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ArbitrLikvidat [17]
3 years ago
14

0.25-kg block oscillates on the end of a spring with a spring constant of 200 N/m. If the oscillations is started by elongating

the spring 0.15 m and giving the block a speed of 3.0 m/s, then the maximum speed of the block is A :

Physics
1 answer:
ValentinkaMS [17]3 years ago
5 0

Answer:

5.2m/s

Explanation:

Plss see attached file

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A long copper rod of diameter 2.0 cm is initially at a uniform temperature of 100°C. It is now exposed to an air stream at 20°C
Lisa [10]

Answer:

t = 4.0 min

Explanation:

given data:

diameter of rod = 2 cm

T_1 = 100 degree celcius

Air stream temperature =  20 degree celcius

heat transfer coefficient = 200 W/m2. K

WE KNOW THAT

copper thermal conductivity = k = 401 W/m °C

copper specific heat Cp = 385 J/kg.°C

density of copper = 8933 kg/m3

charateristic length is given as Lc

Lc = \frac{V}{A_s}

Lc = \frac{\frac{\pi D^2}{4} L}{\pi DL}

Lc = \frac{D}{4}

Lc = \frac{0.02}{6} = 0.005 m

Biot number is given as Bi = \frac{hLc}{k}

Bi = \frac{200*0.005}{401}

Bi = 0.0025

As Bi is greater than 0.1 therefore lumped system analysis is applicable

so we have

\frac{T(t) - T_∞}{Ti - T_∞} = e^{-bt} ............1

where b is given as

b = \frac{ hA}{\rho Cp V}

b = \frac{ h}{\rho Cp Lc}

b = \frac{200}{8933*385*0.005}

b = 0.01163 s^{-1}

putting value in equation 1

\frac{25-20}{100-20} = e^{-0.01163t}

solving for t we get

t = 4.0 min

6 0
3 years ago
A ball is dropped from a height of 20 meters. At what helght does the ball have a velocity of 10 meters/second?
Sunny_sXe [5.5K]

Answer:

5.10 meters.

Explanation:

v²=u²+2gh

or, (10)²=(0)²+2×9.8×h

or, 19.6h=100

or, h=5.10 meters

Hope, this helps you.

8 0
3 years ago
Suppose that the weight (in pounds) of an airplane is a linear function of the amount of fuel (in gallons) in its tank. When car
Inessa [10]

Answer:

W=2219pounds

Explanation:

If the weight  is a linear function of the amount of fuel, the following correlation is fulfilled :

\frac{2153pounds-1999pounds}{46gallons-18gallons} = \frac{W-1999pounds}{58gallons-18galons}

we solve the equation:

W=2219pounds

6 0
3 years ago
Calculate the amount of heat liberated (in kj) from 411 g of mercury when it cools from 88.0°c to 12.0°c.
Katen [24]
You should note that the melting point of mercury is -38.83°C, while the boiling point is at 356.7°C. Then, that means that there is no latent heat involved here. We only compute for the sensible heat.

ΔH = mCpΔT
The Cp of mercury is 0.14 J/g·°C
Thus,
ΔH = (411 g)(0.14 J/g·°C)(88 - 12°C)
<em>ΔH = 4,373.04 J</em>
5 0
3 years ago
Which of the following is correct in order of lowest to highest energy?
Serggg [28]
It is number 3. Because microwaves are the lowest of the sepectum then visible light the the highest of them all is Gamma Ray's. Hope this helps!!
3 0
3 years ago
Read 2 more answers
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