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babunello [35]
3 years ago
9

When tightening a bolt, you push perpendicularly on a wrench with a force of 165 N at a distance of 0.140 m from the center of t

he bolt? (a) How much torque are you exerting in newtons x meters (relative to the center of the bolt) (b) Convert this torque to footpounds
Physics
1 answer:
choli [55]3 years ago
3 0

Answer:

Part a)

\tau = 23.1 Nm

Part b)

\tau = 17.05 Foot pound force

Explanation:

As we know that torque is defined as the product of force and its perpendicular distance from reference point

so here we have

\tau = \vec r \times \vec F

now we have

\tau = (0.140)(165)

\tau = 23.1 Nm

Part b)

Now we know the conversion as

1 meter = 3.28 foot

1 N = 0.225 Lb force

now we have

\tau = 23.1 Nm

\tau = 23.1 (0.225 Lb)(3.28 foot)

\tau = 17.05 Foot pound force

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A 40 kg person sits on top of a 400 kg rock. What is the person’s weight? 390 N
algol13

The weight of the person is given by:

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Plug in and solve for W:

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The mass of the Earth is 6 × 1024 kg, the mass of the Moon is 7 × 1022 kg, and the center-to-center distance is 4 × 108 m. How f
Karolina [17]

Answer:

The center of mass of the Earth–Moon system is 4.613 × 10⁶ m from center of the Earth.

Explanation:

Let the reference point be the center of the Earth

X_{Cm = \frac{M_eX_e +M_mX_m}{M_e+M_m}

Where;

Xcm is the distance from center of the Earth =?

Me is the mass of the Earth = 6 × 10²⁴ kg

Xe is the center mass of the Earth = 0

Mm is the mass of the moon = 7 × 10²² kg

Xm is the center mass of the moon =  4 × 10⁸ m

X_{Cm = \frac{M_eX_e +M_mX_m}{M_e+M_m} =  \frac{M_e(0) +M_mX_m}{M_e+M_m} = \frac{ M_mX_m}{M_e+M_m}}\\\\X_C_m = \frac{7 X 10^{22}*4X10^8}{6X10^{24}+7X10^{22}} =\frac{28 X10^{30}}{607X10^{22}}\\\\  X_C_m = 4.613 X10^6 m

Therefore, the center of mass of the Earth–Moon system is 4.613 × 10⁶ m from center of the Earth.

8 0
3 years ago
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