Which statement best describes how power and work are related?O A. Power is the ability to do more work with less force.
O B. Power is a measure of how quickly work is done.
O C. Power and work have the same unit of measurement
O D. Power is the amount of work needed to overcome friction.
Pls answer quick

Answers

Answer 1

Answer:

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1. An automobile travels along a straight road at 15.65 m/s through a 11.18 m/sspeed zone. A police car observed the automobile. At the instant that the two
vehicles are abreast of each other, the police car starts to pursue the automobile at
a constant acceleration of 1.96 m/s². The motorist noticed the police car in his rear
view mirror 12 s after the police car started the pursuit and applied his brakes and decelerates at 3.05 m/s². (Hint: The police will not go against the law.)
a) Find the total time required for the police car to overtake the automobile.
b) Find the total distance travelled by the police car while overtaking the
automobile.
c) Find the speed of the police car at the time it overtakes the automobile.
d) Find the speed of the automobile at the time it was overtaken by the police car.

Answers

Answer:

a.) Time = 17.13 seconds

b.) 31.88 m

c.) V = 11.18 m/s

d.) V = 7.1 m/s

Explanation:

The initial velocity U of the automobile is 15.65 m/s.

At the instant that the two vehicles are abreast of each other, the police car starts to pursue the automobile with initial velocity U = 0 at a constant acceleration of 1.96 m/s². Because the police is starting from rest.

For the automobile, let us use first equation of motion

V = U - at.

Acceleration a is negative since it is decelerating with a = 3.05 m/s² . And

V = 0.

Substitute U and a into the formula

0 = 15.65 - 3.05t

15.65 = 3.05t

t = 15.65/3.05

t = 5.13 seconds

But the motorist noticed the police car in his rear view mirror 12 s after the police car started the pursuit and applied his brakes and decelerates at 3.05 m/s².

The total time required for the police car to overtake the automobile will be

12 + 5.13 = 17.13 seconds.

b.) Using the third equation of motion formula for the police car at V = 11.18 m/s and a = 1.96 m/s²

V^2 = U^2 + 2aS

Where S = distance travelled.

Substitute V and a into the formula

11.18^2 = 0 + 2 × 1.96 ×S

124.99 = 3.92S

S = 124.99/3.92

S = 31.88 m

c.) The speed of the police car at the time it overtakes the automobile will be in line with the speed zone which is 11.18 m/s

d.) That will be the final velocity V of the automobile car.

We will use third equation of motion to solve that.

V^2 = U^2 + 2as

V^2 = 15.65^2 - 2 × 3.05 × 31.88

V^2 = 244.9225 - 194.468

V = sqrt( 50.4545)

V = 7.1 m/s

Which of the following expressions causes an implicit conversion between types? Assume variable x is an integer, t is a float, and name is a string.Group of answer choices7.5 + (x / 2)x + 2 * x"Hello, " + str(name)print(str(t))

Answers

x + 2 * x is the correct option. The above-selected option demonstrates implicit conversion, which is an automated type of conversion. Thus, option B is correct.

The series of conversions are necessary to change the type of a function call's argument to that of the parameter with the same name in the function declaration is known as an implicit conversion sequence. For each parameter, the compiler tries to identify an implicit conversion sequence.

If both user-defined conversion sequences A and B contain the same user-defined conversion function or constructor, and if the second standard conversion sequence of A is superior to the second standard conversion sequence of B, then user-defined conversion sequence A is preferable to user-defined conversion sequence B.

Learn more about implicit conversion here:

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A Coca Cola can with diameter 62 mm and wall thickness 300 um has an internal pressure of 100 kPa. Calculate the principal stresses at a point on the cylindrical surface of the can far from its ends.

Answers

Answer:

$\sigma _1=10.33MPa$

$\sigma _2=5.16MPa$

Explanation:

Given that

Diameter(d)=62 mm

Thickness(t)= 300 μm=0.3 mm

Internal pressure(P)=100 KPa

Actually there is no any shear stress so normal stress will become principle stress.This is the case of thin cylinder.The stress in thin cylinder are hoop stress and longitudinal stress .

The hoop stress

$\sigma _h=(Pd)/(2t)$

Longitudinal stress

$\sigma _l=(Pd)/(4t)$

Now by putting the values

$\sigma _h=(Pd)/(2t)$

$\sigma _h=(100* 62)/(2* 0.3)$

$\sigma _h=10.33MPa$

$\sigma _l=5.16MPa$

So the principle stress are

$\sigma _1=10.33MPa$

$\sigma _2=5.16MPa$

Q1. A truck traveling at 40 mph is approaching a stop sign. At time ????0 and at a distance of 80ft, the truck begins to slow down by decelerating rate of 12 ft/sec2 . Will the truck be able to stop in time?

Answers

Answer:

The truck will not stop in time. The truck passes the stop sign by about 63.41 ft before it stops.

