3. Which forces would exist in a free body diagram of this car accelerating along the highway?

Raul wants to increase his upper body and cardiovascular strength. He plans to go to the gym and use his backyard swimming pool

Mkey [24]

Answer: An exercise, activity, or workout that targets a specific training need Progression - Gradually increasing your skill level and ability over time Overload -Working your body harder than usual to improve how it functions Reversibility - Losing what you have gained by taking time off Tedium - getting bored by doing the same thing

Explanation:

4 0

3 years ago

Diffraction is observable when the opening is ______ than the wavelength.

masya89 [10]

Diffraction is observable when the is smaller than the wavelength

5 0

3 years ago

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30) A force produces power P by doing work W in a time T. What power will be produced by a force that does six times as much wor

schepotkina [342]

Answer:

A) 12P

Explanation:

The power produced by a force is given by the equation

$P=\frac{W}{T}$

where

W is the work done by the force

T is the time in which the work is done

At the beginning in this problem, we have:

W = work done by the force

T = time taken

So the power produced is

$P=\frac{W}{T}$

Later, the force does six times more work, so the work done now is

$W'=6W$

And this work is done in half the time, so the new time is

$T'=\frac{T}{2}$

Substituting into the equation of the power, we find the new power produced:

$P'=\frac{W'}{T'}=\frac{6W}{T/2}=12\frac{W}{T}=12P$

So, 12 times more power.

4 0

2 years ago

A porter can carry 40 bricks of 10 N load of each. He can carry up to 75m in 40 sec, calculate his power.

alexira [117]

Answer:

750W

Explanation:

40×10= 400N

work done= force × distance

=400 × 75

=30000 J

Power= work done/ time

= 30000 ÷ 40

= 750 W

4 0

1 year ago

A block is released to slide down a frictionless incline of 15∘ and then it encounters a frictional surface with a coefficient o

Elodia [21]

The block's potential energy at the top of the incline (at a height h from the horizontal surface) is equal to its kinetic energy at the bottom of the incline, so that

mgh = 1/2 mv²

where v is its speed at the bottom of the incline. It follows that

v = √(2gh)

If the incline is 20.4 m long, that means the block has a starting height of

sin(15°) = h/(20.4 m) ⇒ h = (20.4 m) sin(15°) ≈ 5.2799 m

and so the block attains a speed of

v = √(2gh) ≈ 10.1728 m/s

The block then slides to a rest over a distance d. Kinetic friction exerts a magnitude F over this distance and performs an amount of work equal to Fd. By the work-energy theorem, this quantity is equal to the block's change in kinetic energy, so that

Fd = 0 - 1/2 mv² ⇒ d = (-1293.58 J)/F

By Newton's second law, the net vertical force on the block as it slides is

∑ F [vertical] = n - mg = 0

where n is the magnitude of the normal force, so that

n = mg = (25 kg) g = 245 N

and thus the magnitude of friction is

F = -0.16 (245 N) = -39.2 N

(negative since it opposes the block's motion)

Then the block slides a distance of

d = (-1293.58 J) / (-39.2 N) ≈ 32.9994 m ≈ 33 m

5 0

2 years ago