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DENIUS [597]
3 years ago
15

10. A person is about to kick a soccer ball. Consider the leg and foot to be a single rigid body and that it rotates about a fix

ed axis through the knee joint center. Immediately prior to impact the leg and foot are in the vertical direction and the distal end of the foot has an acceleration of 10 g in the horizontal direction. The muscle force vector makes an angle of 15 degrees with the vertical and has a moment arm of 5 cm from the knee joint center. Assume the person is 1.7 m tall and has a mass of 75 kg. Find the force in the muscle
Physics
1 answer:
statuscvo [17]3 years ago
4 0

Answer:

i don't know but i hope you get it right

Explanation

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A 5.0-kg box rests on a horizontal surface. The coefficient of kinetic friction between the box and surface is ????K=0.50. A hor
blagie [28]

Answer:

a)W= 1 J

b)W= 1 J

c)W= 0 J

Explanation:

Given that

m = 5 kg

Coefficient of kinetic friction,K=0.5

d= 10 cm

We know that if velocity is constant it means that the acceleration of the system is zero.Or we can say that all forces are balance in the system.

We know that work is the dot product of force and displacement

W= F.d

The friction force on the box

Fr= K m g

Fr= 0.5 x 5 x 10 = 25 N                      ( take g =10 m/s²)

Fr=25 N

a)

Acceleration a= 0

So horizontal force F = Fr

W = Fr.d

W=10 x 0.1 J

W= 1 J

b)

The work done by friction force

W= 1 J

c)

The net force on the system is zero because acceleration is zero.

F= 0

So

W= 0 J

3 0
3 years ago
As you drive in your car at 22.5 mi/h, you see a child’s ball roll into the street ahead of you. You hit the brakes and stop as
evablogger [386]

Answer:

Explanation:

Speed of car =22.5miles/hr

U=22.5miles/hour

Applied brake and come to rest

Final velocity, =0

t, =2sec

Given that,

Speed=distance /time

Then,

Distance, =speed, ×time

Converting mile/hour to m/s

Given that

Use: 1 mile= 1600 m, 1 h= 3600s

22.5miles/hour × 1600m/mile × 1hour/3600s

Therefore, 22.5mile/hour=10m/s

Using speed =10m/s

Distance =speed ×time

Distance=10×2

Distance, =20m

The distance travelled before coming to rest is 20m.

5 0
3 years ago
As rotational speed increases, thrust____?
never [62]
Increases exponentially is your correct answer
6 0
3 years ago
A circular loop of flexible iron wire has an initial circumference of 164cm , but its circumference is decreasing at a constant
Travka [436]

Answer:

emf = 0.02525 V

induced current with a counterclockwise direction

Explanation:

The emf is given by the following formula:

emf=-\frac{\Delta \Phi_B}{\Delta t}=-B\frac{\Delta A}{\Delta t}\ \ =-B\frac{A_2-A_1}{t_2-t_1}   (1)

ФB: magnetic flux =  BA

B: magnitude of the magnetic field = 1.00T

A2: final area of the loop; A1: initial area

t2: final time, t1: initial time

You first calculate the final A2, by taking into account that the circumference of loop decreases at 11.0cm/s.

In t = 4 s the final circumference will be:

c_2=c_1-(11.0cm/s)t=164cm-(11.0cm/s)(4s)=120cm

To find the areas A1 and A2 you calculate the radius:

r_1=\frac{164cm}{2\pi}=26.101cm\\\\r_2=\frac{120cm}{2\pi}=19.098cm

r1 = 0.261 m

r2 = 0.190 m

Then, the areas A1 and A2 are:

A_1=\pi r_1^2=\pi (0.261m)^2=0.214m^2\\\\A_2=\pi r_2^2=\pi (0.190m)^2=0.113m^2

Finally, the emf induced, by using the equation (1), is:

emf=-(1.00T)\frac{(0.113m^2)-(0.214m^2)}{4s-0s}=0.0252V=25.25mV

The induced current has counterclockwise direction, because the induced magneitc field generated by the induced current must be opposite to the constant magnetic field B.

4 0
3 years ago
A point charge Q is placed at the center of a conducting spherical shell (inner radius a, outer radius b). What is the electric
sp2606 [1]

Answer:

a)E=\dfrac{Q}{\varepsilon _o\times 4\pi r^2}\ N/C

b)E=0

c)E=\dfrac{Q}{\varepsilon _o\times 4\pi r^2}\ N/C

Explanation:

Given that

A point charge Q is placed at the center of a conducting spherical shell .Due to this - Q charge will induce on the inner sphere surface and +Q will induce on the outer sphere surface .

a)    r < a

At a radius r ,from gauss theorem

E.ds=\dfrac{q_i}{\varepsilon _o}

E\times 4\pi r^2=\dfrac{Q}{\varepsilon _o}

E=\dfrac{Q}{\varepsilon _o\times 4\pi r^2}\ N/C

b)   a < r < b

E.ds=\dfrac{q_i}{\varepsilon _o}

The total induce in this surface = - Q+ Q =0

E.ds=\dfrac{0}{\varepsilon _o}

E = 0

c)   r > b

E.ds=\dfrac{q_i}{\varepsilon _o}

E\times 4\pi r^2=\dfrac{Q}{\varepsilon _o}

E=\dfrac{Q}{\varepsilon _o\times 4\pi r^2}\ N/C

4 0
3 years ago
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