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Strike441 [17]
2 years ago
7

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

k in half as much time?
A) 12P
B) 6P
C) P
D) P
E) P
Physics
1 answer:
schepotkina [342]2 years ago
4 0

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.

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AleksandrR [38]

Answer:

choosing a material that will show warning before it fails

Explanation:

According to my research on different architectural engineering techniques, I can say that based on the information provided within the question this is an example of choosing a material that will show warning before it fails. By choosing aluminum he can detect certain failures a long time before it actually happens since aluminum shows signs of wear and tear and doesn't just break immediately.

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4 0
2 years ago
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A 200 g hockey puck is launched up a metal ramp that is inclined at a 30° angle. The coefficients of static and kinetic friction
Vaselesa [24]

Answer:

H=1020.12m

Explanation:

From a balance of energy:

\frac{m*Vo^2}{2} -mg*H=-Ff*d   where H is the height it reached, d is the distance it traveled along the ramp and Ff = μk*N.

The relation between H and d is given by:

H = d*sin(30)   Replace this into our previous equation:

\frac{m*Vo^2}{2} -mg*d*sin(30)=-\mu_k*N*d

From a sum of forces:

N -mg*cos(30) = 0    =>  N = mg*cos(30)   Replacing this:

\frac{m*Vo^2}{2} -mg*d*sin(30)=-\mu_k*mg*cos(30)*d   Now we can solve for d:

d = 2040.23m

Thus H = 1020.12m

6 0
2 years ago
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A baseball catcher puts on an exhibition by catching a 0.15-kg ball dropped from a helicopter at a height of 101 m. What is the
yaroslaw [1]

Answer:

The speed of the ball 1.0 m above the ground is 44 m/s (Answer A).

Explanation:

Hi there!

To solve this problem, let´s use the law of conservation of energy. Since there is no air resistance, the only energies that we should consider is the gravitational potential energy and the kinetic energy. Because of the conservation of energy, the loss of potential energy of the ball must be compensated by a gain in kinetic energy.

In this case, the potential energy is being converted into kinetic energy as the ball falls (this is only true when there are no dissipative forces, like air resistance, acting on the ball). Then, the loss of potential energy (PE) is equal to the increase in kinetic energy (KE):

We can express this mathematically as follows:

-ΔPE = ΔKE

-(final PE - initial PE) = final KE - initial KE

The equation of potential energy is the following:

PE = m · g · h

Where:

PE = potential energy.

m = mass of the ball.

g = acceleration due to gravity.

h = height.

The equation of kinetic energy is the following:

KE = 1/2 · m · v²

Where:

KE = kinetic energy.

m = mass of the ball.

v = velocity.

Then:

-(final PE - initial PE) = final KE - initial KE          

-(m · g · hf - m · g · hi) = 1/2 · m · v² - 0     (initial KE = 0 because the ball starts from rest)  (hf = final height, hi = initial height)

- m · g (hf - hi) = 1/2 · m · v²

2g (hi - hf) = v²

√(2g (hi - hf)) = v

Replacing with the given data:

√(2 · 9.8 m/s²(101 m - 1.0 m)) = v

v = 44 m/s

The speed of the ball 1.0 m above the ground is 44 m/s.

3 0
2 years ago
One species of eucalyptus tree, Eucalyptus regnans, grow to heights similar to those attained by California redwoods. Suppose a
mote1985 [20]

Answer:

The tree is 143.325 meters tall

Explanation:

The given parameters of the eucalyptus tree are;

The mass of the eucalyptus tree nut = 1.7 ounces

The speed of the nut at 50.3 m above the ground, v = 42.7 m/s

The equation for free fall is given as follows;

v² = 2·g·h

Where;

v = The velocity after falling through a height, h

g = The acceleration due to gravity = 9.8 m/s²

h = The height through which the seed has already fallen

Therefore, we have;

h = v²/(2·g) = (42.7 m/s)²/(2 × 9.8 m/s²) = 93.025 m

The height through which the seed has already fallen, h = 93.025 m

The height of the tree = h + The height of the seed above ground at the moment it was falling at 42.7 m/s

The height of the tree = 93.025 m + 50.3 m = 143.325 m

The height of the tree = 143.325 m.

4 0
2 years ago
A force F of strength 20 N acts on an object of mass 3 kg as it moves a distance of 4 m. If F is perpendicular to the 4 m displa
forsale [732]
Work = (displacement) x (force in the direction of the displacement)

Since none of the force is in the direction of the displacement, the work it does
is zero. Something else is making the object move.  It's not the 20N force.
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2 years ago
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