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The clouds that occur at the highest altitude are usually

gogolik [260]

The answer is A. Cirrus clouds occur at the highest altitude.

5 0

2 years ago

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

A ball is dropped from a height of 20 meters. At what helght does the ball have a velocity of 10 meters/second?

Sunny_sXe [5.5K]

Answer:

5.10 meters.

Explanation:

v²=u²+2gh

or, (10)²=(0)²+2×9.8×h

or, 19.6h=100

or, h=5.10 meters

Hope, this helps you.

8 0

2 years ago

1. A mass suspended from a spring oscillates vertically with amplitude of 15 cm. At what distance from the equilibrium position

antiseptic1488 [7]

Answer:

The value of the distance is $\bf{14.52~cm}$ .

Explanation:

The velocity of a particle(v) executing SHM is

$v = \omega \sqrt{A^{2} - x^{2}}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~`~(1)$

where, $\omega$ is the angular frequency, $A$ is the amplitude of the oscillation and $x$ is the displacement of the particle at any instant of time.

The velocity of the particle will be maximum when the particle will cross its equilibrium position, i.e., $x = 0$ .

The maximum velocity( $\bf{v_{m}}$ ) is

$v_{m} = \omega A~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(2)$

Divide equation (1) by equation(2).

$\dfrac{v}{v_{m}} = \dfrac{\sqrt{A^{2} - x^{2}}}{A}~~~~~~~~~~~~~~~~~~~~~~~~~~~(3)$

Given, $v = 0.25 v_{m}$ and $A = 15~cm$ . Substitute these values in equation (3).

$&& \dfrac{1}{4} = \dfrac{\sqrt{15^{2} - x^{2}}}{15}\\&or,& A = 14.52~cm$

6 0

3 years ago

Define bond length using your own words

Brut [27]

Answer:

the free encyclopedia. In molecular geometry, bond length or bond distance is defined as the average distance between nuclei of two bonded atoms in a molecule. It is a transferable property of a bond between atoms of fixed types, relatively independent of the rest of the molecule.

Explanation:

6 0

2 years ago

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