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A 46-ton monolith is transported on a causeway that is 3500 feet long and has a slope of about 3.7. How much force parallel to the incline would be required to hold the monolith on this causeway?

Answers

Answer 1

Answer:

The required force parallel to the incline to hold the monolith on this causeway will be "2.9 tons".

Angle and Force

According to the question,

Angle, a = 3.7 degrees or,

Sin a = 0.064

Force, F = 46 tons

We know the relation,

Parallel (tangential), $F_t$ = F Sin a

By substituting the values,

= 46 × 0.064

= 2.9 tons

Thus the response above is appropriate answer.

Find out more information about Force here:

brainly.com/question/25239010

Answer 2

Answer:

2.9tons

Explanation:

Note that On an incline of angle a from horizontal, the parallel and perpendicular components of a downward force F are:

parallel ("tangential"): F_t = F sin a

perpendicular ("normal"): F_n = F cos a

At a=3.7 degrees, sin a is about 0.064 and with F = 46tons:

F sin a ~~ (46 tons)*0.064 ~~ 2.9tons

Also see attached file

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The tension of a string is found to be 4050 N and the mass of 1 m of the string is 0.5 kg. What could be the velocity of a wave traveling in that string?

Answers

Answer:

90 m/s

Explanation:

The formula for velocity of wave in a string is given as,

v' = √(T/m') ................ Equation 1

Where v' = velocity of the string, T = Tension on the string, m' = mass per unit length of the string.

Given: T = 4050 N, and

m' = m/l where m = mass of the string, l = length of the spring.

m = 0.5 kg, l = 1 m

m' = 0.5/1 = 0.5 kg/m

Substitute into equation 1

v' = √(4050/0.5)

v' = √(8100)

v' = 90 m/s.

Hence the velocity of the wave in the string = 90 m/s

A fully loaded, slow-moving freight elevator has a cab with a total mass of 1200 kg, which is required to travel upward 35 m in 3.5 min, starting and ending at rest. The elevator's counterweight has a mass of only 940 kg, so the elevator motor must help pull the cab upward. What average power is required of the force the motor exerts on the cab via the cable

Answers

Answer:

425.1 W

Explanation:

We are given;

Counter mass of elevator; m_c = 940 kg

Cab mass of elevator; m_d = 1200 kg

Distance from rest upwards; d = 35 m

Time to cover distance; t = 3.5 min

Now, this elevator will have 3 forces acting on it namely;

Force due to the counter weight of the elevator; F_c

Force due to the cab weight on the elevator; F_d

Force exerted by the motor; F_m

Now, from Newton's 2nd law of motion,

The force exerted by the motor on the elevator can be given by the relationship;

F_m = F_d - F_c

Now,

F_d = m_d × g

F_d = 1200 × 9.81

F_d = 11772 N

F_c = m_c × g

F_c = 940 × 9.81

F_c = 9221.4 N

Thus;

F_m = 11772 - 9221.4

F_m = 2550.6 N

Now, the average power required of the force the motor exerts on the cab via the cable is given by;

P_m = F_m × v

Where v is the velocity of the elevator.

The velocity is calculated from;

v = distance/time

v = 35/3.5

v = 10 m/min

Converting to m/s gives;

v = 10/60 m/s = 1/6 m/s

Thus;

P_m = 2550.6 × 1/6

P_m = 425.1 W

Air enters an adiabatic compressor at 104 kPa and 292 K and exits at a temperature of 565 K. Determine the power (kW) for the compressor if the inlet volumetric flow rate is 0.15 m3/s. Use constant specific heats evaluated at 300 K.

Answers

Answer:

$\dot W_(in) = 49.386\,kW$

Explanation:

An adiabatic compressor is modelled as follows by using the First Law of Thermodynamics:

$\dot W_(in) + \dot m \cdot c_(p)\cdot (T_(1)-T_(2)) = 0$

The power consumed by the compressor can be calculated by the following expression:

$\dot W_(in) = \dot m \cdot c_(v)\cdot (T_(2)-T_(1))$

Let consider that air behaves ideally. The density of air at inlet is:

$P\cdot V = n\cdot R_(u)\cdot T$

$P\cdot V = (m)/(M)\cdot R_(u)\cdot T$

$\rho = (P\cdot M)/(R_(u)\cdot T)$

$\rho = ((104\,kPa)\cdot (28.02\,(kg)/(kmol)))/((8.315\,(kPa\cdot m^(3))/(kmol\cdot K) )\cdot (292\,K))$

$\rho = 1.2\,(kg)/(m^(3))$

The mass flow through compressor is:

$\dot m = \rho \cdot \dot V$

$\dot m = (1.2\,(kg)/(m^(3)))\cdot (0.15\,(m^(3))/(s) )$

$\dot m = 0.18\,(kg)/(s)$

The work input is:

$\dot W_(in) = (0.18\,(kg)/(s) )\cdot (1.005\,(kJ)/(kg\cdot K))\cdot (565\,K-292\,K)$

$\dot W_(in) = 49.386\,kW$

List Five examples from daily life in which you see periodic motion caused by a pendulum(Marking Brainliest)

Answers

Answer:

by a rocking chair, a bouncing ball, a vibrating tuning fork, a swing in motion, the Earth in its orbit around the Sun, and a water wave.

Explanation:

Analysis of the relationship between the fuel economy (mpg) and engine size (liters) for 35 models of cars produces the regression model ModifyingAbove mpg with caret equals 36.44 minus 3.829 times Engine size. If a car has a 5 liter engine, what does this model suggest the gas mileage would be?

Answers

Answer:

Gas mileage is 17.29

Explanation:

Given data:

The total number of the model is 35

The total size of the engine is 5 ltr

The regression model is given as

$36.44 - 3.829* engine\ size$

From the information given in question we have

Regression equation is : model- mpg $= 36.44 - 3.829* engine\ size$

Therefore for engine capacity of 5 liters;

Gas mileage $= 36.44 - 3.829* 5 = 17.29$

Gas mileage is 17.29

Answer:16.

Explanation:

Determine whether or not each of the following statement is true. If a statement is true, prove it. If the statement is false, provide a counterexample and explain how it constitutes a counterexample. Diagrams can be useful in explaining such things. If the electric potential in a certain region of space is constant, then the charge enclosed by any closed surface completely contained within that region is zero.

Answers

Answer:

True

Explanation:

This is a representation of Gauss law.

Gauss’s law does hold for moving charges, and in this respect Gauss’s law is more general than Coulomb’s law. In words, Gauss’s law states that: The net outward normal electric flux through any closed surface is proportional to the total electric charge enclosed within that closed surface. The law can be expressed mathematically using vector calculus in integral form and differential form, both are equivalent since they are related by the divergence theorem, also called Gauss’s theorem.

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A ​46-ton monolith is transported on a causeway that is 3500 feet long and has a slope of about 3.7. How much force parallel to the incline would be required to hold the monolith on this​ causeway?

Answers

Angle and Force

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A 46-ton monolith is transported on a causeway that is 3500 feet long and has a slope of about 3.7. How much force parallel to the incline would be required to hold the monolith on this causeway?