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qwelly [4]
4 years ago
6

A force of 265.1 N acts on an object to produce an acceleration of 14.52 m/s^2. What is the mass of the object?

Physics
2 answers:
astra-53 [7]4 years ago
6 0

The answer is :


18.26


Hope I helped.

Lady bird [3.3K]4 years ago
4 0

Answer:

The mass of the object is 18.26kg

Explanation:

We know by Newton's second law we know that a force applied on an object is equal to its mass times the acceleration caused by the force:

F=ma

From this equation, solving for the mass:

m=\frac{F}{a}

Plugging in the values:

m=\frac{265.1N}{14.52m/s^2}=18.26kg

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The standard wave format for any wave is _________wave. When depicting ______ wave in standard wave format, the direction of mot
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4 years ago
A merry-go-round with a a radius of R = 1.99 m and moment of inertia I = 194 kg-m2 is spinning with an initial angular speed of
Aleksandr [31]

Answer:

Part 1)

L_1 = 185.2 kg m^2/s^2

Part 2)

L_2 = 663.07 kg m^2/s^2

Part 3)

L = 663.07 kg m^2/s^2

Part 4)

\omega = 1.83 rad/s

Part 5)

F_c = 453.6 N

Explanation:

Part a)

Initial angular momentum of the merry go round is given as

L_1 = I \omega

here we know that

I = 194 kg m^2

\omega = 1.47 rad/s^2

now we have

L_1 = 194 \times 1.47

L_1 = 185.2 kg m^2/s^2

Part b)

Angular momentum of the person is given as

L = mvR

so we have

m = 68 kg

v = 4.9 m/s

R = 1.99 m

so we have

L_2 = (68)(4.9)(1.99)

L_2 = 663.07 kg m^2/s^2

Part 3)

Angular momentum of the person is always constant with respect to the axis of disc

so it is given as

L = 663.07 kg m^2/s^2

Part 4)

By angular momentum conservation of the system we will have

L_1 + L_2 = (I_1 + I_2)\omega

185.2 + 663.07 = (194 + 68(1.99^2))\omega

848.27 = 463.28 \omega

\omega = 1.83 rad/s

Part 5)

Force required to hold the person is centripetal force which act towards the center

so we will have

F_c = m\omega^2 R

F_c = 68(1.83^2)(1.99)

F_c = 453.6 N

3 0
3 years ago
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