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NeTakaya
2 years ago
14

A bike accelerates uniformly from rest to a speed of 7.10 m/s over a distance of 35.4m .determine the acceleration of the bike.

Physics
1 answer:
sergeinik [125]2 years ago
5 0
Given: Change of x is 35.4m, Velocity Final=7.10 m/s, Velocity Initial=0m/s
Find: Acceleration
Analysis:
Vf²=Vi²+2aΔx (Velocity final squared equals Velocity initial squared plus 2 times acceleration times change of x)
(7.10 m²/s)²=(0 m/s)²+2a(35.4 m)
50.41 m/s²=(70.8 m)a
a=0.712 m/s²
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Answer:

C.

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2 years ago
A 75-kilogram hockey player is skating across the
azamat
I think it is (3) 690 N. You can use the theorem of Impulse and momentum where the impulse is equal to the change in momentum or:
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3 years ago
at 1 p.m. a car traveling at a constant velocity of 78 km per hour towards the West it's 34 km to the west of our school how far
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3 0
3 years ago
"The gravity of the Sun causes the planets to move in a circular path."
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3 0
3 years ago
A straight wire of length 0.56 m carries a conventional current of 0.4 amperes. What is the magnitude of the magnetic field made
djverab [1.8K]

Answer:

The magnitude of the magnetic field made by current in the wire is 3.064 x 10⁻⁶ T.

Explanation:

Given;

length of the straight wire, L = 0.56 m

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distance of magnetic field from the wire, r = 2.6 cm = 0.026 m

To determine magnitude of magnetic field made by current in the wire, we will apply Bio-Savart Law;

B = \frac{\mu_o}{4\pi r} \frac{LI}{\sqrt{r^2 +(L/2)^2} } \\\\B = \frac{4\pi *10^{-7}}{4\pi *0.026} \frac{0.56*0.4}{\sqrt{(0.026)^2 +(0.56/2)^2} }\\\\B = 3.846*10^{-6}(0.7966)\\\\B = 3.064*10^{-6} \ T

Therefore, the magnitude of the magnetic field made by current in the wire is 3.064 x 10⁻⁶ T.

3 0
3 years ago
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