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serious [3.7K]
2 years ago
3

A 2.00-kg copper rod is 50.00 cm long at 23°C. If 40 000 J are transferred to the rod by heat, what is its change in length? cco

pper = 387 J/kg·°C and αcopper = 17 × 10-6/°C.
a. 0.022 cm
b. 0.044 cm
c. 0.059 cm
d. More information is needed
Physics
1 answer:
ella [17]2 years ago
3 0

Answer:

0.044 cm

Explanation:

mass of copper, m = 2 kg

length of rod, l = 50 cm

T1 = 23°C

Heat, Q = 40000 J

specific heat of copper, c = 387 J/kg°C

α = 17 x 10^-6 /°C

Let the change in the length of copper is Δl.

heat = mass x specific heat x ΔT

40000 = 2 x 387 x ΔT

ΔT = 51.68

Now use the formula for the linear expansion

Δl = l x α x ΔT

Δl = 50 x 17 x 10^-6 x 51.68

Δl = 0.044 cm

Thus the increase in length of rod is 0.044 cm.

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a. For a spring-mass oscillator, if you double the mass but keep the stiffness the same, by what numerical factor does the perio
Katena32 [7]

Answer:

a) factor b=\sqrt{2}

b) factor b=\frac{1}{2}

c) factor b=1

d) factor b=1

Explanation:

Time period of oscillating spring-mass system is given as:

T=\frac{1}{f}

T={2\pi} \sqrt{\frac{m}{k} }

where:

f= frequency of oscillation

m= mass of the object attached to the spring

k= stiffness constant of the spring

a) <u>On doubling the mass:</u>

  • New mass, m'=2m

<u>Then the new time period:</u>

T'=2\pi\sqrt{\frac{m'}{k} }

T'=2\pi\sqrt{\frac{2m}{k} }

T'=\sqrt{2}\times  2\pi\sqrt{\frac{m}{k} } }

T'=\sqrt{2} \times T

where the factor b=\sqrt{2} as asked in the question.

b) On quadrupling the stiffness constant while other factors are constant:

New stiffness constant, k'=4k

<u>Then the new time period:</u>

T'=2\pi\sqrt{\frac{m}{k'} }\\\\T'=2\pi\sqrt{\frac{m}{4k} }\\\\T'=\frac{1}{2} \times  2\pi\sqrt{\frac{m}{k} } }\\\\T'=\frac{1}{2} \times T

where the factor  b=\frac{1}{2}  as asked in the question.

c) On quadrupling the stiffness constant as well as mass:

New stiffness constant, k'=4k

New mas, m'=4m

<u>Then the new time period:</u>

T'=2\pi\sqrt{\frac{m'}{k'} }\\\\T'=2\pi\sqrt{\frac{4m}{4k} }\\\\T'=1 \times  2\pi\sqrt{\frac{m}{k} } }\\\\T'=1 \times T

where factor b=1 as asked in the question.

d) On quadrupling the amplitude there will be no effect on the time period because T is independent of amplitude as we can observe in the equation.

so, factor b=1

7 0
2 years ago
An green hoop with mass mh = 2.8 kg and radius rh = 0.13 m hangs from a string that goes over a blue solid disk pulley with mass
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The only force on the system is the mass of the hoop F net = 2.8kg*9.81m/s^2 = 27.468 N The mass equal of the rolling sphere is found by: the sphere rotates around the contact point with the table. 
So by applying the theorem of parallel axes, the moment of inertia of the sphere is computed by:I = 2/5*mR^2 for rotation about the center of mass + mR^2 for the distance of the axis of rotation from the center of mass of the sphere. 
I = 7/5*mR^2 M = 7/5*m 
Therefore, linear acceleration is computed by:F/m = 27.468 / (2.8 + 1/2*2 + 7/5*4) = 27.468/9.4 = 2.922 m/s^2 
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2 years ago
An electric winch is used to raise a 40-kg package and a 10-kg package vertically up the side of a building as pictured in the d
just olya [345]

Answer:

Explanation:

40 divided by 10 then which would equal 4. Add the 1.0 , 2 ,and 15 together. Then multply the 60 by 18.0 after you are done dividing the answer is 3 with a remainder of 6.

3 0
2 years ago
A wire with a length of 150 m and a radius of 0.15 mm carries a current with a uniform current density of 2.8 x 10^7A/m^2. The c
Mrac [35]

Answer:

The current is 2.0 A.

(A) is correct option.

Explanation:

Given that,

Length = 150 m

Radius = 0.15 mm

Current densityJ=2.8\times10^{7}\ A/m^2

We need to calculate the current

Using formula of current density

J = \dfrac{I}{A}

I=J\timesA

Where, J = current density

A = area

I = current

Put the value into the formula

I=2.8\times10^{7}\times\pi\times(0.15\times10^{-3})^2

I=1.97=2.0\ A

Hence, The current is 2.0 A.

7 0
2 years ago
In the ENGR 10 lab (E391), there are 50 long light bulbs (P=100 W) and 30 regular bulbs (P=60 W). How much energy is consumed li
Alenkinab [10]

Answer:

Total energy saving will be 0.8 KWH

Explanation:

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We know that energy E=power\times time=6.8\times 3=20.4KWH

Now power of each CFL bulb = 25 W

So power of 80 bulbs = 80×25 = 2000 W = 2 KW

Energy of 80 bulbs = 2×3 = 6 KWH

So total energy saving = 6.8-6 = 0.8 KWH

6 0
2 years ago
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