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Tomtit [17]
3 years ago
9

Make a free-body diagram of someone pushing a refrigerator that shows:

Physics
1 answer:
adoni [48]3 years ago
5 0

Answer:

head fah ueah ik

Explanation:

You might be interested in
Determine the launch speed of a horizontally launched projectile that lands 26.3m from the base of a 19.3m high cliff
mote1985 [20]

The launch speed of a horizontally launched projectile is 13.28 m/s.

<h3>Time of motion of the projectile</h3>

The time of motion is calculated as follows;

t = √(2h / g)

where;

  • h is height of the cliff
  • g is acceleration due to gravity

t = √(2 x 19.3 / 9.8)

t = 1.98 s

<h3>Launch speed of the projectile</h3>

X = Vxt

Vx = X/t

Vx = (26.3)/(1.98)

Vx = 13.28 m/s

Thus, the launch speed of a horizontally launched projectile is 13.28 m/s.

Learn more about horizontal speed here: brainly.com/question/24949996

#SPJ1

5 0
2 years ago
What is the difference between speed and velocity?
alekssr [168]
Speed does not include a direction, but velocity does.

       30 miles per hour north
and
       30 miles per hour west

Same speed.
Different velocity.
4 0
3 years ago
PLEASE HELP AS SOON AS POSSIBLE!!
Naily [24]

Answer:

Saved Energy ?

Explanation:

5 0
3 years ago
Which statement best describes how waves carry energy?
ratelena [41]

Correct answer choice is :


A) Energy moves from one place to another through the wave.


Explanation:


Waves can carry energy over a range without moving matter the complete distance. For instance, an ocean wave can move many kilometers without the water itself traveling many kilometers. The water moves up and down a motion recognized as a disturbance. A wave brings its energy without carrying matter. Waves are observed to travel through an ocean or lake, yet the water perpetually reverts to its rest state. Energy is carried through the medium, yet the water particles are not carried.


7 0
3 years ago
Read 2 more answers
Un Esquiador Inicia un Salto Horizontal con una velocidad Inicial de 25 m/s. La Altura al Final de la Rampa es de 80 m del punto
Evgesh-ka [11]

Responder:

Explicación:

Para que podamos calcular el tiempo que le tomará al esquiador permanecer en el aire, encontraremos el tiempo de vuelo como se muestra;

T = 2Usin theta / 2

theta = 90 grados

U = 25 m / s

T = 25sin90 / 2 (9,8)

T = 25 / 19,62

T = 1,27 segundos

Por lo tanto, los cielos usarán 1.27 segundos en el aire.

La distancia horizontal es el rango;

Rango R = U√2H / g

R = 25√2 (80) /9,8

R = 25√160 / 9,8

R = 25 * √16,326

R = 25 * 4.04

R = 101,02 m

Por tanto, la distancia horizontal recorrida por el esquiador es 101,02 m

5 0
3 years ago
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