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3 years ago

5

What did the protoplanets become?

2 answers:

alexira [117]3 years ago

7 0

What did the protoplanets become?

a. nebulae
b. planets
c. solar nebulae
d. planetesimals

The protoplanets become nebulae. The answer is letter A. The rest of the choices do not answer the question above.

nataly862011 [7]3 years ago

7 0

The answer is B. Planets

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two students are on a balcony 19.6 m above the street. one student throws a ball vertically downward at 14.7 m:ds. at the same i

NARA [144]

A. The difference in the two ball's time in the air is 3 seconds

B. The velocity of each ball as it strikes the ground is 24.5 m/s

C. The balls 0.500 s after they are thrown are 14.7 m apart

<h3>Further explanation</h3>

Acceleration is rate of change of velocity.

$\large {\boxed {a = \frac{v - u}{t} } }$

$\large {\boxed {d = \frac{v + u}{2}~t } }$

<em>a = acceleration ( m/s² )</em>

<em>v = final velocity ( m/s )</em>

<em>u = initial velocity ( m/s )</em>

<em>t = time taken ( s )</em>

<em>d = distance ( m )</em>

Let us now tackle the problem!

<u>Given:</u>

Initial Height = H = 19.6 m

Initial Velocity = u = 14.7 m/s

<u>Unknown:</u>

A. Δt = ?

B. v = ?

C. Δh = ?

<u>Solution:</u>

<h2>Question A:</h2><h3>First Ball</h3>

$h = H - ut - \frac{1}{2}gt^2$

$0 = 19.6 - 14.7t - \frac{1}{2}(9.8)t^2$

$0 = 19.6 - 14.7t - 4.9t^2$

$4.9t^2 + 14.7t - 19.6 = 0$

$t^2 + 3t - 4 = 0$

$(t + 4)(t - 1) = 0$

$(t - 1) = 0$

$\boxed {t = 1 ~ second}$

<h3>Second Ball</h3>

$h = H + ut - \frac{1}{2}gt^2$

$0 = 19.6 + 14.7t - \frac{1}{2}(9.8)t^2$

$0 = 19.6 + 14.7t - 4.9t^2$

$4.9t^2 - 14.7t - 19.6 = 0$

$t^2 - 3t - 4 = 0$

$(t - 4)(t + 1) = 0$

$(t - 4) = 0$

$\boxed {t = 4 ~ seconds}$

The difference in the two ball's time in the air is:

$\Delta t = 4 ~ seconds - 1 ~ second$

$\large {\boxed {\Delta t = 3 ~ seconds} }$

<h2>Question B:</h2><h3>First Ball</h3>

$v^2 = u^2 - 2gH$

$v^2 = (-14.7)^2 + 2(-9.8)(-19.6)$

$v^2 = 600.25$

$v = \sqrt {600.25}$

$\boxed {v = 24.5 ~ m/s}$

<h3>Second Ball</h3>

$v^2 = u^2 - 2gH$

$v^2 = (14.7)^2 + 2(-9.8)(-19.6)$

$v^2 = 600.25$

$v = \sqrt {600.25}$

$\boxed {v = 24.5 ~ m/s}$

The velocity of each ball as it strikes the ground is 24.5 m/s

<h2>Question C:</h2><h3>First Ball</h3>

$h = H - ut - \frac{1}{2}gt^2$

$h = 19.6 - 14.7(0.5) - \frac{1}{2}(9.8)(0.5)^2$

$\boxed {h = 11.025 ~ m}$

<h3>Second Ball</h3>

$h = H + ut - \frac{1}{2}gt^2$

$h = 19.6 + 14.7(0.5) - \frac{1}{2}(9.8)(0.5)^2$

$\boxed {h = 25.725 ~ m}$

The difference in the two ball's height after 0.500 s is:

$\Delta h = 25.725 ~ m - 11.025 ~ m$

$\large {\boxed {\Delta h = 14.7 ~ m} }$

<h3>Learn more</h3>

Velocity of Runner : brainly.com/question/3813437
Kinetic Energy : brainly.com/question/692781
Acceleration : brainly.com/question/2283922
The Speed of Car : brainly.com/question/568302

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Kinematics

Keywords: Velocity , Driver , Car , Deceleration , Acceleration , Obstacle

6 0

3 years ago

A rectangular certificate has a perimeter of 32 inches. Its area is 63 square inches. What are the dimensions of the certificate

irina1246 [14]

Perimeter = 2 ( L + W )
32 = 2 ( L + W )
16 = L + W
L = 16 - W

Area = L W
63 = L W
63 = (16-W) W
63 = 16W - W²
-W² + 16 W - 63 = 0
By factorizing W = 9 or W = 7
So the dimensions are 7 and 9

3 0

3 years ago

a car with a mass of 1200 kilograms is moving around a circular curve at a uniform velocity of 20 meters per second. the centrip

qaws [65]

Well, first of all, a car moving around a circular curve is not moving
with uniform velocity. The direction of motion is part of velocity, and
the direction is constantly changing on a curve.

The centripetal force that keeps an object moving in a circle is

   Force = (mass of the object) · (speed)² / (radius of the circle)

   F = m s² / r

We want to know the radius, to rearrange the formula to give us
the radius as a function of everything else.

      F   = m s² / r

Multiply each side by 'r': F· r = m · s²

Divide each side by 'F':    r = m · s² / F

We know all the numbers on the right side,
so we can pluggum in:

r =       m    · s² /    F

      r = (1200 kg) · (20 m/s)² / (6000 N) .

I'm pretty sure you can finish it up from here.

5 0

3 years ago

A student drops a rock from a bridge to the water 12.7 m below. with what speed does the rock strike the water? the acceleration

kolezko [41]

The rock strike the water with the speed of 15.78 m/sec.

The speed by which rock hit the water is calculated by the formula

v= $\sqrt{2gh}$

v= $\sqrt{2*9.8*12.7}$

v=15.78 m/sec

Hence, the rock strike the water with the speed of 15.78 m/sec.

7 0

3 years ago

A student gives a brief push to a block of dry ice. A moment later, the block moves across a very smooth surface at a constant s

Monica [59]

Hey There,

Question: "<span>A student gives a brief push to a block of dry ice. A moment later, the block moves across a very smooth surface at a constant speed. When drawing the free body diagram for the block of dry ice moving at a constant speed, the forces that should be included are: (select all that apply)"

Answer: C. Force Of Friction
B. Force

If This Helps May I Have Brainliest?</span>

7 0

3 years ago

Read 2 more answers