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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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Which is an example of conduction?
densk [106]

Answer:

Touching the stove and getting burned, ice melting in your hand

Explanation:

7 0
3 years ago
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The theoretical line perpindicular to the surface where a light ray hits a mirror is called the
valkas [14]
That's called the "normal" to the surface at that point.
3 0
3 years ago
A commuter backs her car out of her garage with an acceleration of 1.40 m/s^2. (a) How long does it take her to reach a speed of
jonny [76]

Answer:

a) It takes her 1.43 s to reach a speed of 2.00 m/s.

b) Her deceleration is - 2.50 m/s²

Explanation:

The equation of velocity for an object that moves in a straight line with constant acceleration is as follows:

v = v0 + a · t

Where:

v = velocty.

v0 = initial velocity.

a = acceleration.

t = time.

a) Using the equation of velocity, let´s consider that the car moves in the positive direction. Then:

v = v0 + a · t

2.00 m/s = 0 m/s + 1.40 m/s² · t

t = 2.00 m/s / 1.40 m/s²

t = 1.43 s

It takes her 1.43 s to reach a speed of 2.00 m/s

b) Let´s use again the equation of velocity, knowing that at t = 0.800 s the velocity is 0 m/s:

v = v0 + a · t

0 = 2.00 m/s + a · 0.800 s

-2.00 m/s / 0.800 s = a

a = -2.50 m/s²

Her deceleration is - 2.50 m/s²

7 0
3 years ago
Please help !!!! I’m taking a quiz
icang [17]

Answer: 512 and 16

Explanation:

7 0
3 years ago
A volcano erupts and launches a chunk of hot magma horizontally with a speed of 252 m/s. The magma travels a horizontal distance
ArbitrLikvidat [17]

Answer:

The value is v_y  =  -48.61 \ m/s

Explanation:

From the question we are told that

   The horizontal speed is  u_x  = 252 \  m/s

    The horizontal distance is  d = 1250 \ m

Generally the time taken by the hot magma in air before landing is mathematically represented as

       t = \frac{d}{u_x}

=>    t = \frac{ 1250 }{252}

=>    t = 4.96 \  s

Generally the initial vertical velocity of the magma when it was lunched is  

    u_y = 0 \ m/ s

Then the final velocity of the magma when it hits the ground is mathematically represented s

       - v_y  =  u_y + gt

Here the negative sign mean that the direction of the velocity is towards the negative y -axis

So  

        - v_y  =  48.61 \ m/s

=>     v_y  =  -48.61 \ m/s

7 0
3 years ago
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