Explanation:

The distance that the truck starts slowing down = 80 ft from the stop sign

Using equations of motion, we can calculate the distance it will take the truck to stop, then check of it is less than or more than 80 ft.

u = initial velocity of the truck = 40 mph = 58.667 ft/s

v = final velocity of the truck = 0 ft/s (since it comes to a stop eventually)

x = horizontal distance covered during the deceleration

a = Deceleration = -12 ft/s² (it'll have a negative sign, since it is negative acceleration

v² = u² + 2ax

0² = 58.667² + 2(-12)(x)

24x = 3441.816889

x = 143.41 ft

143.41 ft > 80 ft; hence, the truck will not stop in time. The truck passes the stop sign by about 63.41 ft before it stops.

Corrected Question:

A truck traveling at 40 mph is approaching a stop sign. At time t₀ and at a distance of 80 ft, the truck begins to slow down by decelerating at 12 ft/s2, will the truck be able to stop in time?

Answer:

The truck will not be able to stop in time.

Explanation:

==> First lets convert all variables to SI units

1 mph = 0.45m/s

40mph = 40 miles per hour = 40 x 0.45 m/s

40mph = 18m/s

1 ft = 0.3048m

80 ft = 80 x 0.3048m

80 ft = 24.38m

Also;

12ft/s² = 12 x 0.3048m/s²

12ft/s² = 3.66m/s²

==> Now, consider one of the equations of motion as follows;

v² = u² + 2as -----------------(i)

Where;

v = final velocity of motion

u = initial velocity of motion

a= acceleration/deceleration of motion

s = distance covered during motion

Using this equation, lets calculate the distance, s, covered during the acceleration;

We know that;

v = 0 [since the truck comes to a stop]

u = 40mph = 18m/s

a = -12ft/s² = -3.66m/s² [the negative sign shows that the truck decelerates]

Substitute these values into equation (i) as follows;

0² = 18² + 2 (-3.66)s

0 = 324 - 7.32s

7.32s = 324

s = $(324)/(7.32)$

s = 44.26m

The distance from where the truck starts decelerating to where it eventually stops is 44.26m which is past the stop sign (which is at 80ft = 24.38m). This means that the truck stops, 44.26m - 24.38m = 19.88m, after the stop sign. Therefore, the truck will not be able to stop in time.

A hydraulic jump is induced in an 80 ft wide channel. The water depths on either side of the jump are 1 ft and 10 ft. Please calculate: a) The velocity of the faster moving flow. b) The flow rate (discharge). c) The Froude number of the sub-critical flow. d) The flow energy dissipated in the hydraulic jump (expressed as percentage of the energy prior to the jump). e) The critical depth.

Answers

Answer:

a) The velocity is 42.0833 ft/s

b) The flow rate is 3366.664 ft³/s

c) The Froude number is 0.2345

d) The flow energy dissipated (expressed as percentage of the energy prior to the jump) is 18.225 ft

e) The critical depth is 3.8030 ft

Explanation:

Given data:

80 ft wide channel, L

1 ft and 10 ft water depths, d₁ and d₂

Questions: a) Velocity of the faster moving flow, v = ?

b) The flow rate (discharge), q = ?

c) The Froude number, F = ?

d) The flow energy dissipated, E = ?

e) The critical depth, dc = ?

a) For the velocity:

$(d_(2) )/(d_(1) ) =(1)/(2) (\sqrt{1+8F^(2) } -1)$

$10*2=\sqrt{1+8F^(2) } -1$

Solving for F:

F = 7.4162

$v=F\sqrt{gd_(1) }$

Here, g = gravity = 32.2 ft/s²

$v=7.4162*√(32.2*1) =42.0833ft/s$

b) The flow rate:

$q=v*L*d_(1) =42.0833*80*1=3366.664ft^(3) /s$

c) The Froude number:

$v_(2) =(q)/(L*d_(2) ) =(3366.664)/(80*10) =4.2083ft/s$

$F=\frac{v_(2)}{\sqrt{gd_(2) } } =(4.2083)/(√(32.2*10) ) =0.2345$

d) The flow energy dissipated:

$E=((d_(2)-d_(1))^(3) )/(4d_(1)d_(2)) =((10-1)^(3) )/(4*1*10) =18.225ft$

e) The critical depth:

$d_(c) =((((q)/(L))^(2) )/(g) )^(1/3) =((((3366.664)/(80))^(2) )/(32.2) )^(1/3) =3.8030ft$

Which outcome most accurately portrays the future for the timber company in the following scenario?A timber company that cuts and sells its own wood has been in high demand for the past three years. The company has gone through many acres of forest in its area, but it is sure to plant tree seeds after it bulldozes a section.

The timber company will lose business soon because not many people will need to build houses and buildings anymore.
The timber company will run out of trees quickly because the seedlings will not have enough time to become full-grown timber.
The timber company will go out of business due to the rising number of buildings using cement and concrete rather than timber.
The timber company will continue to grow because of its good business practices, and it will become the number one timber company in the country.

Answers

Answer: The timber company will run out of trees quickly because the seedlings will not have enough time to become full-grown timber

Explanation: I got it right on Edge!!!! Hope this helps

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