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3 years ago

Paco was driving his scooter west with an initial velocity of 4 m/s. He accelerates at 0.5 m/s2 for 30 seconds.

Physics

2 answers:

KonstantinChe [14]3 years ago

6 0

Answer:

V = 19m/s

Explanation:

Given the following data;

Initial velocity, U = 4m/s

Acceleration, a = 0.5m/s²

Time, t = 30 seconds

To find the final velocity, we would use the first equation of motion;

V = U + at

Where;

V is the final velocity.

U is the initial velocity.

a is the acceleration.

t is the time measured in seconds.

V = 4 + 0.5*30

V = 4 + 15

V = 19m/s

Therefore, his final velocity is 19 meters per seconds.

Norma-Jean [14]3 years ago

4 0

Answer:

its C)19

Explanation:

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Vince weighs 160 pounds, and his friend Nadir weighs 140 pounds. Nadir calculated that his weight on another planet would be abo

prisoha [69]

Answer:

C) 64lb

Explanation:

use the linearity method to find the weight of nadir on another planet, it is applied as follows.

Nadir Weight in earth ⇒ Nadir weight in another planet

Vince Weigh in eart ⇒ X

our goal is to find the weight of vince in another planet (X), for this we multiply the diagonal that continents the data and divide among the remaining

140pounds ⇒ 56lb

160pounds ⇒ X

X= $\frac{(160)(56)}{140} =64lb$

Vince weigh on the other planet is C) 64lb

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3 years ago

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Two cars start from rest at a red stop light. When the light turns green, both cars accelerate forward. The blue car accelerates

saul85 [17]

First, create an illustration of the motion of the two cars as shown in the attached picture. The essential equations used is

For constant acceleration:
a = v,final - v,initial /t

The solutions is as follows:

a = v,final - v,initial /t
3.8 = (v - 0)/2.8 s
v = 10.64 m/s
After 2.8 seconds, the speed of the blue car is 10.64 m/s.

4 0

3 years ago

An 8-kg mass is in free fall. What is the velocity of the mass after 9 seconds

zimovet [89]

The velocity of the mass after 9 second is 88 m/s

7 0

3 years ago

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Brainliest please help tell me if am right if not correct me please

REY [17]

Answer:

See the answers below

Explanation:

To solve this problem we must use the following equation of kinematics.

$v_{f}=v_{o}+a*t$

where:

Vf = final velocity [m/s]

Vo = initial velocity [m/s]

a = acceleration [m/s²]

t = time [s]

First case

Vf = 6 [m/s]

Vo = 2 [m/s]

t = 2 [s]

$6=2+a*2\\4=2*a\\a=2[m/s^{2} ]$

Second case

Vf = 25 [m/s]

Vo = 5 [m/s]

a = 2 [m/s²]

$25=5+2*t\\t = 10 [s]$

Third case

Vo =4 [m/s]

a = 10 [m/s²]

t = 2 [s]

$v_{f}=4+10*2\\v_{f}=24 [m/s]$

Fourth Case

Vf = final velocity [m/s]

Vo = initial velocity [m/s]

a = acceleration [m/s²]

t = time [s]

$v_{f}=5+8*10\\v_{f}=85 [m/s]$

Fifth case

Vf = final velocity [m/s]

Vo = initial velocity [m/s]

a = acceleration [m/s²]

t = time [s]

$8=v_{o}+4*2\\v_{0}=8-8\\v_{o}=0$

8 0

3 years ago

When a rocket is 4 kilometers high, it is moving vertically upward at a speed of 400 kilometers per hour. At that instant, how f

Y_Kistochka [10]

Answer:

The angle of elevation of the rocket is increasing at a rate of 48.780º per second.

Explanation:

Geometrically speaking, the distance between the rocket and the observer ( $r$ ), measured in kilometers, can be represented by a right triangle:

$r = \sqrt{x^{2}+y^{2}}$ (1)

Where:

$x$ - Horizontal distance between the rocket and the observer, measured in kilometers.

$y$ - Vertical distance between the rocket and the observer, measured in kilometers.

The angle of elevation of the rocket ( $\theta$ ), measured in sexagesimal degrees, is defined by the following trigonometric relation:

$\tan \theta = \frac{y}{x}$ (2)

If we know that $x = 5\,km$ , then the expression is:

$\tan \theta = \frac{y}{5}$

And the rate of change of this angle is determined by derivatives:

$\sec^{2}\theta \cdot \dot \theta = \frac{1}{5}\cdot \dot y$

$\frac{\dot \theta}{\cos^{2}\theta} = \frac{\dot y}{5}$

$\frac{\dot \theta\cdot (25+y^{2})}{25} = \frac{\dot y}{5}$

$\dot \theta = \frac{5\cdot \dot y}{25+y^{2}}$

Where:

$\dot \theta$ - Rate of change of the angle of elevation, measured in sexagesimal degrees.

$\dot y$ - Vertical speed of the rocket, measured in kilometers per hour.

If we know that $y = 4\,km$ and $\dot y = 400\,\frac{km}{h}$ , then the rate of change of the angle of elevation is:

$\dot \theta = 48.780\,\frac{\circ}{s}$

The angle of elevation of the rocket is increasing at a rate of 48.780º per second.

3 0

3 years ago