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Katarina [22]
3 years ago
6

The triceps muscle in the back of the upper arm is primarily used to extend the forearm. Suppose this muscle in a professional b

oxer exerts a force of 1.9 × 103 N with an effective perpendicular lever arm of 2.85 cm, producing an angular acceleration of the forearm of 150 rad/s2?
Physics
1 answer:
Romashka-Z-Leto [24]3 years ago
4 0

Answer:

I=0.0361 kg.m^2

Explanation:

Torque is the rotational equivalent of a force

Torque= perpendicular distance r X Force F

Torque T = I(moment of inertia) X α (angular acceleration)

T= Iα

r= 0.0285m

F= 1.9 x 10^3

T=0.0285 x 1.9 x 10^3

T= 54.15Nm

I=T/α

I=54.15/150

I=0.361 kg.m^2

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I need help with a physics worksheet about Newtons Laws.
Viktor [21]

The diagram of the object on the inclined plane is shown below

The mass of 7kg is exerted on the plane surface. It is acted upon by gravity. The force exerted on the surface is calculated by applying the formula,

F = mass x acceleration due to gravity

Assumme acceleration due to gravity is 9.8m/s^2, then

Force = 7 x 9.8 = 68.6N

The surface exerts an opposite force of the same magnitude with the force of the object but in opposite direction. Since it is inclined at an angle of 36.9 degrees,

Normal force = mgCosθ = 68.6Cos36.9 = 54.86N

Recall, frictional force = normal reaction x coefficient of friction

Given that coefficient of kinetic friction = 0.35,

Frictional force = 54.86 x 0.35 = 19.201N

This is the force that must be overcome to keep the object moving.

The force pulling the object upwards along the inclined plane is

mgSinθ = 7 x 9.8 x Sin36.9 = 41.19N

Since the velocity is constant, it means that there is no acceleration. The net force is zero. The force required to pull the mass and make it move at constant velocity must be equal to the sum of the exerted force and the the frictional force(Since it involves coeffricient of kinetic friction, it would cause the object to keep moving)

Thus,

Required force = 19.2 + 41.19 = 60.4N

Option E is correct

8 0
1 year ago
A man walking on a tightrope carries a long a pole which has heavy items attached to the two ends. If he were to walk the tight-
katen-ka-za [31]

Answer:

 I_weight = M L²

this value is much larger and with it it is easier to restore balance.I

Explanation:

When man walks a tightrope, he carries a linear velocity, this velocity is related to the angular velocity by

            v = w r

For man to maintain equilibrium needs the total moment to be zero

             ∑τ = I α

              S  τ = 0

The forces on the home are the weight of the masses, the weight of the man and the support on the rope, the latter two are zero taque the distance to the center of rotation is zero.

Therefore the moment of the masses and the open is the one that must be zero.

If the man carries only the bar, we could approximate it by two open one on each side of the axis of rotation formed by the free of the rope

              I = ⅓ m L² / 4

As the length of half the length of the bar and the mass of the bar is small, this moment is small, therefore at the moment if there is some imbalance it is difficult to recover.

If, in addition to the opening, each of them carries a specific weight, the moment of inertia of this weight is

             I_weight = M L²

this value is much larger and with it it is easier to restore balance.

5 0
3 years ago
Velocity question...is it A B Cor D
Ket [755]
In a constant acceleration of 3m per second, after 10 seconds,

3 x 10 = 30

B. 30m/s is your answer

hope this helps :D
6 0
3 years ago
Read 2 more answers
In a 49 s interval, 595 hailstones strike a glass window of an area of 0.954 m at an angle of 25° to the window surface. Each ha
eduard

Average  force on the window: 0.32 N

Explanation:

The average force exerted on the window is given by Newton's second law

F=\frac{\Delta p}{\Delta t}

where

\Delta p is the net change in momentum of the hailstones in a time interval of \Delta t

In order to find the change in momentum, we have to consider only the component of the hailstone's momentum perpendicular to the window, therefore:

p_i =m u sin \theta is the initial momentum of one hailstone, with

m = 7 g = 0.007 kg is the mass

u=4.5 m/s is the initial speed

\theta=25^{\circ} is the angle with the window

The final momentum is

p_f = mv sin \theta

where

v = 4.5 m/s is the final speed (the  collision is elastic so the speed is equal, while the direction changes)

\theta=-25^{\circ} (after the rebound, the direction has changed)

So the change in momentum of 1 hailstone is

\Delta p = mv sin(-25^{\circ})-mu sin(25^{\circ})=-2mu sin(25^{\circ})=-0.0266 kg m/s

We are interested only in the magnitude, so

\Delta p = 0.0266 kg m/s

There are 595 hailstones hitting the window in 49 s, so the total change in momentum is

\Delta p = 595\cdot 0.0266 = 15.8 kg m/s

And therefore, the average force on the window is

F=\frac{\Delta p}{\Delta t}=\frac{15.8}{49}=0.32 N

Learn more about  force:

brainly.com/question/8459017

brainly.com/question/11292757

brainly.com/question/12978926

#LearnwithBrainly

3 0
3 years ago
A 70 kg student stands on top of a 5.0 m platform diving board . how much gravitational potential energy does the student have?
Marianna [84]

Answer:

a. P.E = 3430Joules.

b. Workdone = 3430Nm

Explanation:

<u>Given the following data;</u>

Mass = 70kg

Distance = 5m

We know that acceleration due to gravity is equal to 9.8m/s²

To find the potential energy;

Potential energy = mgh

P.E = 70*9.8*5

<em>P.E = 3430J</em>

b. To find the workdone;

Workdone = force * distance

But force = mass * acceleration

Force = 70*9.8

Force = 686 Newton.

Workdone = 686 * 5

<em>Workdone = 3430Nm</em>

6 0
2 years ago
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