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3 years ago

Two very quick questions!!

2 answers:

8 0

The weight lifted by a machine to the applied force on a machine is called mechanical advantage. This is written as Mechanical advantage, M. A, = load(weight)/effort. So for 1) M.A = 2 and load = 2, 000lb = 8896.446N. So 2 = 8896.446/ effort Effort = 8896.446/2 = 4448.48 Similarly for M.A of 2, 000, 000 we have Effort = 8896.446/ 2, 000, 000 = 0.004448

lana [24]3 years ago

5 0

Explanation:

1. Mechanical advantage of a pulley system, m = 2

Mechanical advantage of a machine is defined as the ratio of load to the effort force i.e.

$m=\dfrac{F_L}{F_E}$

Here, $F_L=2000\ lb$

$F_E=\dfrac{F_L}{m}$

$F_E=\dfrac{2000\ lb}{2}$

$F_E=1000\ lb$

Hence, 1000 lb effort will be needed to move the car.

2. Mechanical advantage of a pulley system, m = 2,000,000

Value of load, $F_L=2000\ lb$

Mechanical advantage, $m=\dfrac{F_L}{F_E}$

$F_E=\dfrac{F_L}{m}$

$F_E=\dfrac{2000\ lb}{2000000}$

$F_E=0.001\ lb$

Hence, 0.001 lb effort would be needed to move the car.

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Which car is accelerating?

Oliga [24]

Answer:

The car that is accelerating is B a car that rounds a curve at a constant speed

Explanation:

Although all of the cars are at a constant speed or not moving acceleration is the change in speed or the change of directions therefore making the only car changing directions your answer.

8 0

2 years ago

Read 2 more answers

A 8.8 cm diameter circular loop of wire is in a 1.04 T magnetic field. The loop is removed from the field in 0.30 s . Assume tha

denis23 [38]

Answer:

0.021 V

Explanation:

The average induced emf (E) can be calculated usgin the Faraday's Law:

$E = - \frac{N*\Delta \phi}{\Delta t}$

Where:

N = is the number of turns = 1

ΔΦ = ΔB*A

Δt = is the time = 0.3 s

A = is the loop of wire area = πr² = πd²/4

ΔB: is the magnetic field = (0 - 1.04) T

Hence the average induced emf is:

$E = - \frac{\Delta B*A}{\Delta t} = - \frac{(0- 1.04 T) \pi (0.088 m)^{2}}{4*0.3 s} = 0.021 V$

Therefore, the average induced emf is 0.021 V.

I hope it helps you!

8 0

3 years ago

From an h = 53 feet observation tower on the coast, a Coast Guard officer sights a boat in difficulty. The angle of depression o

maksim [4K]

Answer:

757,93 feets

Explanation:

We can make a right triangle between the boat (A), the Coast Guard officer (B) and the base of the observation tower (C), like in the graph attached. Now, you could also made a rectangle, adding the horizontal at the height of the Coast Guard, starting in B and ending in D, the vertex opossing C.

The angle of depression, its O in the graph.

Now, as we got an rectangle, of course, the segment AD its the same length as CB, and CA, the distance from the boat to shoreline, its the same length as DB.

ADB its an right triangle, with AB, the hypothenuse, and BD and DA, the catheti (or legs).

Now, we know the lenght BC, the height of the tower, 53 feets, so we also know the lenght of DA. DA its the opposite cathetus to the angle O. We wish to know the length AC, equal to the lenght DB, the adjacent cathetus of the angle O.

Know, the trigonometric function that connects the adjacent cathetus with the opossite cathetus its the tangent.

$tangent( O ) = \frac{opposite}{adjacent}$

We can take that the angle O = 4 °, and knowing that the opossite cathetus its 53 feets, we got:

$tangent( 4) = \frac{53 feets}{DB}$

$DB= \frac{53 feets}{tangent( 4)}$

$DB= 757,93 feets$

This its equal to the distance from the boat to the shoreline.

4 0

2 years ago

Two students measure the time constant of an RC circuit. The first student charges the capacitor using a 12 V battery, then lets

luda_lava [24]

Answer:

The answer is: c.) Both students get the same time constant, since the time constant does not depend on the charge on the capacitor

Explanation:

Both students, because the time constant is not dependent on the capacitor charge. We can express the equation of the time constant as follows:

Time constant = RC

In this equation it is observed that the time constant is equal to the multiplication of the resistance (R) multiplied by the capacitance (C)

8 0

3 years ago

Prove 1/P = 1/P1 + 1/P2

Sati [7]

Answer:

Proof:

|----------(R1)-------------(R2)---------|

Let the resistances be R1 , R2 connected in series. And the potential difference be V ;

We know that :

P1 = V²/R1 ..........(1) and

P2 = V²/R2 ...........(2)

In series combination , the net resistance is the algebraic addition of individual resistance :

R = R1 + R2

=> V²/P = V²/P1 + V²/P2

=> 1/P = 1/P1 + 1/P2

Explanation:

Copied it from a different brainly question that is this question. Please try to find it on brainly instead of asking it'd make it much easier.

7 0

3 years ago