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3 years ago

7

A pool ball moving 1.83 m/s strikes an identical ball at rest. Afterward, the first ball moves 1.15 m/s at a 23.3° angle. What i

s the x-component of the velocity of the second ball?

1 answer:

Oksi-84 [34.3K]3 years ago

6 0

Answer:

v_{1fy} = - 0.4549 m / s

Explanation:

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What is the net charge of a copper atom if it gains 2 electrons?

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If an atom gains electrons, it develops a negative charge equal to the number of electrons gained.
So the net charge on the copper atom which gained 2 electrons will be -2.

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Answer: make objective observations.

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An unmarked police car traveling a constant 95 km/h is passed by a speeder. Precisely 1.00 s after the speeder passes, the polic

Montano1993 [528]

Answer:

Speed of the speeder will be 28 m/sec

Explanation:

In first case police car is traveling with a speed of 90 km/hr

We can change 90 km/hr in m/sec

So $90km/hr=\frac{90\times 1000m}{3600sec}=25m/sec$

Car is traveling for 1 sec with a constant speed so distance traveled in 1 sec = 25×1 = 25 m

After that car is accelerating with $a=2m/sec^2$ for 7 sec

So distance traveled by car in these 7 sec

$S=ut+\frac{1}{2}at^2=25\times 7+\frac{1}{2}\times 2\times 7^2=224m$

So total distance traveled by police car = 224 m

This distance is also same for speeder

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3 years ago

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A 150 kg line backer sacks the 120 kg quarterback. With what force is the quarterback sacked if the line backer has an accelerat

Gekata [30.6K]

Answer:

The force required to move the quarterback with linebacker is <u>1215 N</u>

Explanation:

$\text { Mass of linebacker } \mathrm{m}_{2}=150 \mathrm{kg}$

$\text { Mass of quarterback } \mathrm{m}_{2}=120 \mathrm{kg}$

$\text { Moved at an acceleration }(a)=4.5 \mathrm{m} / \mathrm{s}^{2}$

Using Newton's second law, it is established that F = Ma

Where F is net force acting on the system, a is the acceleration and M is mass of the two object $\left(m_{1}+m_{2}\right)$

Now consider both $\mathrm{m}_{1} \text { and } \mathrm{m}_{2}$ as a system, so net force acting on the system is $\text { Force }=\left(m_{1}+m_{2}\right) a$

Substitute the given values in the above formula,

$\text { Force }=(150+120) \mathrm{kg} \times 4.5 \mathrm{m} / \mathrm{s}^{2}$

$\text { Force }=270 \mathrm{kg} \times 4.5 \mathrm{m} / \mathrm{s}^{2}$

Force = 1215 N

<u>1215 N </u>is the force required to move the quarterback with linebacker.

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3 years ago

Alex pushes on a 2.0 kg book, resulting in a net force of 6.0 N on the book.

Yakvenalex [24]

Answer:

<h2>3.0 m/s²</h2>

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

$a = \frac{f}{m} \\$

From the question we have

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