Item 13You have four $10 bills and eighteen $5 bills in your piggy bank. How much money do you have?

What force (in N) must be exerted on the master cylinder of a hydraulic lift to support the weight of a 2400 kg car (a large car

makvit [3.9K]

Answer:

Fm= 91.88 N

Explanation:

Pascal principle

The pressure acting on one side is transmitted to all the molecules of the liquid because the liquid is incompressible.

The pressure is definited like this:

P=F/A

Where:

P: Pressure in pascals (Pa)

F: Force acting in the area (N)

A : Area where the force acts (m²)

Pascal principle

Pm=Ps

Fm/ Am= Fs/ As Formula (1)

Where :

Pm : Pressure on the master piston

Ps : Pressure on the slave piston

Fm : Force on the master piston (N)

Fs: Force on the slave piston ((N)

Am: master piston area (m²)

As: slave piston area (m²)

Area Formula (A)

A= π*R²

R : piston radius

Calculation of the weight of the car (W)

W= m*g= 2400 kg*9.8m/s²= 23520 N

W = Fs

Data

Fs = 23520 N

Dm = 1.5 cm

Ds = 24 cm

Rm = 0.75 cm

Rs = 12 cm

Am = π*Rm² = π*(0.75)²

As = π*Rs² = π*(12)²

Force exerted on the master cylinder

We replace data in the formula (1)

$\frac{F_{m} }{A_{m} } = \frac{F_{s} }{A_{s} }$

$F_{m} = \frac{F_{s}*A_{m} }{A_{s}}$

$F_{m} = \frac{(23520 N)*(\pi *(0.75)^{2})(cm^{2})}{(\pi *(12)^{2})(cm^{2})}$

$F_{m} = (23520 N)*\frac{(0.75)^{2} }{(12)^{2} }$

Fm= 91.88 N

8 0

2 years ago

During a testing process, a worker in a factory mounts a bicycle wheel on a stationary stand and applies a tangential resistive

olganol [36]

Answer:

The force is $F_c = 789.03 \ N$

Explanation:

From the question we are told that

The tangential resistive force is $F_t = 115 \ N$

The mass of the wheel is m = 1.80 kg

The diameter of the wheel is $d = 50.0 cm = 0.5 \ m$

The diameter of the sprocket is $d_c = 8.50 \ cm =0.085 \ m$

The angular acceleration considered is $\alpha = 4.30\ rad/s^2$

Generally the radius of the wheel is

$r = \frac{d}{2}$

=> $r = \frac{0.5}{2}$

=> $r = 0.25 \ m$

Generally the radius of the sprocket is

$r_c = \frac{d_c}{2}$

=> $r_c = \frac{0.085}{2}$

=> $r_c = 0.0425 \ m$

Generally the moment of inertia of the wheel is mathematically represented as

$I = m * r^2$

=> $I = 1.80 * 0.25^2$

=> $I = 1.1125 \ kg \cdot m^2$

Generally the torque experienced by the wheel due to the forces acting on it is mathematically represented as

$\tau = F_c * r_c - F_t * r$

Here $F_c$ is the force acting on the sprocket

So

$\tau = F_c * 0.0425 - 115 * 0.25$

$\tau = 0.0425F_c - 28.75$

Generally the torques that will cause the wheel to move with $\alpha = 4.30\ rad/s^2$ is mathematically represented as

$\tau = I * \alpha$

So

$0.0425F_c - 28.75 = I * \alpha$

$0.0425F_c - 28.75 = 1.1125 *4.30$

$F_c = 789.03 \ N$

5 0

3 years ago

"On a movie set, an alien spacecraft is to be lifted to a height of 32.0 m for use in a scene. The 260.0-kg spacecraft is attach

Lisa [10]

Answer:

The time interval required to lift the spacecraft to this specified height is 123.94 seconds

Explanation:

Height through which the spacecraft is to be lifted = 32.0 m

Mass of the spacecraft = 260.0 kg

Four crew member each pull with a power of 135 W

18.0% of the mechanical energy is lost to friction.

work done in this situation is proportional to the mechanical energy used to move the spacecraft up

work done = (weight of spacecraft) x (the height through which it is lifted)

but the weight of spacecraft = mg

where m is the mass,

and g is acceleration due to gravity 9.81 m/s

weight of spacecraft = 260 x 9.81 = 2550.6 N

work done on the space craft = weight x height

==> work = 2550.6 x 32 = 81619.2 J

this is equal to the mechanical energy delivered to the system

18.0% of this mechanical energy delivered to the pulley is lost to friction.

this means that

0.18 x 81619.2 = 14691.456 J is lost to friction.

Total useful mechanical energy = 81619.2 J - 14691.456 J = 66927.74 J

Total power delivered by the crew to do this work = 135 x 4 = 540 W

But we know tat power is the rate at which work is done i.e

$P = \frac{w}{t}$

where p is the power

where w is the useful work done

t is the time taken to do this work

imputing values, we'll have

540 = 66927.74/t

t = 66927.74/540

time taken t = 123.94 seconds

8 0

3 years ago

A skateboader is accelerating forward how would his acceleration change if he had more mass

Masteriza [31]

Answer: If the force stays the same, the acceleration would decrease

6 0

2 years ago

A body having uniform velocity has zero acceleration? give reason

Sonja [21]

Answer:

A body having uniform velocity has zero acceleration because

there is not change in velocity.

5 0

3 years ago

Read 2 more answers