All categories

3 years ago

A hydraulic system is used to lift a 20,000 N vehicle in an auto garage. If the vehicle sits on

Physics

1 answer:

ollegr [7]3 years ago

7 0

Answer:

Explanation:

Area covered by the vehicle on the piston A1 = 0.5 m2 Weight of the vehicle, F1 = 2000 kg × 9.8 m s-2 Area on which force F2 is applied, A = 0.03 m2 P1 = P2 ; F 1 A 1 F1A1 = F 2 A 2 F2A2 and F2 = F 1 A 1 F1A1 A2 ; F2 = (2000 × 9.8) 0.03 0.5 0.030.5 = 1176 NRead more on Sarthaks.com - https://www.sarthaks.com/957542/hydraulic-system-used-lift-2000-vehicle-in-an-auto-garage-if-the-vehicle-sits-on-piston-area

You might be interested in

Desde una altura de 120 m se deja caer un cuerpo. Calcular a los 2,5 s i) la rapidez que lleva; ii) cuánto ha descendido; iii) c

stira [4]

Answer:

i) 24.5 m/s

ii) 30,656 m

iii) 89,344 m

Explanation:

Desde una altura de 120 m se deja caer un cuerpo. Calcule a 2.5 s i) la velocidad que toma; ii) cuánto ha disminuido; iii) cuánto queda por hacer

i) Los parámetros dados son;

Altura inicial, s = 120 m

El tiempo en caída libre = 2.5 s

De la ecuación de caída libre, tenemos;

v = u + gt

Dónde:

u = Velocidad inicial = 0 m / s

g = Aceleración debida a la gravedad = 9.81 m / s²

t = Tiempo de caída libre = 2.5 s

Por lo tanto;

v = 0 + 9.8 × 2.5 = 24.5 m / s

ii) El nivel que el cuerpo ha alcanzado en 2.5 segundos está dado por la relación

s = u · t + 1/2 · g · t²

= 0 × 2.5 + 1/2 × 9.81 × 2.5² = 30.656 m

iii) La altura restante = 120 - 30.656 = 89.344 m.

6 0

3 years ago

A 55.0-kg skydiver drew falls for a period of time before opening his parachute. what is his kinetic energy when he reaches a ve

Jlenok [28]

Mass (m)=55kg

acceleration (a)=9.81 m/s^2, this is the acceleration due to gravity.

initial velocity=0m/s. The skydiver doesn’t start with any speed because she is on the plane or helicopter.

final velocity=16m/s This is the velocity (speed) the skydiver reaches

The equation we use is KE=.5mv^2
Kinetic energy=.5 mass x velocity^2

KE=.5(55kg)(16m/s)^2
KE=.5(55kg)(256m/s)
KE=.5(14080J)

J=Joules

KE=7040J

Kinetic energy is 7040 Joules (J)

Hope this helps

3 0

4 years ago

A rocket sled accelerates from rest for a distance of 645 m at 16.0 m/s2. A parachute is then used to slow it down to a stop. If

inessss [21]

Answer:

the stopping distance is greater than the free length of the track, the vehicle leaves the track before it can brake

Explanation:

This problem can be solved using the kinematics relations, let's start by finding the final velocity of the acceleration period

v² = v₀² + 2 a₁ x

indicate that the initial velocity is zero

v² = 2 a₁ x

let's calculate

v = $\sqrt {2 \ 15.0 \ 645}$

v = 143.666 m / s

now for the second interval let's find the distance it takes to stop

v₂² = v² - 2 a₂ x₂

in this part the final velocity is zero (v₂ = 0)

0 = v² - 2 a₂ x₂

x₂ = v² / 2a₂

let's calculate

x₂ = $\frac{ 143.666^2 }{2 \ 18.2}$

x₂ = 573 m

as the stopping distance is greater than the free length of the track, the vehicle leaves the track before it can brake

3 0

3 years ago

plz i need help fast................a motion along a straight line with a constant increase in velocity is ?

Monica [59]

Answer:

uniform acceleration

Explanation:

The definition for uniform acceleration is:

if an object travels in a straight line and its velocity increases or decreases by equal amounts in equal intervals of time, then the acceleration is said to be uniform.

Hope this helps.

Good Luck

8 0

4 years ago

Read 2 more answers

1. An object (m = 500 g) with an initial speed of 0.2 m/s collides with another object (m = 1.5 kg) which was at rest before the

Anettt [7]

Answer:

Explanation:

1 )

We shall apply conservation of momentum law to solve the problem.

mv = ( M +m) V , m and M are masses of small and large object , v is the velocity of small object before collision and V is the velocity of both the objects together after collision .

.5 x .2 = (1.5 + .5)V

V = .05 m /s

2 ) We shall use formula for velocity of object after elastic collision as follows

v₁ = $\frac{(m_1-m_2)}{(m_1+m_2)} u_1+\frac{2m_2u_2}{(m_1+m_2)}$

m₁ and m₂ are masses of first and second object u₁ and u₂ are their initial velocity and v₁ and v₂ are their final velocity.

Putting the values

= $\frac{(200-1000)}{(1000+200)} \times 1 +\frac{2 \times1000\times0}{(1000+200)}$

= - .66 m /s

Since the sign is negative so it will be in opposite direction .

4 0

3 years ago