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natta225 [31]
3 years ago
12

An elevator accelerates uniformly from rest to a speed of 2.5 m/s in 12s. What is the distance the elevator travels during this

time? must show work.
Physics
1 answer:
Aleks04 [339]3 years ago
7 0

We use the kinematic equation,

s= u t +\frac{1}{2}a t^2                      

Here, s is the distance traveled by elevator u is the initial velocity and t is time.

As elevator accelerates uniformly from rest to a speed of 2.5 m/s in 12 s, so

a = \frac{v_{f}- v_{i}  }{t} = \frac{2.5 m/s-0}{12 s} = 0.208 \ m/s^2

Thus, the distance the elevator travels during 12 s time is

s= 0+\frac{1}{2} 0.208 \ m/s^2(12)^2 = 14.97 m \simeq  15 m


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The concept required to solve this problem is associated with potential energy. Recall that potential energy is defined as the product between mass, gravity, and change in height. Mathematically it can be described as

U = mg\Delta h

Here,

\Delta h = Change in height

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g = Acceleration due to gravity

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\Delta h = h_f - h_i

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h_f = -16.1m

The initial height will be the zero point of our system of reference,

\Delta h = -16.1m-0m

\Delta h = -16.1m

Replacing all this values we have,

U = mg\Delta h

U = (68.1kg)(9.8m/s^2)(-16.1m)

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Since the final position of the heroine is located below the ground, there will net loss of gravitational potential energy of 10744.81J

4 0
3 years ago
(a) What is the fluid speed in a fire hose with a 9.00-cm diameter carrying 80.0 L of water per second? (b) What is the flow rat
son4ous [18]

Answer:

12.5752053801 m/s

80\times 10^{-3}\ m^3/s

No.

Explanation:

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d = Diameter of pipe = 9 cm

A = Area = \dfrac{\pi}{4}d^2

Volume flow rate is given by

Q=Av\\\Rightarrow v=\dfrac{Q}{A}\\\Rightarrow v=\dfrac{80\times 10^{-3}}{\dfrac{\pi}{4} (9\times 10^{-2})^2}\\\Rightarrow v=12.5752053801\ m/s

Velocity of fluid is 12.5752053801 m/s

The volume flow rate in m³/s is 80\times 10^{-3}\ m^3/s

The flow of fluid does not depend on the type of water used. Hence the answers would be same. If Q is constant v will be the same irrespective of the type of water used.

8 0
3 years ago
A simple pendulum has a period of 3.45 second, when the length of the pendulum is shortened by 1.0m, the period is 2.81 second c
den301095 [7]

Answer:

Original length = 2.97 m

Explanation:

Let the original length of the pendulum be 'L' m

Given:

Acceleration due to gravity (g) = 9.8 m/s²

Original time period of the pendulum (T) = 3.45 s

Now, the length is shortened by 1.0 m. So, the new length is 1 m less than the original length.

New length of the pendulum is, L_1=L-1

New time period of the pendulum is, T_1=2.81\ s

We know that, the time period of a simple pendulum of length 'L' is given as:

T=2\pi\sqrt{\frac{L}{g}}-------------- (1)

So, for the new length, the time period is given as:

T_1=2\pi\sqrt{\frac{L_1}{g}}------------ (2)

Squaring both the equations and then dividing them, we get:

\dfrac{T^2}{T_1^2}=\dfrac{(2\pi)^2\frac{L}{g}}{(2\pi)^2\frac{L_1}{g}}\\\\\\\dfrac{T^2}{T_1^2}=\dfrac{L}{L_1}\\\\\\L=\dfrac{T^2}{T_1^2}\times L_1

Now, plug in the given values and calculate 'L'. This gives,

L=\frac{3.45^2}{2.81^2}\times (L-1)\\\\L=1.507L-1.507\\\\L-1.507L=-1.507\\\\-0.507L=-1.507\\\\L=\frac{-1.507}{-0.507}=2.97\ m

Therefore, the original length of the simple pendulum is 2.97 m

4 0
3 years ago
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d1i1m1o1n [39]

The two ladybugs have same rotational (angular) speed

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2 years ago
Water has a specific heat of 4.184
LuckyWell [14K]

Answer:

Water.

Explanation:

This means:

1) For the temperature of water to raise at any point to the next degree by 1°C, will require a specific heat capacity of  4.184  J/Kg°C

2) For the temperature of wood to raise at any point to the next degree by 1°C, will require a specific heat capacity of  1.760  J/Kg°C

Note that: specific heat is directly proportional to energy, therefore the higher the heat capacity, the higher the energy.

4.184  J/Kg°C is higher than 1.760  J/Kg°C, hence WATER needs more energy.

8 0
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