a race car is traveling at a speed of 80.0m/s on a circular race track of radius 450m what is the centripetal acceleration
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Answer:
The answer of this question is =1.258*10-4
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wire has a resistance of 1.3 ohms. The crosssectional
area of this wire is
(1) 7.9 × 10^−8 m2 (3) 4.6 × 10^0 m2
(2) 1.1 × 10^−7 m2 (4) 1.3 × 10^7 m2
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The cross-sectional area of the copper wire is which has 1.3 ohm at 20 degree Celsius during laboratory experiment by student is 7.9×10⁻⁸ ohm.
What is the resistance?
Resistance is obstacle for the current flow in the circuit. It is the measure of reverse act to current flow in through a material. It can be given as,
Here, (ρ) is the specific resistance (l) is the length of the wire and (A) is the cross-sectional area of the wire.
During a laboratory experiment, the student finds that at 20° Celsius, a 6.0-meter length of copper wire has a resistance of 1.3 ohms.
The value of resistivity of copper wire is 1.72×10⁻⁸ ohm-m. Put these values in the above formula as,
Thus, the cross-sectional area of the copper wire is which has 1.3 ohm at 20 degree Celsius during laboratory experiment by student is 7.9×10⁻⁸ ohm.
Learn more about the resistance here;
ρ = resistivity of copper at 20 °C = 1.72 x 10⁻⁸ ohm-m
R = resistance of the copper wire = 1.3 ohm
L = length of the copper wire = 6 meter
A = area of cross-section of the copper wire = ?
Resistivity of copper wire is given as
R = ρL/A
inserting the values in the above equation
1.3 = (1.72 x 10⁻⁸) (6)/A
A = 7.9 x 10⁻⁸ m²
Hence the correct choice is
(1) 7.9 x 10⁻⁸ m²
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Impulse = change in momentum
Impulse = (force ) x (time applied)
Momentum = (mass) x (speed)
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Momentum of the ball after the kick = (mass)x(speed) = 0.43kg x 25 m/s = 10.75 kg-m/s .
Impulse exerted by the kicker's toe = 10.75 kg-m/s .
(force) x (time) = 10.75 kg-m/s .
Force x 0.001 sec = 10.75 kg-m/s .
Divide each side by 0.001 sec :
Force = (10.75 kg-m/sec) / (0.001 sec) = 10,750 kg-m/s² = 10,750 newtons.
That's about 2,416 pounds from the kicker's toe, during that ¹/₁₀₀₀ of a second.
(1.21 tons !)
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Answer:
The centripetal acceleration that the moon experiences will be almost equal to the gravitational force that the Earth does in the moon,
Now, remember these two things:
F = m*a
and Fg = G*M1*M2/r^2
the first equation says that the force applied to something is equal to the mass of the object times the acceleration.
The second equation is for the gravitational force, where G is a constant, M1 and M2 are the masses of both objects, in this case, the Earth and the moon, and r is the distance.
We know that the acceleration in the surface of the Earth is:
a = Fg/M2 = g = G*M1/(RE)^2
now, for the moon we will have:
a = G*M1/(60RE)^2 = (G*M1/(RE)^2) *(1/60^2)
Here the term in the left is equal to g, so we have:
(G*M1/(RE)^2) *(1/60^2) = g*(1/60^2)
So the centripetal acceleration of the moon is 60^2 = 3600 times smaller than g.