A 1.3-kg ball is attached to the end of a 0.8-m string to form a pendulum. This pendulum is released from rest with the string h
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The force exerted by a magnetic field on a wire carrying current is:
where I is the current, L the length of the wire, B the magnetic field intensity, and
the angle between the wire and the direction of B.
In our problem, the force is F=0.20 N. The current is I=1.40 A, while the length of the wire is L=35.0 cm=0.35 m. The angle between the wire and the magnetic field is
, so we can re-arrange the formula and substitute the numbers to find B:
where I is the current, L the length of the wire, B the magnetic field intensity, and
In our problem, the force is F=0.20 N. The current is I=1.40 A, while the length of the wire is L=35.0 cm=0.35 m. The angle between the wire and the magnetic field is
Answer:
Explanation:
According to “Newton's second law”
“Force” is “mass” times “acceleration”, or F = m× a. This means an object with a larger mass needs a stronger force to be moved along at the same acceleration as an object with a small mass
Force = mass × acceleration
Given that,
Mass = 5.32 kg
F = 12.7N
Normal force = mg + F sinx,
“m” being the object's "mass",
“g” being the "acceleration of gravity",
“x” being the "angle of the cart"
To find normal force substitute the values in the formula,
Normal force = 5.32 × 9.8 + 12.7 × sin(-28.7)
Normal force = 52.136 + 12.7 × 0.480
Normal force = 52.136 + 6.096
Normal force = 58.232 N
<u>Acceleration of the cart</u>: