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hammer [34]
3 years ago
7

You are on roller blades on top of a small hill. Your potential energy is equal to 1,000.0 joules. The last time

Physics
1 answer:
irinina [24]3 years ago
8 0
It is -59 i think or just ask someone esle for help
You might be interested in
3. A bottle of vitamin C contains 100 tablets and weighs 80 g. If the
Elan Coil [88]

Answer:

a

0.57g

b

20.52

c

28.5

Explanation:

a

The bottle weighs 80g with tablets

If the bottle alone weighs 23g, the tablets weigh 57g

100 tablets weigh 57g, 1 tablet weighs 0.57g

b

0.57g is one tablet, so to achieve 36 tablets we must multiply by 36

0.57 multiplied by 36 is 20.52

c

To find 50 tablets we can use the same method we used before or a slightly faster method

100 tablets is 57g, so all we have to do is halve to find 50

57 divided by 2 is 28.5

4 0
2 years ago
Suppose a 4,000-kg elephant is hoisted 20 m above Earth’s surface. Use a calculator and follow the steps below to find the eleph
stiv31 [10]
GPE = 78,380 J
w = 39,240 N

First list what you know. You know the elephants mass and it’s height. You also know gravity on Earth. I will use g = 9.81.
m = 4,000 kg
h = 20 m
g = 9.81 m/s^2

You need to find the elephants weight. Weight = mass x gravity
w = mg
w = (4000 kg)(9.81 m/s^2)
w = 39,240 N (N = newtons)

Now, knowing the elephants weight, you can calculate its GPE.
Gravitational Potential Energy = weight x height

GPE = wh
GPE = (39,240N)(20m)
GPE = 78,380 J (J = joules)
4 0
3 years ago
Four weightlifters (A-D) enter a competition. The mass, distance, and time of their lifts are shown in the table.
siniylev [52]

Let Pa, Pb, Pc, and Pd be the powers delivered by weightlifters A, B, C, and D, respectively.

Use this equation to determine each power value:

P = W÷Δt

P is the power, W is the work done by the weightlifter, and Δt is the elapsed time.

A) Determining Pa:

Pa = W÷Δt

The weightlifter does work to lift the weight up a certain distance. Therefore the work done is equal to the weight's gain in gravitational potential energy. The equation for gravitational PE is

PE = mgh

PE is the potential energy, m is the mass of the weight, g is the acceleration of objects due to earth's gravity, and h is the distance the weight was lifted.

We can equate W = PE = mgh, therefore we can make the following substitution:

Pa = mgh÷Δt

Given values:

m = 100.0kg

g = 9.81m/s²

h = 2.25m

Δt = 0.151s

Plug in the values and solve for Pa

Pa = 100.0×9.81×2.25÷0.151

<u>Pa = 14600W</u> (watt is the SI derived unit of power)

B) Determining Pb:

Let us use our new equation derived in part A to solve for Pb:

Pb = mgh÷Δt

Given values:

m = 150.0kg

g = 9.81m/s²

h = 1.76m

Δt = 0.052s

Plug in the values and solve for Pb

Pb = 150.0×9.81×1.76÷0.052

<u>Pb = 49800W</u>

C) Determining Pc:

Pc = mgh÷Δt

Given values:

m = 200.0kg

g = 9.81m/s²

h = 1.50m

Δt = 0.217s

Plug in the values and solve for Pc

Pc = 200.0×9.81×1.50÷0.217

<u>Pc = 13600W</u>

D) Determining Pd:

Pd = mgh÷Δt

Given values:

m = 250.0kg

g = 9.81m/s²

h = 1.25m

Δt = 0.206s

Plug in the values and solve for Pd

Pd = 250.0×9.81×1.25÷0.206

<u>Pd = 14900W</u>

Compare the following power values:

Pa = 14600W, Pb = 49800W, Pc = 13600W, Pd = 14900W

Pc is the lowest value.

Therefore, weightlifter C delivers the least power.

7 0
2 years ago
Why are latitude and longitude included on maps
Veseljchak [2.6K]

Answer:

to locate places on earth

3 0
2 years ago
The Event Horizon Telescope needs a 22 micro-arcsecond resolution to view the event horizon regions around black holes. If the a
likoan [24]

Answer:

14869817.395 m

Explanation:

\theta=22 microarcsecond

λ = Wavelength = 1.3 mm

Converting to radians we get

22\times 10^{-6}\frac{\pi}{180\times 3600}\ radians

From Rayleigh Criterion

\theta=1.22\frac{\lambda}{D}\\\Rightarrow D=1.22\frac{\lambda}{\theta}\\\Rightarrow D=1.22\frac{1.3\times 10^{-3}}{22\times 10^{-6}\frac{\pi}{180\times 3600}}\\\Rightarrow D=14869817.395\ m

Diameter of the effective primary objective is 14869817.395 m

It is not possible to build one telescope with a diameter of 14869817.395 m. But, we need this type of telescope. So, astronomers use an array of radio telescopes to achieve a virtual diameter in order to observe objects that are the size of supermassive black hole's event horizon.

7 0
3 years ago
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