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2 years ago

Define constant speed

Physics

2 answers:

mash [69]2 years ago

7 0

It is the speed at which a body moves without any interuptions (external forces etc)

docker41 [41]2 years ago

5 0

Constant speed is the speed in which stays the same speed over the time that it covers. It is a consistent speed that does not change over time.

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A person weighing 0.9 kN rides in an elevator that has a downward acceleration of 1.9 m/s^2. The acceleration of gravity is 9.8

Keith_Richards [23]

Answer:

The magnitude of the force is 0.7255kN

Explanation:

The elevator floor acts on the person with a force that is due to the gravitational acceleration less the downward acceleration of the elevator:

(force of floor F) = (mass of person m) x [ (grav. acceleration g) - (elevator acceleration a) ]

in other words, considering the elevator floor as a reference frame in the Earth's gravitational field, the person's weight decreases due to the downward acceleration, as follows:

$F = m\cdot(g-a)$

We are given the person's weight at rest, 0.9kN, from which the mass can be determined as:

$900 N = m\cdot g \implies m = \frac{900N}{9.8 \frac{m}{s^2}}$

$F = \frac{900N}{9.8 \frac{m}{s^2}}\cdot(9.8-1.9)\frac{m}{s^2}\approx 725.5N=0.7255kN$

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2 years ago

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Irina18 [472]

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2 years ago

Why does it takes a long time to heat a room with a high ceiling?

vlada-n [284]

Answer:

High ceilings make a room feel large and open, but they can be difficult to cool and heat. Because hot air rises, the challenge becomes trying to keep the hot air where you want it and preventing if from being wasted where you don't.

Explanation:

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2 years ago

Read 2 more answers

You can use any coordinate system you like in order to solve a projectile motion problem. To demonstrate the truth of this state

posledela

Answer:

a) y₂ = 49.1 m , t = 1.02 s , b) y = 49.1 m , t= 1.02 s

Explanation:

a) We will solve this problem with the missile launch kinematic equations, to find the maximum height, at this point the vertical speed is zero

$v_{y}$ ² = $v_{oy}$ ² - 2 g (y –yo)

The origin of the coordinate system is on the floor and the ball is thrown from a height

y-yo = $v_{oy}² /2 g
 y- 0 = 10.0²/2 9.8
 y - 0 = 5.10 m
 
The height from the ground is the height that rises from the reference system plus the depth of the ground from the reference system
 y₂ = 5.1 + 44
 y₂ = 49.1 m
Let's use the other equation to find the time
 [tex]v_{y}$ = $v_{oy}$ - g t

t = $v_{oy}$ / g

t = 10 / 9.8

t = 1.02 s

b) the maximum height

y- 44.0 = $v_{y}$ ² / 2 g

y - 44.0 = 5.1

y = 5.1 +44.0

y = 49.1 m

The time is the same because it does not depend on the initial height

t = 1.02 s

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3 years ago