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3 years ago

14

a body is moving with uniform acceleration, has initial velocity 45km/hr. and acceleration 20cm/s^2. find its velocity after 25

seconds

1 answer:

Eddi Din [679]3 years ago

7 0

That’s hard wow!!!!!!

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When one does twice the work in twice the time, the power expended is

devlian [24]

Answer:

the same.

hope this helps!

8 0

3 years ago

An aircraft, traveling northward, lands on a runway with a speed of 79 m/s. Once it touches down, it slows to 6.0 m/s over 785 m

Musya8 [376]

Answer:

$-3.95m/s^2$ , South

Explanation:

We are given that

Initial speed of aircraft,u=79 m/s

Final speed of aircraft,v=6 m/s

Distance,s=785 m

$v^2-u^2=2as$

Substitute the values

$(6)^2-(79)^2=2a(785)$

$36-6241=1570a$

$-6205=1570a$

$a=-\frac{6205}{1570}=-3.95m/s^2$

Where negative sign indicates that the direction of acceleration is opposite to direction of motion.

Therefore, the direction of acceleration is south.

5 0

4 years ago

Can someone help me with #7

bekas [8.4K]

The ratio of (new volume / old volume) is the same as the ratio of (new temperature / old temp).

Just remember that you have to use the ABSOLUTE temperatures.

An absolute temperature is (Celsius + 273).

7 0

4 years ago

Help i d u m b ples!!!!!!!!1111

andrew-mc [135]

Answer: b and c are the same so that leaves you with a and d I believe it's d not sure hope this helps ®:]©™

Explanation:

7 0

3 years ago

On average, both arms and hands together account for 13 % of a person's mass, while the head is 7.0% and the trunk and legs acco

lianna [129]

Answer:

$\dot n_{f} = 85.177\,rpm$

Explanation:

The expression for the moment of inertia of the person is:

Arms outstretched

$I = \frac{1}{12}\cdot (0.13)\cdot (61\,kg)\cdot (1.40\,m)^{2} + \frac{1}{2}\cdot (0.87)\cdot (61\,kg)\cdot (0.35\,m)^{2}$

$I = 4.546\,kg\cdot m^{2}$

Arms parallel to the trunk

$I = \frac{1}{2}\cdot (61\,kg)\cdot (0.35\,m)^{2}$

$I = 3.736\,kg\cdot m^{2}$

The final angular speed is found by means of the Principle of Angular Momentum Conservation:

$I_{o}\cdot \dot n_{o} = I_{f}\cdot \dot n_{f}$

$\dot n_{f} = \frac{I_{o}}{I_{f}}\cdot \dot n_{o}$

$\dot n_{f} = \left(\frac{4.546\,kg\cdot m^{2}}{3.736\,kg\cdot m^{2}}\right)\cdot (70\,rpm)$

$\dot n_{f} = 85.177\,rpm$

8 0

3 years ago