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natima [27]
3 years ago
7

Approximately how many kelvins are equal to 60°f? a. 333 b. 323 c. 413 d. 289

Physics
2 answers:
bearhunter [10]3 years ago
8 0

Answer:

The answer is D.289K

Explanation:

There is a simple way in which you can convert from Fahrenheit to Kelvin, which is a simple calculation that does the conversion directly, where the following formula is used:

K = (y ° F + 459.67) x 5/9

So we can replace the 60°f in the formula and we would have:

K = (60f + 459.67) * \frac{5}{9} \\\\K = 288.70

Afina-wow [57]3 years ago
3 0
D. 289
Take the formula:
K=5/9(Fahrenheit-32)+273
Plug in Fahrenheit
K=5/9 (60-32)+273
From here it is simple math and you can plug it into your calculator getting 288.5555556 and round to 289
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A plant cell is no longer capable of capturing energy from sunlight and converting it into chemical energy. Which organelle is m
Nadya [2.5K]

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8 0
9 months ago
1. A student lifts a box of books that weighs 185 N. The box is
aksik [14]

1)  148 J

When lifting an object, the work done on the object is equal to its change in gravitational potential energy. Mathematically:

W = \Delta U = (mg) \Delta h

where

mg is the weight of the object

\Delta h is the change in height

For the box in this problem,

mg = 185 N

\Delta h = 0.800 m

Substituting into the equation, we find:

W=(185)(0.800)=148 J

2) (a) 28875 J

The work done by a force applied parallel to the direction of motion of the object is given by

W=Fd

where

F is the magnitude of the force

d is the displacement

In this problem,

F = 825 N is the force applied by the two students together

d = 35 m is the displacement of the car

Substituting,

W=(825)(35)=28875 J

2) (b) 57750 J

As seen previously, the equation that gives the work done by the force is

W=Fd

We see that the work done is proportional to the magnitude of the force: therefore, if the force is doubled, then the work done is also doubled.

The work done previously was

W = 28875 J

Now the force is doubled, so the new work done will be

W' = 2(28875)=57750 J

3) 4.4 J

In this case, the force acting on the ball is the force of gravity, whose magnitude is:

F = mg

where

m = 0.180 kg is the mass of the ball

g=9.8 m/s^2 is the acceleration of gravity

Solving the equation,

F=(0.180)(9.8)=1.76 N

Now we find the work done by gravity using the same formula applied before:

W=Fd

where d = 2.5 m is the displacement of the ball. We can apply this version of the formula since the force is parallel to the displacement. Substituting,

W=(1.76)(2.5)=4.4 J

4) 595.2 kg

In this case, we have the work done on the box:

W = 7.0 kJ = 7000 J

And we also know the change in height of the box:

\Delta h = 1.2 m

As we stated in part a), the work done on the box is equal to its change in gravitational potential energy:

W=mg \Delta h

Solving for m, we find

m=\frac{W}{g \Delta h}

And substituting the numerical values, we find the mass of the box:

m=\frac{7000}{(9.8)(1.2)}=595.2 kg

5) They do the same work

In fact, the net work done by each person on the box is equal to the change in gravitational potential energy of the box:

W=mg \Delta h

Where \Delta h is the difference in height between the final position and the initial position of the box.

This means that the work done on the box depends only on its initial and final position, not on the path taken. The two men carry the box along different paths, however the reach at the end the same position, and they started from the same position: this means that the value of \Delta h is the same for both of them, so the work they have done is exactly the same.

5 0
3 years ago
Calculate the speed of the car at each checkpoint by
lana66690 [7]

Answer: The speed at the first quarter checkpoint is 0.74 m/s. The speed at the second quarter checkpoint is 1.40 m/s. The speed at the third quarter checkpoint is 1.61 m/s. The speed at the finish line is 1.89 m/s.

Explanation: I did the assignment and got it correct :)

3 0
2 years ago
If the torque required to loosen the nut that is holding a flat tire in a place on a car has a magnitude of 35 N•m what minimum
VashaNatasha [74]

Answer: F = 130 N

Explanation: Solution:

Convert first 27 cm to m.

27 cm  x 0.01 m / 1 cm = 0.27 m

Calculate the torque using T = Fd

Derive to find force F

F = T /d

  = 35 N.m / 0.27 m

  = 130 N

4 0
3 years ago
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A 95.0kg weightlifter pushes on a 800,000 kg wall for 400s but it does not move. How much work does he do on the wall?
Soloha48 [4]
<span>The weightlifter does no work. Although he has exerted force, work is the product of force over distance. Since he has not moved the wall he has done no work.</span>
7 0
3 years ago
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