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Alborosie
3 years ago
13

Please help.Just True or False​

Physics
1 answer:
True [87]3 years ago
6 0

Answer:

1. T

2. F

3.F

4.T

thank you

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A ball of mass m is thrown straight upward from ground level at speed v0. At the same instant, at a distance D above the ground,
n200080 [17]

Answer:

a. t = \frac{v_{0}  +/- \sqrt{v_{0} ^{2} - gD} }{g}  b. D = v₀²/2g

Explanation:

Here is the complete question

A ball is thrown straight up from the ground with speed v₀ . At the same instant, a second ball is dropped from rest from a height D , directly above the point where the first ball was thrown upward. There is no air resistance

Find the time at which the two balls collide.

Express your answer in terms of the variables D ,v₀ , and appropriate constants..

t = ?!

Part B

Find the value of D in terms of v₀ and g so that at the instant when the balls collide, the first ball is at the highest point of its motion.

Express your answer in terms of the variables v₀ and g .

D =?!

Solution

The distance moved by the ball dropped from distance,D with velocity v₀, H₁ = D - (v₀t - gt²/2) = D + v₀t + gt²/2.

The distance moved by the ball thrown straight upward with velocity v₀ is H₂ = v₀t - gt²/2.

The two balls collide when their vertical distances are equal. That is H₁ = H₂

So, D - v₀t + gt²/2 = v₀t - gt²/2

Collecting like terms

D + gt²/2 + gt²/2 = v₀t + v₀t

D +gt² = 2v₀t

gt² - 2v₀t + D = 0.

Using the quadratic formula,

t = \frac{-(-2v_{0} ) +/- \sqrt{(-2v_{0} )^{2} - 4 X g XD} }{2g} = \frac{2v_{0}  +/- \sqrt{4v_{0} ^{2} - 4gD} }{2g} = \frac{v_{0}  +/- \sqrt{v_{0} ^{2} - gD} }{g}

B. At its highest point, the velocity of the first ball, v = 0. Using v² = u² - 2gs where s = highest point of first ball when they collide and u = v₀.

0 = v₀² - 2gs

s = v₀²/2g.

Also, the time it takes the first ball to reach its highest point is gotten from v = u - gt. At highest point, v = 0 and u = v₀. So,

 0 = v₀ - gt₀

t₀ = v₀/g

Also H = s₁ + s where s₁  = distance moved by second ball in time t₀ for collision = v₀t₀ - gt₀²/2.

So, H = v₀t₀ - gt₀²/2 + v₀²/2g = v₀(v₀/g) - g(v₀/g)²/2 + v₀²/2g = v₀²/2g - v₀²/2g + v₀²/2g = v₀²/2g

6 0
3 years ago
4. A desk with a mass of 10 kg is pushed across a floor with an acceleration of 4 m/s?. What
tatiyna
40 N. If you’re using F=MA MxA would be 10x4 which would get you 40 N
6 0
2 years ago
Read 2 more answers
A slanted vector has a magnitude of 41 N and is at an angle of 23 degrees north of east. What are the magnitude and direction of
jenyasd209 [6]

Answer:

The horizontal component is 37.74 N to east

The vertical component is 16.02 N to North

Explanation:

Any slant vector has two components

→ Horizontal component = R cos Ф

→ Vertical component = R sin Ф

→ R is the magnitude of the vector

→ Ф is the direction of the vector with positive part of the horizontal axis

A slanted vector has a magnitude of 41 N and is at an angle of 23

degrees north of east

23° north of east means the angle between the vector and the

east direction is 23° (east is the positive horizontal direction)

That means R is 41 N and Ф is 23°

→ R = 41 N , Ф = 23°

→ The horizontal component = 41 × cos(23) = 37.74 N east

→ The vertical component = 41 × sin(23) = 16.02 N North

<em>The horizontal component is 37.74 N to east</em>

<em>The vertical component is 16.02 N to North</em>

7 0
3 years ago
When a liquid is cooled, the kinetic energy of the particles . The force of attraction between the particles , the space between
anzhelika [568]

Answer:

For the first blank, the answer is decreases. For the second blank, the answer is increases. And finally for the third blank, the answer is decreases.

Explanation:

For the first blank, the answer is decreases. For the second blank, the answer is increases. And finally for the third blank, the answer is decreases.

When a liquid is cooled, the kinetic energy of the particles decreases. The force of attraction between the particles increases, the space between the particles decreases, and the matter changes its state to solid.

4 0
3 years ago
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Brianna has difficulty focusing in school and has struggled to learn to read. How would this disorder be classified in the DSM-5
-BARSIC- [3]
D
Neurodevelopmental Disorder
8 0
3 years ago
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