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valkas [14]
4 years ago
12

When the palmaris longus muscle in the forearm is flexed, the wrist moves back and forth. If the muscle generates a force of 45.

5 N and it is acting with an effective lever arm of 2.45 cm, what is the torque that the muscle produces on the wrist?
Physics
1 answer:
SpyIntel [72]4 years ago
3 0

Answer:

Torque on the rocket will be 1.11475 N -m

Explanation:

We have given that muscles generate a force of 45.5 N

So force F = 45.5 N

This force acts on the is acting on the effective lever arm of 2.45 cm

So length of the lever arm d = 2.45 cm = 0.0245 m

We have to find torque

We know that torque is given by \tau =F\times d=45.5\times 0.0245=1.11475N-m

So torque on the rocket will be 1.11475 N -m

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Two blocks with rough surfaces are stacked on top of a slippery horizontal surface. The top block has a mass of m, and the botto
lapo4ka [179]

Answer:

Please refer to the attached schematic for free-body diagrams of the blocks.

The contact forces N1 = mg, N2 = 3mg.

The friction force fs = mg/2.

The tension force T = 3mg/2.

The distance y = (gt^2)/4

Explanation:

Starting with the third block which is declining a time t. Newton's Second Law implies that

F = (3m)a\\3mg - T = 3ma\\T = 3mg - 3ma

There are two unknowns in one equation: T and a. Therefore, we need to find more equations to solve these unknown variables.

Let's look at free-body diagram 1:

Newton's Second Law:

F = ma\\f_s = ma

That is another equation with one more unknown: fs.

Free-body diagram 2:

F = (2m)a\\T - f_s = 2ma

That is our third equation. We now have three equations with three unknowns. Combining equation 2 and 3 gives:

T - ma = 2ma\\T = 3ma

Plugging this equation into the first equation gives

T = 3mg - 3ma\\3ma = 3mg - 3ma\\3mg = 6ma\\a = g/2

Following this, we can find the other unknowns:

T = 3ma = 3m(g/2) = \frac{3mg}{2}

f_s = ma = mg/2

The contact forces are N1 and N2.

N_1 = mg\\N_2 = 2mg + N1 = 3mg

Finally, using the acceleration and equation of kinematics, we can find the distance the hanging weight drops in a time t.

y - y_0 = v_0t + \frac{1}{2}at^2\\y = \frac{1}{2}(\frac{g}{2})t^2 = \frac{gt^2}{4}

6 0
4 years ago
Derive an equation for the acceleration of block 3 for any arbitrary values of m3 and m2. Express your answer in terms of m3, m2
dybincka [34]

Complete question is;

Block 1 is resting on the floor with block 2 at rest on top of it. Block 3, at rest on a smooth table with negligible friction, is attached to block 2 by a string that passes over a pulley, as shown in the attachment below. The string and pulley have negligible mass.

Block 1 is removed without disturbing block 2.

Derive an equation for the acceleration of block 3 for any arbitrary values of m3 and m2. Express your answer in terms of m3, m2, and physical constants as appropriate.

Answer:

a = (m2)g/(m3 + m2)

Explanation:

Looking at the attached image, if we consider the free body diagram for block 3, by using Newton's first law of motion, we will arrive at the formula;

T = (m3)a - - - (eq 1)

where;

T is the tension in the string

a is acceleration

m3 is mass of block 3

Meanwhile doing the same with Block 2, the free body diagram would give us the formula; (m2)g - T = (m2)a

Making T the subject gives us;

T = (m2)g - (m2)a - - - (eq 2)

where;

g is acceleration due to gravity

T is the tension in the string

a is acceleration

m2 is mass of block 2

To solve for the acceleration, we will just substitute (m3)a for T in eq 2.

Thus;

(m3)a = (m2)g - (m2)a

(m3)a + (m2)a = (m2)g

a(m3 + m2) = (m2)g

a = (m2)g/(m3 + m2)

3 0
3 years ago
A climber pulls down on a rope causing his body to rise up with the rope? Which law of motion is it?
12345 [234]
This would be called the law of action-reaction. This states that every action will have an equal and opposite reaction. The action in the example is pulling down on the rope. The opposite and equal reaction is the climers body moving upward. The same law can be applied to a rocket. The action is the engines pushing down and the reaction is the rocket going up. :D
4 0
3 years ago
At what condition the screw gauge dors not have zero error
lana66690 [7]
<span>The screw gauge doesn’t have a zero error when the zero of the main scale coincides with the zero of the vernier scale or circular scale.
 In case they don’t coincide then the instrument screw gauge is said to have a zero error which can be positive zero error or a negative zero error. In order to get an exact measurement, the error is added or subtracted from the value measured.</span>
6 0
4 years ago
Longitudinal seismic waves are known as
Helga [31]
Its “A“ primary waves!!!!!!
3 0
3 years ago
Read 2 more answers
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