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3 years ago

Until a train is a safe distance from the station, it must travel at 5 m/s. Once ti

Physics

1 answer:

LuckyWell [14K]3 years ago

3 0

Answer:

Explanation:

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A conveyor belt is used to move sand from one place to another in a factory. The conveyor is tilted at an angle of 18° above the

brilliants [131]

Answer:

x = 2.044 m

Explanation:

given data

initial vertical component of velocity = Vy = 2sin18

initial horizontal component of velocity = Vx = 2cos18

distance from the ground yo = 5m

ground distance y = 0

from equation of motion

$y = yo+ V_y t +\frac{1}{2}gt^2$

$0 = 5 + 2sin18+ \frac{1}{2}*9.8t^2$

solving for t

t = 1.075 sec

for horizontal motion

$x = V_x t$

x = 2cos18*1.075

x = 2.044 m

8 0

3 years ago

A liquid of density 830 kg/m3 flows through a horizontal pipe that has a cross-sectional area of 1.20 x 10-2 m2 in region A and

kicyunya [14]

Answer:

A) volume flow rate = 0.047 m3/s

B) mass flow rate = 39.01 kg/s

Explanation:

Detailed explanation and calculation is shown in the image below

3 0

2 years ago

Un móvil se encuentra en la posición inicial x0 = 22 m, y se mueve con velocidad 5 m/s. Calcula la posición en la que se encuent

Brilliant_brown [7]

Answer:

La posición en la que se encuentra el móvil en el instante t = 30 s es 172 m.

Explanation:

El movimiento rectilíneo uniforme (MRU) es el movimiento que describe un cuerpo o partícula a través de una línea recta a velocidad constante.

La distancia recorrida, x , por un móvil que tiene un MRU con un velocidad v durante el intervalo de tiempo t es:

x= x0 + v*t

donde x0 es la posición inicial.

En este caso:

x0= 22 m
v= 5 m/s
t= 30 s

Reemplazando:

x= 22 m + 5 m/s* 30 s

Resolviendo:

x= 22 m + 150 m

x= 172 m

La posición en la que se encuentra el móvil en el instante t = 30 s es 172 m.

8 0

2 years ago

Whats 6 3/7 ×1 5/9.

aniked [119]

6 3/7 * 1 5/9
45/7 * 14/9
630/63
10

7 0

3 years ago

Read 2 more answers

a 2100-kg car drives with a speed of 18 m/s onb a flat road around a curve that has a radius of curvature of 83m. The coefficien

Law Incorporation [45]

Answer:

The magnitude of the friction force is 8197.60 N

Explanation:

Using the definition of the centripetal force we have:

$\Sigma F=ma_{c}=m\frac{v^{2}}{R}$

Where:

m is the mass of the car
v is the speed
R is the radius of the curvature

Now, the force acting in the motion is just the friction force, so we have:

$F_{f}=m\frac{v^{2}}{R}$

$F_{f}=2100\frac{18^{2}}{83}$

$F_{f}=8197.60 \: N$

Therefore the magnitude of the friction force is 8197.60 N

I hope it helps you!

7 0

3 years ago