Do Brain cells act and look like skin cells? Why or why not?

The destruction of the ozone layer primarily occurs in the mesosphere. thermosphere. stratosphere. troposphere.

Fofino [41]

I believe the answer would be Troposphere.

4 0

3 years ago

In Fig. 4-41, a ball is thrown up onto a roof, landing 4.00 s later at height h ???? 20.0 m above the release level. The ball’s

Yanka [14]

"Fig is attacted with answer"

Answer:

a) d = 33.72 m

b) $v_{i}$ = 26 m/s

c) β = 71.08°

Explanation:

a)

When an object is thrown into the air under the effect of the gravitational force, the movement of the projectile is observed. Then it can be considered as two separate motions, horizontal motion and vertical motion. Both motions are different, so that they can be handled independently.

Given data:

time = t = 4.00 s

Height = h = 20 m

Angle = θ = 60°

Horizontal distance = d = ?

Using 2nd equation of motion

$h = v_{y_{f}}t + \frac{1}{2}gt^{2}$

-20 = $v_{y_{f}}$ (4) + 0.5(-9.8)(4)²

$v_{y_{f}}$ (4) = 58.4

$v_{y_{f}}$ = 14.6 m/s

This is vertical component of velocity when the ball is on the roof. To calculate the Final velocity and horizontal component, we use

$v_{f}$ = $v_{y_{f}}$ / sinθ

$v_{f}$ = 14.6 / sin 60

$v_{f}$ = 16.86 m/s

$v_{x_{f}}$ = $v_{f}$ cosθ

$v_{x_{f}}$ = 16.86 cos 60

$v_{x_{f}}$ = 8.43 m/s

To calculate the horizontal distance

d = $v_{y_{f}}$ t

d = (8.43)(4)

d = 33.72 m

b)

We know the values of Landing angle, height of roof, time of flight. In part a, We calculate the landing velocity of the ball and also its horizontal and vertical component. As the ball followed the projectile path, and we know that in projectile motion the horizontal component of the velocity remain constant throughout his motion. So there is no acceleration along horizontal path.

So,

$v_{x_{f}}$ = $v_{x_{i}}$

but the vertical component of velocity vary with and there is an acceleration along vertical direction which is equal to gravitation acceleration g.

So,

g = ( $v_{y_{f}}$ - $v_{y_{i}}$ ) / t

9.8 = 14.6 - $v_{y_{i}}$ ) / 4

$v_{y_{i}}$ = 24.6 m/s

$v_{i}$ = $\sqrt{v_{x_{i}}^{2}+v_{y_{i}}^{2} }$

$v_{i}$ = $\sqrt{8.43^{2}+24.6^{2}}$

$v_{i}$ = 26 m/s

c)

cos β = $v_{x_{i}}$ / $v_{i}$

β = cos⁻¹ (8.43 / 26)

β = 71.08°

3 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

Please answer me my question

Angelina_Jolie [31]

3 - b) weight decreases
4 -a) 50 kg as mass is same everywhere
5 - b) 200 dynes

5 0

3 years ago

A 20-liter container contains 2.0 moles of oxygen at a pressure of 92 kpa. the average kinetic energy of translation of oxygen m

Vikentia [17]

From ideal gas law, PV=nRT

where P is the pressure, V is the volume of the container, n is number of moles, R is the gas constant and T is the temperature.

Hence, $T=\frac{PV}{nR}=\frac{92*10^{3}*2.0*10^{-3}}{2*3.314}$

T= 110.65 k

Kinetic Energy = $\frac{3}{2}KT=\frac{3}{2} (1.38*10^{-23})(110.65)$

K.E= $2.2*10^{-21}J$

<h3>What is a kinetic energy? </h3>

The energy an object has as a result of motion is known as kinetic energy.

A force must be applied to an object in order to accelerate it. We must put in effort in order to apply a force. After the work is finished, energy is transferred to the item, which then moves at a new, constant speed. Kinetic energy is the type of energy that is transferred and is dependent on the mass and speed attained.

Kinetic energy can be converted into other types of energy and transported between objects. A flying squirrel may run into a chipmunk that is standing still, for instance. Some of the squirrel's initial kinetic energy may have been transferred to the chipmunk or changed into another kind of energy after the collision.

To know more about kinetic energy, visit:

brainly.com/question/22174271

#SPJ4

4 0

1 year ago