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wolverine [178]
3 years ago
10

Nick added five yards to the maximum distance that he can throw a football which component has he most likely added to his train

ing regimen​
Physics
1 answer:
aleksandr82 [10.1K]3 years ago
6 0

I believe the answer is: C.)weight lifting twice a week

Weight lifting increase the muscle composition in your body. When you have higher muscle composition, your body would be able to generate more power which might resulted an increase in throwing distance.

Running every day would only increase endurance and stretching is only benefited to increase flexibility and prevent injuries.

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A meter stick whose mass is 290 grams lies on ice. You pull at one end of the meter stick, at right angles to the stick, with a
katen-ka-za [31]

Answer

given,

mass of the stick = 290 grams = 0.29 Kg

Force on the stick on one side = F =  9 N

force acting perpendicular to stick.

magnitude of acceleration

rate of change of angular momentum is equal to Force

rate of change of angular momentum = 9 N

F = m a

a = \dfrac{F}{m}

a = \dfrac{9}{0.29}

a = 31.034 m/s²

Direction of motion will in the direction of force application or in the direction of change of velocity

5 0
4 years ago
Usain Bolt (m = 86.2 kg)
guapka [62]

Answer: 138 N

Explanation:

6 0
3 years ago
One end of a 34-m unstretchable rope is tied to a tree; the other end is tied to a car stuck in the mud. The motorist pulls side
anyanavicka [17]

Answer:

Fc = 89.67N

Explanation:

Since the rope is unstretchable, the total length will always be 34m.

From the attached diagram, you can see that we can calculate the new separation distance from the tree and the stucked car H as follows:

L1+L2=34m

L1^2=L2^2=L^2=2^2+(H/2)^2  Replacing this value in the previous equation:

\sqrt{2^2+H^2/4}+ \sqrt{2^2+H^2/4}=34  Solving for H:

H=\sqrt{52}

We can now, calculate the angle between L1 and the 2m segment:

\alpha = atan(\frac{H/2}{2})=60.98°

If we make a sum of forces in the midpoint of the rope we get:

-2*T*cos(\alpha ) + F = 0  where T is the tension on the rope and F is the exerted force of 87N.

Solving for T, we get the tension on the rope which is equal to the force exerted on the car:

T=Fc=\frac{F}{2*cos(\alpha) } = 89.67N

7 0
3 years ago
A Tale f Two Elephants
Murljashka [212]

Answer:

u need it now?

Explanation:

I'll answer it

3 0
4 years ago
After a 0.320-kg rubber ball is dropped from a height of 19.0 m, it bounces off a concrete floor and rebounds to a height of 15.
GarryVolchara [31]

Answer:

Imp = 11.666\,\frac{kg\cdot m}{s}

Explanation:

Speed experimented by the ball before and after collision are determined by using Principle of Energy Conservation:

Before collision:

(0.32\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot (19\,m) = \frac{1}{2}\cdot (0.320\,kg)\cdot v_{A}^{2}

v_{A} \approx 19.304\,\frac{m}{s}

After collision:

\frac{1}{2}\cdot (0.320\,kg)\cdot v_{B}^{2} = (0.32\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot (15\,m)

v_{B} \approx 17.153\,\frac{m}{s}

The magnitude of the impulse delivered to the ball by the floor is calculated by the Impulse Theorem:

Imp = (0.32\,kg)\cdot [(17.153\,\frac{m}{s} )-(-19.304\,\frac{m}{s} )]

Imp = 11.666\,\frac{kg\cdot m}{s}

5 0
3 years ago
Read 2 more answers
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