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vazorg [7]
4 years ago
7

A meter stick whose mass is 290 grams lies on ice. You pull at one end of the meter stick, at right angles to the stick, with a

force F= 9 newtons. The ensuing motion of the meter stick is quite complicated, but what are the initial magnitude and direction of the rate of change of the momentum of the stick, dPtot/d⁢t, when you first apply the force?
Magnitude N
Direction leftdownrightup
What is the magnitude of the initial acceleration of the center of the stick?
m/s2
Physics
1 answer:
katen-ka-za [31]4 years ago
5 0

Answer

given,

mass of the stick = 290 grams = 0.29 Kg

Force on the stick on one side = F =  9 N

force acting perpendicular to stick.

magnitude of acceleration

rate of change of angular momentum is equal to Force

rate of change of angular momentum = 9 N

F = m a

a = \dfrac{F}{m}

a = \dfrac{9}{0.29}

a = 31.034 m/s²

Direction of motion will in the direction of force application or in the direction of change of velocity

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Answer:

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Explanation:

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4 0
4 years ago
Read 2 more answers
If 2.2 lbs = 1.0 kg, and Megan Progress weighs 130 lbs, what is her weight in newtons? W = N (Round your answer to nearest whole
lord [1]

as it is given that

2.2 lbs = 1 kg

here we know that Megan Progress weighs 130 lbs

so its mass in kg is given as

m = 130 * \frac{1}{2.2} = 59.1 kg

now to find the weight in Newton unit we can say

W = mg

W = 59.1* 9.8

W = 579 N

so 130 lbs weighs as 579 N in SI units

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