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3 years ago

12

A 12 g plastic ball is dropped from a height of 2.5 m and is moving at 3.2 m/s just before it hits the floor. How much mechanica

l energy was lost during the fall?

1 answer:

Nataly [62]3 years ago

4 0

Answer:

0.24

Explanation:

Mass of ball= 12g=0.012Kg

height of ball= 2.5m

velocity of ball before falling= 3.2m/s

potential energy of the ball=mgh= 0.012*10*2.5=0.3J

kinetic energy of the ball=0.5*m $v^{2}$ =0.5*0.012*3.2*3.2=0.6J

Loss in mechanical energy during the fall= potential energy- Kinetic energy= 0.3-0.06=0.24J

note: During the fall, the potential energy of the ball is converted to kinetic energy. the loss in energy is due to air resistance.

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Cerebrospinal fluid (CSF) is a clear fluid that surrounds the brain and spinal cord.

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A small block with mass 0.0350kg slides in a vertical circle of radius 0.525m on the inside of a circular track. During one of t

kirill115 [55]

Answer:

w = -0.475N

Explanation:

$K.E_{a} + P.E_{a} + W_{fr} = K.E_{b} + P.E_{b}\\K.E = 0.5mv^{2} \\Normal force at point B, N_{B} = 0.665N\\Normal force at point A, N_{A} = 3.85N\\$

To get Va and Vb

$F = mv_{A} ^{2} /R................(1)\\F = N_{A} - mg.........................(2)\\mv_{A} ^{2} /R = N_{A} - mg\\v_{A} ^{2} = R (N_{A}/m - g)\\v_{A} = \sqrt{ R (N_{A}/m - g)}$

R = 0.525 m

m = 0.0350 kg

g = 9.8 m/s²

$v_{A}= \sqrt{0.525(3.85 /0.0350 - 9.8)} \\v_{A} = 7.25 m/s$

K.Ea = 0.5 * 0.035 * 7.25²

K.Ea = 0.92 J

Since point A is at the bottom of the path, h = 0 m

P.Ea = 0 m

For Vb

$F = mv_{B} ^{2} /R................(1)\\F = N_{B} - mg.........................(2)\\mv_{B} ^{2} /R = N_{B} - mg\\v_{B} ^{2} = R (N_{B}/m - g)\\v_{B} = \sqrt{ R (N_{B}/m - g)}$

$N_{B} = 0.665N$

$v_{B}= \sqrt{0.525(0.665 /0.0350 - 9.8)} \\v_{B} = 2.198 m/s$

$K.E_{B} = 0.5* 0.035 * 2.198^{2} \\K.E_{B} = 0.085 J$

$P.E_{B} = mgh_{B} \\h_{B} = A diameter = 2R = 2 * 0.525\\h_{B} = 1.05 m\\P.E_{B} = 0.035 * 9.8 * 1.05\\P.E_{B} =0.36 J$

$from K.E_{a} + P.E_{a} + W_{fr} = K.E_{b} + P.E_{b}$

$0.92 + w_{fr} + 0 = 0.085 + 0.36\\ w_{fr} = -0.475J$

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4 years ago

A current of 17.0 mA is maintained in a single circular loop of 2.00 \mathrm{~m} circumference. A magnetic field of 0.800T is di

poizon [28]

The magnetic moment is -

M = 5.5 x $10^{-3}$ A $m^{2}$ .

We have current carrying single circular loop placed in a magnetic field

parallel to the plane of the loop.

We have to determine the Magnetic moment of the loop.

<h3>What is the formula to calculate the magnetic moment of loop?</h3>

The formula to calculate the magnetic moment of the loop is -

M = NIA

where -

N - Number of turns

I - Current in the loop

A - Area of the loop

According to the question, we have -

I = 17mA = 0.017A

Circumference (C) = 2 meters

B = 0.8 T

Now -

C = 2

2 $\pi$ r = 2

$\pi$ r = 1

r = $\frac{1}{\pi }$

r = 0.32

Using the formula -

M = NIA

M = $\pi$ NI $r^{2}$

M = 3.14 x 1 x 0.017 x 0.32 x 0.32

M = 3.14 x 4 x 0.017

M = 5.5 x $10^{-3}$ A $m^{2}$

Hence, the magnetic moment is -

M = 5.5 x $10^{-3}$ A $m^{2}$

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1 year ago

A carpenter builds an exterior house wall with a layer of wood 3.0 cm thick on the outside and a layer of Styrofoam insulation 2

Leno4ka [110]

Answer:

A. T=15.54 °C

B. Q/A= 0.119 W/m2

Explanation:

To solve this problem we need to use the Fourier's law for thermal conduction:

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Here, the rate of flow per square meter must be the same through the complete wall. Therefore, we can use it to find the temperature at the plane where the wood meets the Styrofoam as follows:

$\frac{Q}{A} =\frac{T_1-T_0}{d_w}k_w=\frac{T_2-T_1}{d_s}k_s\\T_1(\frac{k_w}{d_w}+\frac{k_s}{d_s})=T_2\frac{k_s}{d_s}+T_0\frac{k_w}{d_w}\\T_1=\frac{T_2\frac{k_s}{d_s}+T_0\frac{k_w}{d_w}}{\frac{k_w}{d_w}+\frac{k_s}{d_s}}\\T_1= 15.54 \°C$

Then, to find the rate of heat flow per square meter, we have:

$\frac{Q}{A}=\frac{T_1-T_0}{d_w}k_w=0.119 \frac{W}{m^2}\\\frac{Q}{A}=\frac{T_2-T_1}{d_s}k_s= 0.119 \frac{W}{m^2}$

$T_0: Temperature \ in \ the \ house\\T_1: Temperature \ at \ the \ plane \ between \ wood \ and \ styrofoam\\T_2: Temperature \ outside\\k_w: k \ for \ wood\\d_w: wood \ thickness\\k_s: k \ for \ styrofoam\\d_s: styrofoam \ thickness$

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3 years ago