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3 years ago

6

If a ball is thrown vertically upward from the roof of "64 foot building" with a velocity of 80 ft/sec, its height after tt seco

nds is s(t)=64+80t−16t^2. what is the maximum height the ball reaches

1 answer:

daser333 [38]3 years ago

8 0

The maximum height the ball reaches is 100ft.

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An object is given an initial velocity. What will happen to the object if no other forces act on it?

Phoenix [80]

Answer:

The object will travel at a constant rate in along a straight line.

Explanation:

In the given situation, it is mentioned that there is no external force acting on the given object. Thus, it will retain its initial velocity along a straight path.

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3 years ago

Question 12 (1 point) Question 12 Unsaved

Bezzdna [24]

Slope is your answer

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3 years ago

Explain how atomic mass and molecular mass are determined

GaryK [48]

Molecular mass may be calculated by taking the atomic mass of each element present and multiplying it by the number of atoms of that element in the molecular formula. Then, the number of atoms of each element is added together. This value may be reported as a decimal number or as 16.043 Da or 16.043 amu.

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3 years ago

A 51.0 kg crate, starting from rest, is pulled across a floor with a constant horizontal force of 225 N. For the first 10.0 m th

shepuryov [24]

Answer:

The final speed of the crate is 12.07 m/s.

Explanation:

For the first 10.0 meters, the only force acting on the crate is 225 N, so we can calculate the acceleration as follows:

$F = ma$

$a = \frac{F}{m} = \frac{225 N}{51.0 kg} = 4.41 m/s^{2}$

Now, we can calculate the final speed of the crate at the end of 10.0 m:

$v_{f}^{2} = v_{0}^{2} + 2ad_{1}$

$v_{f} = \sqrt{0 + 2*4.41 m/s^{2}*10.0 m} = 9.39 m/s$

For the next 10.5 meters we have frictional force:

$F - F_{\mu} = ma$

$F - \mu mg = ma$

So, the acceleration is:

$a = \frac{F - \mu mg}{m} = \frac{225 N - 0.17*51.0 kg*9.81 m/s^{2}}{51.0 kg} = 2.74 m/s^{2}$

The final speed of the crate at the end of 10.0 m will be the initial speed of the following 10.5 meters, so:

$v_{f}^{2} = v_{0}^{2} + 2ad_{2}$

$v_{f} = \sqrt{(9.39 m/s)^{2} + 2*2.74 m/s^{2}*10.5 m} = 12.07 m/s$

Therefore, the final speed of the crate after being pulled these 20.5 meters is 12.07 m/s.

I hope it helps you!

7 0

3 years ago

Tina is driving her sports car down US1, traveling 27m/s. She sees her friend Rita up ahead, waiting to cross the street at the

mrs_skeptik [129]

Answer:

434 Hz

Explanation:

According to the Doppler effect, when a source of a wave is moving towards an observer at rest, then the observer will observe an apparent frequency which is higher than the original frequency of the source.

In this situation, Tina is driving towards Rita. Tina is the source of the sound wave (the horn), while RIta is the observer. Since the original frequency of the sound is 400 Hz, Rita will hear a sound with a frequency higher than this value.

The only choice which is higher than 400 Hz is 434 Hz, so this is the frequency that Rita will hear.

3 0

3 years ago