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3 years ago

9

A meterstick is placed on a pivot point of 42.5cm and a 45g mass is hung at the 20cm mark. When released the meterstick remains

in static equilibrium. What is the mass of the meterstick

2 answers:

Vikki [24]3 years ago

6 0

Hey i dont have an answer but i need the points for finals today. Thank you

nordsb [41]3 years ago

3 0

I don’t know the answer sorry I need points

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A 34n force is applied to a 213kg mass how much does the mass accelerate

motikmotik

We Know, F = m*a
Here, F = 34 N
m = 213 Kg

Substitute their values in the equation,
34 = 213 * a
a = 34/213
a = 0.159 m/s²

So, your final answer & the acceleration of the object would be 0.159 m/s²

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what is the magnitude of g at a height above earth surface where free fall acceleration equal 6.5m/s2

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3 years ago

HELP ME PLEASE!!!!!!!!!

Stolb23 [73]

As per the given Figure attached here we know that both charges q1 and q2 will apply same force on charge q3 and hence the resultant force due to both charges will be along Y axis vertically upwards

So here we know that

$F = \frac{kq_1q_3}{d_{13}^2}$

now from the above equation

$F = \frac{(9\times 10^9)(2\times 10^{-6})(4 \times 10^{-6})}{0.5^2}$

$F = 0.288 N$

so both of the charges will apply 0.288 N force on q3 charge along the line joining them

now the net force due to vector sum is given by

$F_{net} = 2Fcos\theta$

here we know that angle is

$\theta = 37 degree$

now we have

$F_{net} = 2\times 0.288 cos37$

$F_{net} = 0.46 N$

so net force on q3 is 0.46 N vertically upwards along +Y axis

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As populations in North America grow, so does the need for wood to build houses, stores, and other buildings. Select the stateme

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question 1, A

Question 2, D

Question 3, B

hope this helps :)

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