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alexandr402 [8]
3 years ago
13

A mechanic turns a wrench using a force of 27.00 N at a distance of 16.00 cm from the rotation axis. The force is perpendicular

to the wrench handle. What magnitude torque does she apply to the wrench?
Physics
1 answer:
lutik1710 [3]3 years ago
3 0

Answer:4.32Nm

Explanation:

The magnitude of the torque will be the product of the force and its perpendicular distance from the force.

Force = 27N

Perpendicular distance = 16cm = 0.16m

Torque = 27×0.16

Torque = 4.32Nm

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What is thermodynamics?
Kryger [21]
D. The flow of energy,heat, and work
8 0
2 years ago
ASK YOUR TEACHER A meter stick is found to balance at the 49.7-cm mark when placed on a fulcrum. When a 51.5-gram mass is attach
Simora [160]

Answer:

0.114 kg or 114 g

Explanation:

From the diagram attaches,

Taking the moment about the fulcrum,

sum of clockwise moment = sum of anticlockwise moment.

Wd = W'd'

Where W = weight of the mass, W' = weight of the meter rule, d = distance of the mass from the fulcrum, d' = distance of the meter rule.

make W'  the subject of the equation

W' = Wd/d'................ Equation 1

Given: W = mg = 0.0515(9.8) = 0.5047 N, d = (39.2-16) = 23.2 cm, d' = (49.7-39.2) = 10.5 cm

Substitute these values into equation 1

W' = 0.5047(23.2)/10.5

W' = 1.115 N.

But,

m' = W'/g

m' = 1.115/9.8

m' = 0.114 kg

m' = 114 g

6 0
3 years ago
A mass of 4.8 kg is dropped from a height of 4.84 meters above a vertical spring anchored at its lower end to the floor. If the
krek1111 [17]

Answer:

45.6 cm

Explanation:

Let x (m) be the length that the spring is compressed. We know that when we drop the mass from 4.84 m above and compress the springi, ts gravitational energy shall be converted to spring potential energy due to the law of energy conservation

E_g = E_p

mgh = 0.5kx^2

where h = 4.84 + x is the distance from the dropping point the the compressed point, and k = 24N/cm = 2400N/m is the spring constant, g = 9.81 m/s2 is the gravitational acceleration constant. And m = 4.8 kg is the object mass.

4.8*9.81(4.84 + x) = 0.5 * 2400 * x^2

47.088x + 227.906 = 1200x^2

1200x^2 - 47.088x - 227.906 = 0

x = 0.456m or 45.6 cm

5 0
4 years ago
Two people, one with mass m1 and the other with mass m2, stand on a stationary sled with mass M on a frozen lake. Assume that th
lozanna [386]

Answer:

Part a)

Velocity of sled

v = \frac{m_1 s}{m_1 + m_2 + M}

velocity of first man who jump off

v_1 = -\frac{(m_2 + M) s}{m_1 + m_2 + M}

Part b)

Velocity of sled

v_f = (\frac{m_1 s}{m_1 + m_2 + M}) + (\frac{m_2}{m_2 + M})s

Also the speed of second person is given as

v_2 = (\frac{m_1 s}{m_1 + m_2 + M}) - \frac{Ms}{m_2 + M}

Part c)

change in kinetic energy of sled + two people is given as

KE = \frac{1}{2}Mv_f^2 + \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2

Explanation:

As we know that here we we consider both people + sled as a system then there is no external force on it

So here we can use momentum conservation

since both people + sled is at rest initially so initial total momentum is zero

now when first people will jump with relative velocity "s" then let say the sled + other people will move off with speed v

so by momentum conservation we have

0 = m_1(v - s) + (m_2 + M)v

v = \frac{m_1 s}{m_1 + m_2 + M}

so velocity of the sled + other person is

v = \frac{m_1 s}{m_1 + m_2 + M}

velocity of first man who jump off

v_1 = \frac{m_1 s}{m_1 + m_2 + M} - s

v_1 = -\frac{(m_2 + M) s}{m_1 + m_2 + M}

Part b)

now when other man also jump off with same relative velocity

so let say the sled is now moving with speed vf

so by momentum conservation we have

(m_2 + M)(\frac{m_1 s}{m_1 + m_2 + M}) = m_2(v_f - s) + Mv_f

(m_2 + M)(\frac{m_1 s}{m_1 + m_2 + M}) + m_2s = (m_2 + M)v_f

Now we have

v_f = (\frac{m_1 s}{m_1 + m_2 + M}) + (\frac{m_2}{m_2 + M})s

Also the speed of second person is given as

v_2 = (\frac{m_1 s}{m_1 + m_2 + M}) + (\frac{m_2}{m_2 + M})s - s

v_2 = (\frac{m_1 s}{m_1 + m_2 + M}) - \frac{Ms}{m_2 + M}

Part c)

change in kinetic energy of sled + two people is given as

KE = \frac{1}{2}Mv_f^2 + \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2

here we know all values of speed as we found it in part a) and part b)

4 0
3 years ago
A 591 μF capacitor is discharged through a resistor, whereby its potential difference decreases from its initial value of 88.5 V
Stels [109]

Answer:

2.6 kilo Ohm

Explanation:

Capacitance, C = 591 μF = 591 x 10^-6 F

Vo = 88.5 V

V = 11.9 V

t = 3.09 s

Let the resistance is R.

V = V_{0}e^{\frac{-t}{RC}}

\frac{11.9}{88.5} = e^{\frac{-t}{RC}}

0.135 = e^{\frac{-t}{RC}}

Take natural log on oth the sides

ln 0.135 = - 3.09 / RC

RC = 1.545

R = 1.545 / ( 591 x 10^-6)

R = 2614.2 ohm

R = 2.6 kilo Ohm

Thus the resistance is  2.6 kilo Ohm.

8 0
3 years ago
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