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Lunna [17]
3 years ago
9

A man (mass=68 kg) on a parachute is falling at terminal velocity (v=59 m/s)

Physics
1 answer:
raketka [301]3 years ago
3 0

Explanation:

It is given that,

Mass of the man, m = 68 kg

Terminal velocity of the man, v = 59 m/s

We need to find the rate at which the internal energy of the man and of the air around him increase. The gravitational potential energy of the man is given by :

E=mgh

Differentiating equation (1) wrt t as :

\dfrac{dE}{dt}=mg\dfrac{dh}{dt}

Since, v=\dfrac{dh}{dt}=59\ m/s

\dfrac{dE}{dt}=68\times 9.8\times 59

\dfrac{dE}{dt}=39317.6\ J/s

So, the internal energy of the man and the air around him is increasing at the rate of 39317.6 J/s. Hence, this is the required solution.

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What is endurance? a). the ability to run faster b). a combination of balance and coordination c). how much you can stretch d).
Alex787 [66]

Answer:

D. the ability to exercise for longer periods of time

Explanation:

For example, when someone does endurance training, they are stretching their body's ability to do a certain exercise for longer times as opposed to increasing strength.

8 0
3 years ago
Q. No. 9 A body falls freely from the top of a tower and during the last second of its fall, it falls through 25m. Find the heig
HACTEHA [7]

Answer:

45.6m

Explanation:

The equation for the position y of an object in free fall is:

y=-\frac{1}{2} gt^2+v_0t+y_0

With the given values in the question the equation has one unknown v₀:

v_0=\frac{y-y_0}{t}+\frac{1}{2}gt

Solving for t=1:

1) v_0=y-y_0+\frac{g}{2}

To find the hight of the tower you can use the concept of energy conservation:

The energy of the body 1 sec before it hits the ground:

2) E=\frac{1}{2}m{v_0}^2+mgy_0

If h is the height of the tower, the energy on top of the tower:

3) E=mgh

Combining equation 2 and 3 and solving for h:

4) h=\frac{{v_0}^2}{2g}+y_0

Combining equation 1 and 4:

h=\frac{{(y-y_0+\frac{g}{2}})^2}{2g}+y_0

4 0
3 years ago
A compass in a magnetic field will line up __________.
Anastaziya [24]
Line up in a direction parallel to the magnetic field lines<span />
5 0
3 years ago
Which of the following is an example of projectile
jonny [76]

Answer:

The answer is choice A.

Explanation:

Assuming you are in a situation with a gravitational field. You can divide the motion of the bullet into two components. One horizontal and the other in the vertical.

7 0
3 years ago
A weight of 30.0 N is suspended from a spring that has a force constant of 220 N/m. The system is undamped and is subjected to a
Nimfa-mama [501]

Answer:

F_0 = 393 N

Explanation:

As we know that amplitude of forced oscillation is given as

A = \frac{F_0}{ m(\omega^2 - \omega_0^2)}

here we know that natural frequency of the oscillation is given as

\omega_0 = \sqrt{\frac{k}{m}}

here mass of the object is given as

m = \frac{W}{g}

\omega_0 = \sqrt{\frac{220}{\frac{30}{9.81}}}

\omega_0 = 8.48 rad/s

angular frequency of applied force is given as

\omega = 2\pi f

\omega = 2\pi(10.5) = 65.97 rad/s

now we have

0.03 = \frac{F_0}{3.06(65.97^2 - 8.48^2)}

F_0 = 393 N

6 0
3 years ago
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