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Pie
3 years ago
15

In tae-kwon-do, a hand is slammed down onto a target at a speed of 10 m/s and comes to a stop during the 7.0 ms collision. Assum

e that during the impact the hand is independent of the arm and has a mass of 0.90 kg. What are the magnitudes of the (a) impulse and (b) average force on the hand from the target
Physics
1 answer:
wlad13 [49]3 years ago
5 0

Answer:

(A) Impulse = 9Ns

(B) F = 1286N

Explanation:

Impulse = change in momentum = m(v-u)

v = 0 (the hand comes to a stop)

u = -10m/s

Mass = 0.9kg

Impulse = 0.9 ×(0- (-10))

= 9Ns

(B) F×t = Impulse

F = Impulse/ t

t = 7ms = 7×10-³

F = 9/ (7×10-³)

F = 1286N.

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Which planet has the greatest mass?
Firlakuza [10]

Answer:

Jupiter

Explanation:

The largest planet in our solar system by far is Jupiter, which beats out all the other planets in both mass and volume. Jupiter's mass is more than 300 times that of Earth, and its diameter, at 140,000 km, is about 11 times Earth's diameter.

Note:

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5 0
3 years ago
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Use the information and diagram to answer the following question.
Mumz [18]

Answer:b

Explanation:

Because I don’t know

8 0
3 years ago
A .183 kg ball is moving 18.8 m/s when it runs into a spring of spring constant 86.9 N/m. How much KE does the ball have when it
Lemur [1.5K]

Answer:

The value is  KE_B  = 20.59 \  J

Explanation:

From the question we are told that

   The mass of the ball is  m  =  183 \  kg

   The initial  speed of the ball is  u  =  18.8 \  m/s

    The spring constant is  k  =  86.9 \ N/m

     The compression distance is  x =  0.520 \ m

Generally the energy stored in the string is mathematically represented as

        E =  \frac{1}{2}  *  k  * x^2

=>     E =  \frac{1}{2}  *  86.9  * 0.520 ^2

=>      E =   11.75 \  J

Generally the kinetic energy of the ball is mathematically represented as

          KE_b  =  \frac{1}{2} * m * u^2

=>      KE_b  =  \frac{1}{2}  *  0.183 * (18.8 )^2

        KE_b  = 32.34 \  J

Generally the KE   the ball have when it has compressed the spring is mathematically represented as

          KE_B  =  KE_b -  E

=>        KE_B  =  32.34 - 11.75

=>        KE_B  = 20.59 \  J

7 0
3 years ago
If a spring is stretched 4m from its starting length when 20n of force is applied, then how much work (in joules) is done by the
lys-0071 [83]

ANSWER:

250 J

STEP-BY-STEP EXPLANATION:

F = 20N is required to stretch the spring by 4 meters

We know that the force is equal to:

F=k\cdot x

We solve for k (spring constant):

k=\frac{F}{x}=\frac{20}{4}=5\text{ N/m}

The work done in stretching the spring is given by the following equation (in this case the stretch is 10 meters:

\begin{gathered} W=\frac{1}{2}k\cdot x^2 \\ \text{ Replacing} \\ W=\frac{1}{2}\cdot5\cdot10^2 \\ W=250\text{ J} \end{gathered}

The work required is 250 joules.

5 0
2 years ago
A kangaroo kicks downward with a 1000N force. According to Newton's Law the kangaroo is propelled into the air by:
bogdanovich [222]
<h3>Answer: B) his muscles</h3>

Explanation:

Specifically his leg muscles. As the leg muscles expand, they push down on the ground. Newton's 3rd law says that for any action, there's an opposite and equal reaction. That means a downward push into the ground will have the ground push back, more or less, and that's why the kangaroo will jump. The ground (and the earth entirely) being much more massive compared to the animal means that the ground doesn't move while the kangaroo does move. Perhaps on a very microscopic tiny level the ground/earth does move but it's so small that we practically consider it 0.

This experiment can be done with a wall as well. Go up to a wall and lean against it with your hands. Then do a pushup to move further away from the wall, but you don't necessarily need to lose contact with the wall's surface. As you push against the wall, the wall pushes back, and that causes you to move backward. If the wall was something flimsy like cardboard, then you could easily push the wall over and you wouldn't move back very much. It all depends how much mass is in the object you're pushing on.  

5 0
3 years ago
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