PHYSICS COLLEGE

Ball 1 is launched with an initial vertical velocity v1 = 146 ft/sec. Ball 2 is launched 2.3 seconds later with an initial vertical velocity v2. Determine v2 if the balls are to collide at an altitude of 234 ft. At the instant of collision, is ball 1 ascending or descending?

Answers

Answer 1

Answer:

Answer:

Initial velocity of ball 2 = 38.21 m/s = 125.36 ft/s

At the time of collision velocity of ball one is descending.

Explanation:

Velocity of ball 1 = 146 ft/sec = 44.50m/s

The balls are to collide at an altitude of 234 ft

H = 234 ft = 71.32 m

We have equation of motion

v² = u² + 2as

v² = 44.50² + 2 x (-9.81) x 71.32

v = ±24.10 m/s.

Time for each velocity can be calculated using equation of motion

v = u + at

24.10 = 44.50 - 9.81 t , t = 2.07 s

-24.10 = 44.50 - 9.81 t , t = 6.99 s

Since the second ball throws after 2.3 seconds we can avoid case with t = 2.07 s.

So at the time of collision velocity of ball one is descending.

The collision occurs at t = 6.99 s.

Time of flight of ball 2 = 6.99 - 2.3 = 4.69 seconds.

Height traveled by ball 2 = 71.32 m

We need to find velocity

We have

s = ut + 0.5 at²

71.32 = u x 4.69 - 0.5 x 9.81 x 4.69²

u = 38.21 m/s = 125.36 ft/s

Initial velocity of ball 2 = 38.21 m/s = 125.36 ft/s

Answer 2

Answer:

Answer:

v2=139 ft

Explanation:

First we just look at the motion of the first particle. It is moving vertically in a gravitational field so is decelerating with rate g = 9.81 m/s^2 = 32.18 ft/s^2. We can write it's vertical position as a function of time.

h_1=v_1*t-(a*t/2)

We set this equal to 234 ft to find when the body is passing that point, a solve the quadratic equation for t.

t_1,2=v_1±(√v_1^2-4*a/2*h_1)/a=2.57 s, 7.44 s

Since we know the second ball was launched after 2.3 seconds, we know that the time we are looking for is the second one, when the first ball is descending. The second ball will have 2.3 seconds less so the time we further use is t_c = 7.44 - 2.3 = 5.14 s. With this the speed of the second ball needed for collision at given height, can be found.

Solving a similar equation, but this time for v2 to obtain the result.

h_2=234 ft=v2*t_c-(a*t_c^2/2)--->v2=139 ft

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What best describes the bromide ion that forms

Answers

Answer:

it A

Explanation:

Its a negative ion that hss one less valence electron than a netural bromine atom

A spectroscope:measures light from distant objects
makes object far away look closer
receives radio signals from objects in space

Answers

Answer:

Option A

Measures light from distant objects

Explanation:

A spectroscope is used to measure the use of light from a distant object to work out the object is made of.

It could be the single-most powerful tool astronomers use.

Professor Fred Watson from the Australian Astronomical Observatory says that "It lets you see the chemicals being absorbed or emitted by the light source"

An indestructible bullet 2.00cm long is fired straight through a board that is 10.0cm thick. The bullet strikes the board with a speed of 470 m/s and emerges with a speed of 270 m/s. (To simplify, assume that the bullet accelerates only while the front tip is in contact with the wood.) a). What is the average acceleration of the bullet through the board? ________m/s^2b). What is the total time that the bullet is in contact with the board? (Enter the total time for the bullet to completely emerge from the board.) _________s
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Answers

Answer:

a) $a=-7.4* 10^(-5)\ m/s^2$

b) $t=0.27* 10^(-3)\ s$

c)s=14.92 cm

Explanation:

Given that

u= 470 m/s

v = 270 m/s

s= 10 cm

We know that

$v^2=u^2+2as$

$270^2=470^2+2* a* 0.1$

$a=-7.4* 10^(-5)\ m/s^2$

v= u + a t

$270=470-7.4* 10^(-5)* t$

$t=0.27* 10^(-3)\ s$

To stop the bullet it means that the final velocity will be zero.

$v^2=u^2+2as$

$0^2=470^2-2* 7.4* 10^(-5) * s$

s=14.92 cm

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Answers

Answer:

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Explanation:

Garza travels at a speed of 5 m/s. How long will it take him to travel 640 m?

Answers

Answer:

128 s

Explanation:

The distance, speed and time are related as;

$Distance=Speed* Time$

Given that the speed = 5 m/s

Distance = 640 m

Time = ?

So,

$Distance=Speed* Time$

$640\ m=5\ m/s* Time$

$Time=\frac {640\ m}{5\ m/s}=128\ s$

Thus, Garza takes 128 s to travel 640 m at 5 m/s speed.

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