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ki77a [65]
2 years ago
5

To exercise, a man attaches a 4.0 kg weight to the heel of his foot. When his leg is stretched out before him, what is the torqu

e exerted by the weight about his knee, 40 cm away from the weight? Use g = 10 m/s2. A. 160 Nâm B. 16 Nâm C. 1600 Nâm D. 1.6 Nâm
Physics
1 answer:
Masja [62]2 years ago
8 0

Answer:

B. τ = 16 Nm

Explanation:

In order to find the torque exerted by the weight attached to the heel of man's foot, when his leg is stretched out. We use following formula:

τ = Fd

here,

τ = Torque = ?

F = Force exerted by the weight = Weight = mg

F = mg = (4 kg)(10 m/s²) = 40 N

d = distance from knee to weight = 40 cm = 0.4 m

Therefore,

τ = (40 N)(0.4 m)

<u>B. τ = 16 Nm</u>

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Essam is abseiling down a steep cliff. How much gravitational potential energy does he lose for every metre he descends? His mas
Dafna11 [192]

Answer:

720 J

Explanation:

The gravitational potential energy that Essam loses for every metre is given by:

\Delta U=mg \Delta h

where

m=72 kg is Essam's mass

g=10 N/kg is the gravitational field strength

\Delta h=1 m is the difference in height

By substituting the numbers into the formula, we find

\Delta U=(72 kg)(10 N/kg)(1 m)=720 J

5 0
2 years ago
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A gas laser has a cavity length of 1/3 m and a single oscillation frequency of 9.0 x 1014 Hz. What is the cavity mode number?
Eddi Din [679]

Answer:

2 × 10⁶

Explanation:

Data provided in the question:

Cavity length, L = \frac{1}{3}m

Oscillation frequency, f_m = 9.0 × 10¹⁴ Hz

Now,

we know,

f_m=\frac{c}{\lambda_m}

here,

c is the speed of light = 3 × 10⁸ m/s

\lambda_m = Wavelength of mode m inside the laser cavity

m is the cavity mode number

Thus,

9.0\times10^{14}=\frac{3\times10^8}{\lambda_m}

or

\lambda_m = \frac{1}{3} × 10⁻⁶

Also,

m\lambda_m = 2L

Therefore,

m × \frac{1}{3} × 10⁻⁶ = 2 × \frac{1}{3}

or

m = 2 × 10⁶

5 0
2 years ago
A11) A solenoid of length 18 cm consists of closely spaced coils of wire wrapped tightly around a wooden core. The magnetic fiel
vitfil [10]

Answer:

A

Explanation:

From a Solenoid we know that a magnetic fiel is always inversely proportional to lenght L or BL = constant

B= frac{\mu_0}{2R}

As I is constant

\frac{B2}{B1} = \frac{R1}{R2}

B2 = 2mT*\frac{18}{21}

B2 = 1.714mT

7 0
2 years ago
wheel rotates from rest with constant angular acceleration. Part A If it rotates through 8.00 revolutions in the first 2.50 s, h
Alex73 [517]

Answer:

x2 = 64 revolutions.

it rotate through 64 revolutions in the next 5.00 s

Explanation:

Given;

wheel rotates from rest with constant angular acceleration.

Initial angular speed v = 0

Time t = 2.50

Distance x = 8 rev

Applying equation of motion;

x = vt +0.5at^2 ........1

Since v = 0

x = 0.5at^2

making a the subject of formula;

a = x/0.5t^2 = 2x/t^2

a = angular acceleration

t = time taken

x = angular distance

Substituting the values;

a = 2(8)/2.5^2

a = 2.56 rev/s^2

velocity at t = 2.50

v1 = a×t = 2.56×2.50 = 6.4 rev/s

Through the next 5 second;

t2 = 5 seconds

a2 = 2.56 rev/s^2

v2 = 6.4 rev/s

From equation 1;

x = vt +0.5at^2

Substituting the values;

x2 = 6.4(5) + 0.5×2.56×5^2

x2 = 64 revolutions.

it rotate through 64 revolutions in the next 5.00 s

5 0
2 years ago
You are driving on a road where rain has left large pools of water, and you have driven through water that was several inches de
makvit [3.9K]
In would say that you may have water in your brakes which may have gotten in the brake lines or in the brake discs so that could cause the brakes to malfunction due to driving through the pools of water so the brakes should be examined as soon as possible.
3 0
2 years ago
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