass m over the edge of a bridge of height h. She drops the banana and it falls to the river below. Use

conservation of energy to show that the speed of the banana just before hitting the water is v
Physics
1 answer:
bearhunter [10]3 years ago
5 0

Answer:

The mass of the banana is m and it is at height h.

Applying the Law of Conservation of Energy

              Total Energy before fall = Total Energy after fall

                                E_{i}  = E_{f}

Here, total energy is the sum of kinetic energy and potential energy

K.E_{i} + P.E_{i} = K.E_{f} + P.E_{f}       (a)

When banana is at height h, it has

                 K.E_{i} = 0    and    P.E_{i} = mgh          

and when it reaches the river, it has

       K.E_{f}  = 1/2mv^{2}    and   P.E_{f}  = 0

Putting the values in equation (a)

                              0 + mgh = 1/2mv^{2} + 0

                                      mgh = 1/2mv^{2}

<em>cutting 'm' from both sides</em>

<em>                                           </em>gh = 1/2v^{2}

                                          v = \sqrt{2gh}

Hence, the velocity of banana before hitting the water is

                                          v = \sqrt{2gh}

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What is the strength of the electric field inside the membrane just before the action potential?
gtnhenbr [62]

Answer:

Incomplete question, check attachment for the graph needed to solve problem.

A 8.1nm........

Explanation:

Electric Field is given as

E=V/d

Where V is voltage

And d is the distance apart

E is the electric field

The voltage V just before action of potential is -70mV,

The value d=8.1nm

d=8.1×10^-9m

E=V/d

E=-70×10^-3/8.1×10^-9

E=-8.6×10^6 N/C

Then the magnitude of the electric field is 8.6×10^6N/C

5 0
3 years ago
The moment of inertia of a thin uniform rod of mass M and length L about an Axis perpendicular to the rod through its Centre is
sleet_krkn [62]

Answer:

I = I₀ + M(L/2)²

Explanation:

Given that the moment of inertia of a thin uniform rod of mass M and length L about an Axis perpendicular to the rod through its Centre is I₀.

The parallel axis theorem for moment of inertia states that the moment of inertia of a body about an axis passing through the centre of mass is equal to the sum of the moment of inertia of the body about an axis passing through the centre of mass and the product of mass and the square of the distance between the two axes.

The moment of inertia of the body about an axis passing through the centre of mass is given to be I₀

The distance between the two axes is L/2 (total length of the rod divided by 2

From the parallel axis theorem we have

I = I₀ + M(L/2)²

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If the accepted value of a wave is 121 m/s, who has the most accurate method of measuring the speed of a wave?
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How do the tension of the cord and the force of gravity affect a pendulum?
LuckyWell [14K]

Answer:

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6 0
3 years ago
Read 2 more answers
Consider two wheels with fixed hubs. The hub cannot move, but the wheel can rotate about it. The hubs are fixed to a stationary
kykrilka [37]

Answer:

Magnitude of force on wheel B is 4 N

Explanation:

Given that

I=mr^2

For wheel A

m= 1 kg

d= 1 m,r= 0.5 m

F=1 N

We know that

T= F x r

T=1 x 0.5 N.m

T= 0.5 N.m

T= I α

Where I is the moment of inertia and α is the angular acceleration

I_A=1 \times 0.5^2\ kg.m^2

I_A=0.25\ kg.m^2

T= I α

0.5= 0.25 α

\alpha = 2\ rad/s^2

For Wheel B

m= 1 kg

d= 2 m,r=1 m

I_B=1 \times 1^2\ kg.m^2

I_B=1 \ kg.m^2

Given that angular acceleration is same for both the wheel

\alpha = 2\ rad/s^2

T= I α

T= 1 x 2

T= 2 N.m

Lets force on wheel is F then

T = F x r

2 = F x 1

So F= 2 N

Magnitude of force on wheel B is 2 N

3 0
3 years ago
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