Ask question

Ask question

All categories

3 years ago

6

.You have always been impressed by the speed of the elevators in your apartment building. You wonder about the maximum accelerat

ion for these elevators during normal operation, so you decide to measure it by using your bathroom scale. While the elevator is at rest on the ground floor,you get in, put down your scale, and stand on it. The scale reads 50 kg. You continue standing on the scale when the elevator goes up, carefully watching the reading. During the trip to the 10th floor, the greatest scale reading was

1 answer:

AlexFokin [52]3 years ago

3 0

Answer:

5.51 m/s^2

Explanation:

Initial scale reading = 50 kg

assume the greatest scale reading = 78.09 kg

<u>Determine the maximum acceleration for these elevators</u>

At rest the weight is = 50 kg

Weight ( F ) = mg = 50 * 9.81 = 490.5 N<u> </u>

<u> </u>At the 10th floor weight = 78.09 kg

Weight at 10th floor ( F ) = 78.09 * 9.81 = 766.11 N

F = change in weight

Change in weight( F ) = ma = 766.11 - 490.5 (we will take the mass as the starting mass as that mass is calculated when the body is at rest)

50 * a = 275.61

Hence the maximum acceleration ( a ) = 275.61 / 50 = 5.51 m/s^2

You might be interested in

One train (22,680 kg) moving east at 170 km/h collides with a car (1,200 kg) that was moving north at 5 km/h. They get attached

bezimeni [28]

The final velocity of the objects is 229.82km/hr

Since energy was lost after the collision, the type of collision that occurs is an elastic collision.

The required force is 1,090,918,000N

The amount of Kinetic energy lost is 12,401,976.656Joules

According to the law of collision'

m1u1 + m2u2 = (m1+m2)v

Given the following:

m1 = 22,680kg

u1 =170km/h

m2 = 1200kg

u2 = 5km/hr

Get the final velocity "v"

22680(170) + 1200(5) = (22680 + 1200)v

3855600 + 6000 = 12880v

3,861,600 = 12880v

v = 3,861,600/12,880

v = 229.82km/hr

Hence the final velocity of the objects is 229.82km/hr

Since energy was lost after the collision, the type of collision that occurs is an elastic collision.

According to Newton's second law, the formula for calculating the force is expressed as;

F = ma

F = m(v-u)/t

F = 22680+1200(229.82-175)/t

Ft = 23880(54.82)

F = 1,309,101.6/0.0012

F = 1,090,918,000N

Hence the required force is 1,090,918,000N

KE lost = Kinetic energy after collision - Kinetic energy before collision

Kinetic energy after collision = 1/2 * 12880 * 229.82²

Kinetic energy after collision = 340,142,976.656 Jolues

Kinetic energy before collision = 1/2 * 22680(170)²+ 1/2*1200(5)²

Kinetic energy before collision = 327,726,000 + 15000

Kinetic energy before collision = 327,741,000

Kinetic energy lost = 340,142,976.656 - 327,741,000 = 12,401,976.656Joules

Learn more here: brainly.com/question/9537044

8 0

2 years ago

Is it possible to produce a continuous and oriented aramid fiber–epoxy matrix composite having longitudinal and transverse modul

gregori [183]

Answer:

It is not possible to produce continued and oriented fiber.

Explanation:

To solve the problem it is necessary to take into account the concepts related to Fiber volume ratio. The amount of fiber in a fiber reinforced compound corresponds directly to the mechanical properties of the compound. Given the fiber volume fraction, the theoretical elastic properties of a compound can be determined. The elastic modulus of a compound in the fiber direction of a unidirectional compound can be calculated using the following equation:

$E = (1-V_f)E_m+V_fE_f$

Where,

E is the longitudinal modulus of Elasticity

$V_f$ is the fiber volume ratio

$E_m$ is the elastic modulus of the matrix

$E_f$ is the elastic modulus of the fibers

We need to consult the table of characteristics of Fibers and Reinforcements of Materials, in which they specify that the modulus of elasticity of the aramid fiber-epoxy is

$E_f = 131Gpa$

Moreover from the statement,

$E = 35Gpa$

$E_m = 3.4Gpa$

Replacing in the previous equation,

$35 = 3.4 (1-V_f)+131V_f$

$V_f = 0.25 \rightarrow longitudinal$

To make the comparison we now calculate the Fiber volume ratio through the transverse elastic modulus,

$E = \frac{E_mE_f}{(1-V_f)E_f+V_fE_f}$

Our values are given in this case as:

$E = 5.17Gpa\\E_m = 3.4Gpa \\E_f = 131Gpa$

Replacing,

$5.17 = \frac{3.4*131}{(1-V_f)(131)+V_f*3.4}$

$V_f = 0.351 \rightarrow transversal$

From both cases it is possible to conclude that it is not possible to produce a fiber of the specified material in a continuous and oriented manner, as long as the volume fraction is different in the different cases.

5 0

2 years ago

By what factor must we increase the amplitude of vibration of an object at the end of a spring in order to double its maximum sp

strojnjashka [21]

Answer:

$A'=2A$

Explanation:

According to the law of conservation of energy, the total energy of the system can be expresed as the sum of the potential energy and kinetic energy:

$E=U+K=\frac{kA^2}{2}\\E=\frac{kx^2}{2}+\frac{mv^2}{2}=\frac{kA^2}{2}$

When the spring is in its equilibrium position, that is $x=0$ , the object speed its maximum. So, we have:

$\frac{k(0)^2}{2}+\frac{mv_{max}^2}{2}=\frac{kA^2}{2}\\A^2=\frac{mv_{max}^2}{k}\\A=\sqrt{\frac{mv_{max}^2}{k}}$

In order to double its maximum speed, that is $v'{max}=2v_{max}$ . We have:

$A'=\sqrt{\frac{m(v'_{max})^2}{k}}\\A'=\sqrt{\frac{m(2v_{max})^2}{k}}\\A'=\sqrt{\frac{4mv_{max}^2}{k}}\\A'=2\sqrt{\frac{mv_{max}^2}{k}}\\A'=2A$

6 0

2 years ago

25. How long does it take a child on a swing to complete one swing if her center of gravity is 4.00 m below the pivot?

Lina20 [59]

Length=l=4m

$\\ \rm\rightarrowtail T=2\pi\sqrt{\dfrac{l}{g}}$

$\\ \rm\rightarrowtail T=2\pi\sqrt{\dfrac{4}{9.8}}$

$\\ \rm\rightarrowtail T=2\pi(0.63887)$

$\\ \rm\rightarrowtail T=1.27774\pi$

$\\ \rm\rightarrowtail T=4.012s$

4 0

2 years ago

The heart working with the blood vessels to pump blood is which body system?

LiRa [457]

I think it’s the cardiovascular system

6 0

3 years ago