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mezya [45]
3 years ago
5

A cylindrical Nickel rod (9 mm diameter, 50 m long) is pulled in tension with a load of 6,283 N. What would the elongation of th

e rod be under this load?
Physics
1 answer:
Norma-Jean [14]3 years ago
6 0

Answer:

0.29 m

Explanation:

9 mm = 0.009 m in diameter

Cross-sectional area A = \pi d^2/4 = \pi * 0.009^2/4 = 6.36\times 10^{-5} m^2

Let the tensile modulus of Nickel E = 170 \times 10^9Pa.

The elongation of the rod can be calculated using the following formula:

\Delta L = \frac{F L}{A E} = \frac{6283*50}{6.36\times 10^{-5} * 170 \times 10^9} = \frac{314150}{1081200} = 0.29 m

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AlexFokin [52]

If <em>the isotherms</em> are spaced closely together over some portion of the map, there is a drastic temperature change over that portion.

6 0
3 years ago
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We have two solenoids: solenoid 2 has twice the diameter, half the length, and twice as many turns as solenoid 1. The current in
leva [86]

Answer:

the field at the center of solenoid 2 is 12 times the field at the center of solenoid 1.

Explanation:

Recall that the field inside a solenoid of length L, N turns, and a circulating current I, is given by the formula:

B=\mu_0\, \frac{N}{L} I

Then, if we assign the subindex "1" to the quantities that define the magnetic field (B_1) inside solenoid 1, we have:

B_1=\mu_0\, \frac{N_1}{L_1} I_1

notice that there is no dependence on the diameter of the solenoid for this formula.

Now, if we write a similar formula for solenoid 2, given that it has :

1) half the length of solenoid 1 . Then L_2=L_1/2

2) twice as many turns as solenoid 1. Then N_2=2\,N_1

3) three times the current of solenoid 1. Then I_2=3\,I_1

we obtain:

B_2=\mu_0\, \frac{N_2}{L_2} I_2\\B_2=\mu_0\, \frac{2\,N_1}{L_1/2} 3\,I_1\\B_2=\mu_0\, 12\,\frac{N_1}{L_1} I_1\\B_2=12\,B_1

5 0
3 years ago
I have a voltage source of 12V but a light that only burns at 5V. The lamp works on 18 mA. Calculate the resistance that you EXT
ratelena [41]

Answer:

The resistance that will provide this potential drop is 388.89 ohms.

Explanation:

Given;

Voltage source, E = 12 V

Voltage rating of the lamp, V = 5 V

Current through the lamp, I = 18 mA

Extra voltage or potential drop, IR =  E- V  

                                                    IR = 12 V - 5 V = 7 V

The resistance that will provide this potential drop (7 V) is calculated as follows:

IR = V

R = \frac{V}{I} = \frac{7 \ V}{18 \times 10^{-3} A} \ = 388.89 \ ohms

Therefore, the resistance that will provide this potential drop is 388.89 ohms.

7 0
3 years ago
a car starts from the rest and accelerates at 9.54m/s for 6.5 seconds. what is the distance covered by the car​
uysha [10]

Answer:

= 201.53 meters

Explanation:

A car started from rest and accelerated at 9.54 m/s^2 for 6.5 seconds. How much distance was covered by the car?

Use the formula  d = \frac{at^{2} }{2} ,

where d is the distance, t is the time and "a" is the acceleration.
d=\frac{9*54*6*5^{2} }{2} = 201.53 m

3 0
2 years ago
Help me please I need help?
sweet [91]
Your answer is c. homologous structures
4 0
3 years ago
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