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3 years ago

PLEASE ANSWER ASAP!!! ITS DUE IN 10 MINUTES!!!!!!

Physics

1 answer:

forsale [732]3 years ago

5 0

Explanation:

2. Aerobic exercise. ...

Strength building. ...

Balance Training. ...

Endurance. ...

Flexibility. ...

Moderate intensity exercise. ...

Vigorous exercise.

3.Brisk walking.

Running.

Jogging

Bear crawls.

Swimming.

Water aerobics.

Cycling/bicycling.

4. Running or Jogging. ...

Outdoor Cycling. ...

Walking.

5. The type of exercise to build muscle is strength training, although cardiovascular activity can also provide benefits.

Hope it helps..

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A car (mass = 1200 kg) is traveling at 31.1 m/s when it collides head-on with a sport utility vehicle (mass = 2830 kg) traveling

IrinaVladis [17]

Answer:

13.18 m/s

Explanation:

Let the velocity of sports utility car is

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mc = 1200 kg, uc = 31.1 m/s

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Using conservation of momentum

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4 years ago

A 60 cm diameter wheel accelerates from rest at a rate of 7 rad/s2. After the wheel has undergone 14 rotations, what is the radi

Mamont248 [21]

Answer:

$a=368.97\ m/s^2$

Explanation:

Given that,

Initial angular velocity, $\omega=0$

Acceleration of the wheel, $\alpha =7\ rad/s^2$

Rotation, $\theta=14\ rotation=14\times 2\pi =87.96\ rad$

Let t is the time. Using second equation of kinematics can be calculated using time.

$\theta=\omega_it+\dfrac{1}{2}\alpha t^2\\\\t=\sqrt{\dfrac{2\theta}{\alpha }} \\\\t=\sqrt{\dfrac{2\times 87.96}{7}} \\\\t=5.01\ s$

Let $\omega_f$ is the final angular velocity and a is the radial component of acceleration.

$\omega_f=\omega_i+\alpha t\\\\\omega_f=0+7\times 5.01\\\\\omega_f=35.07\ rad/s$

Radial component of acceleration,

$a=\omega_f^2r\\\\a=(35.07)^2\times 0.3\\\\a=368.97\ m/s^2$

So, the required acceleration on the edge of the wheel is $368.97\ m/s^2$ .

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3 years ago

A bat strikes a 0.145-kg baseball. Just before impact, the ball is traveling horizontally to the right at 60.0 m/s , and it leav

nata0808 [166]

Answer:

F = -307.4 N

Explanation:

It is given that,

Mass of the baseball, m = 0.145 kg

Initial speed of the baseball, u = 60 m/s

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Time of contact, $t=1.75\ ms=1.75\times 10^{-3}\ s$

(a) It is assumed to find the horizontal component of average force. It is given by :

$F=m\dfrac{v-u}{t}$

$F=0.145\dfrac{56.29-60}{1.75\times 10^{-3}}$

F = -307.4 N

So, the horizontal component of average force is 307.4 N. Hence, this is the required solution.

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