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boyakko [2]
3 years ago
8

An artist is creating a gigantic mobile using old cars. On one end of the mobile, a 2100 lb Plymoth is suspended 30 feet from th

e fulcrum. In order for the mobile to be balanced, where should the artist suspend a 1400 lb Volkswagon?
a
20 feet from the fulcrum

b
7 feet from the fulcrum

c
45 feet from the fulcrum

d
on the fulcrum
Engineering
2 answers:
ozzi3 years ago
7 0

Answer:

  c  45 feet from the fulcrum

Explanation:

The moment at the fulcrum must be the same for each car. If the distance is d, then the artist must have ...

  (2100 lb)(30 ft) = (1400 lb)(d)

  2100·30/1400 ft = d = 45 ft

The Volkswagen should be 45 ft from the fulcrum.

Tresset [83]3 years ago
6 0

Answer:

<h2> C. <u>4</u><u>5</u><u> </u><u>feet</u><u> </u><u>from</u><u> </u><u>the</u><u> </u><u>fulcrum</u></h2>

  • <em>That</em><em> </em><em>artist</em><em> </em><em>should</em><em> </em><em>suspend</em><em> </em><em>a</em><em> </em><em>1</em><em>4</em><em>0</em><em>0</em><em> </em><em>lb</em><em>/</em><em>pounds</em><em> </em><em>Volkswagen</em><em>.</em>
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Drag each tile to the correct box.
Trava [24]

Answer:

Bluray

DVD

CD

Explanation:

Blu ray can hold 25gb per layer

Dvd can hold 4.7GB on a single layer

Cd can hold around 737 mb

Also, dvds can go up to 2 layers

Blu ray can go up to 4

6 0
3 years ago
Indicate the correct statement about the effect of Reynolds number on the character of the flow over an object.
sergeinik [125]

Answer:

If Reynolds number increases the extent of the region around the object that is affected by viscosity decreases.

Explanation:

Reynolds number is an important dimensionless parameter in fluid mechanics.

It is calculated as;

R_e__N} = \frac{\rho vd}{\mu}

where;

ρ is density

v is velocity

d is diameter

μ is viscosity

All these parameters are important in calculating Reynolds number and understanding of fluid flow over an object.

In aerodynamics, the higher the Reynolds number, the lesser the viscosity plays a role in the flow around the airfoil. As Reynolds number increases, the boundary layer gets thinner, which results in a lower drag. Or simply put, if Reynolds number increases the extent of the region around the object that is affected by viscosity decreases.

5 0
3 years ago
Assume the work done compressing the He gas is -63 kJ and the internal energy change of the gas is 79 kJ. What is the heat loss
klemol [59]

Answer:

Heat gain of 142 kJ

Explanation:

We can see that job done by compressing the He gas is negative, it means that the sign convention we are going to use is negative for all the work done by the gas and positive for all the job done to the gas. With that being said, the first law of thermodynamics equation will help us to solve this problem.

ΔU = Q + W ⇒ Q = ΔU -W

Q = 79 - (-63) = 142 kJ

Therefore, the gas gained heat by an amount of 142 kJ.

3 0
3 years ago
Technician A says high resistance causes an increase in current flow. Technician B says a higher than
Kryger [21]

The correct statement is: a higher than a normal voltage drop could indicate high resistance. Technician B is correct.

<h3>Ohm's law</h3>

Ohm's law states that the current flowing through a metallic conductor is directly proportional to the voltage provided all physical conditions are constant. Mathematically, it is expressed as

V = IR

Where

V is the potential difference

I is the current

R is the resistance

<h3>Technician A</h3>

High resistance causes an increase in current flow

V = IR

Divide both side by I

R = V / I

Thus, technician A is wrong as high resistance suggest low current flow

<h3>Technician B</h3>

Higher than normal voltage drop could indicate high resistance

V = IR

Thus, technician B is correct as high voltage indicates high resistance

<h3>Conclusion </h3>

From the above illustration, we can see that technician B is correct

Learn more about Ohm's law:

brainly.com/question/796939

8 0
2 years ago
The current entering the positive terminal of a device is i(t)= 6e^-2t mA and the voltage across the device is v(t)= 10di/dtV.
liberstina [14]

Answer:

a) 2,945 mC

b) P(t) = -720*e^(-4t) uW

c) -180 uJ

Explanation:

Given:

                           i (t) = 6*e^(-2*t)

                           v (t) = 10*di / dt

Find:

( a) Find the charge delivered to the device between t=0 and t=2 s.

( b) Calculate the power absorbed.

( c) Determine the energy absorbed in 3 s.

Solution:

-  The amount of charge Q delivered can be determined by:                      

                                       dQ = i(t) . dt

                  Q = \int\limits^2_0 {i(t)} \, dt = \int\limits^2_0 {6*e^(-2t)} \, dt = 6*\int\limits^2_0 {e^(-2t)} \, dt

- Integrate and evaluate the on the interval:

                   = 6 * (-0.5)*e^-2t = - 3*( 1 / e^4 - 1) = 2.945 C

- The power can be calculated by using v(t) and i(t) as follows:

                 v(t) = 10* di / dt = 10*d(6*e^(-2*t)) /dt

                 v(t) = 10*(-12*e^(-2*t)) = -120*e^-2*t mV

                 P(t) = v(t)*i(t) = (-120*e^-2*t) * 6*e^(-2*t)

                 P(t) = -720*e^(-4t) uW

- The amount of energy W absorbed can be evaluated using P(t) as follows:

                 W = \int\limits^3_0 {P(t)} \, dt = \int\limits^2_0 {-720*e^(-4t)} \, dt = -720*\int\limits^2_0 {e^(-4t)} \, dt

- Integrate and evaluate the on the interval:

                  W = -180*e^-4t = - 180*( 1 / e^12 - 1) = -180uJ

6 0
3 years ago
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