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vitfil [10]
3 years ago
13

A rubber band has a spring constant of 75 N/m. How much force is required to stretch the rubber band 0.02 m past its natural len

gth?
A. 2.3 N
B. 1.5 N
C. 75 N
D. 3750 N
Physics
1 answer:
ale4655 [162]3 years ago
7 0

Answer:

(75 x 0.02) = 1.5 N

Explanation:

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3 years ago
What is 7.4 in it's simplest form?
Iteru [2.4K]
Assuming that you mean to ask <em>. . . "What is 7.4 </em><u><em>as a fraction</em></u><em> in simplest form</em>?"

7.4 = 7 and 4/10

4/10 can be reduced to 2/5

7 represents = (7*5)/5 = 35/5

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or
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4 0
3 years ago
A surveyor is using a magnetic compass 6.4 m below a power line in which there is a steady current of 120 A. (a) What is the mag
Mademuasel [1]

Answer:

Explanation:

Magnetic field due to a long current carrying conductor

μ₀ / 4π x 2i / r  ( i = current , r = distance of point from wire )

= 10⁻⁷ x 2 x 120 / 6.4  ( i = 120 A , r = 6.4m )

= 37.5 x 10⁻⁷ T .

= 3. 75 X 10⁻⁶ T .

= 3.75 µT.

b )

The direction of this field will be horizontal hence it will affect magnetic needle.

6 0
3 years ago
Read 2 more answers
Calculate the speed of a proton after it accelerates from rest through a potential difference of 380 v .
nikitadnepr [17]

The magnitude of the change in potential energy is equal to the kinetic energy,

\left | \Delta U \right |= K

or                        

qV=\frac{1}{2} mv^{2}    

Here, V is potential difference, q is charge, m is mass and v is velocity.

We can also write,

v=\sqrt{\frac{2qV}{m} }

Given  V=380 V

Substituting this value with mass of proton, m=1.672\times10^{-27}  kg and charge of proton,q= 1.6\times10^{-19} C we get

v=\sqrt{\frac{2\times1.6\times10^{-19}C \times 380V}{1.672\times10^{-27}  kg} }\\\\v= 727.27\times10^{8} m/s =7.27\times10^{6}m/s

Therefore, the speed of proton is 7.27\times10^{6}m/s.


7 0
3 years ago
Read 2 more answers
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