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wariber [46]
3 years ago
8

PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!! I CANNOT RETAKE THIS!!

Physics
1 answer:
Tasya [4]3 years ago
3 0

Answer:

98 N

Explanation:

The weight of an object is given by:

W=mg

where

m is the mass of the object

g=9.8 m/s^2 is the acceleration due to gravity

In this problem, the mass of the lemur is m=10 kg, so its weight is

W=mg=(10 kg)(9.8 m/s^2)=98 N

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A large truck is crossing a bridge 1 mile long. The bridge can only hold 14000 lbs, which is the exact weight of the truck. The
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Depends on the weight of the bird.
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3 years ago
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A student sits at rest on a piano stool that can rotate without friction. the moment of inertia of the student-stool system is 4
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<span>As we Know ω=v/r I2 = m*R² = 1.5*0.35² = 0.18375 kgâ™m² I = Is + I2 = 4.28375 kgâ™m² w = L/I = (v*R*m)/I = (2.8*0.35*1.5)/4.28375 = 0.343 rad/sec So the answer is 0.343 rd/sec</span>
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4 years ago
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8 0
3 years ago
A spring with spring constant k is suspended vertically from a support and a mass m is attached. The mass is held at the point w
garik1379 [7]

Answer:

The oscillation frequency of the spring is 1.66 Hz.

Explanation:

Let m is the mass of the object that is suspended vertically from a support. The potential energy stored in the spring is given by :

E_s=\dfrac{1}{2}kx^2

k is the spring constant

x is the distance to the lowest point form the initial position.

When the object reaches the highest point, the stored potential energy stored in the spring gets converted to the potential energy.

E_P=mgx

Equating these two energies,

\dfrac{1}{2}kx^2=mgx

\dfrac{k}{m}=\dfrac{2g}{x}.............(1)

The expression for the oscillation frequency is given by :

f=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}

f=\dfrac{1}{2\pi}\sqrt{\dfrac{2g}{x}} (from equation (1))

f=\dfrac{1}{2\pi}\sqrt{\dfrac{2\times 9.8}{0.18}}

f = 1.66 Hz

So, the oscillation frequency of the spring is 1.66 Hz. Hence, this is the required solution.

8 0
4 years ago
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