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kompoz [17]
3 years ago
7

A temperature of 200°F is equivalent to approximately

Physics
2 answers:
Blizzard [7]3 years ago
8 0

93.3* C is correct, I just took the test and it was correct. Good luck

aev [14]3 years ago
7 0
<span>93.3°C
A temperature in Fahrenheit (°F) can be converted to Celsius (°C), using the formula
[°C] = ([°F] − 32) ×  5⁄9. Here we have to convert a temperature of 200°F in to Celsius. Thus Subtract 32 from Fahrenheit and multiply by 5 then divide by 9 . That is (200°F - 32) × 5/9=168 × 5/9
                                          =840/9
                                          =93.333333333°C
                                          = 93.3°C</span>
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In order to place a satellite into orbit, it requires enough fuel to supply the necessary mechanical energy. Into what types of
nexus9112 [7]

When a satellite is revolving into the orbit around a planet then we can say

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F_g = F_c

\frac{GM_pM_s}{r^2} = \frac{M_s v^2}{r}

now we can say that kinetic energy of satellite is  given as

KE = \frac{1}{2}M_s v^2

KE = \frac{GM_sM_p}{2r}

also we know that since satellite is in gravitational field of the planet so here it must have some gravitational potential energy in it

so we will have

U = -\frac{GM_sM_p}{r}

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6 0
3 years ago
A car driving at 20 m/s accelerates continuously 2m/s2​ ​ for 3 seconds. What is its final velocity
Lesechka [4]

Answer: V= u+ at

V= final velocity

u=initial velocity

a=acceleration

t=time taken

V= 20 + 2*3

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Explanation:

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3 years ago
A penny rolls along the table for a distance of 1.3 meters. Jackie pushes it 40 centimeters further in the same direction.
brilliants [131]
40 meters times 1 meter over 100 centimeters equals 0.4 meters. 1.3 meters + 40 centimeters =. 1.3 m + 0.4 m = 1.7 m. The answer is 1.7 meters
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4 years ago
The force of gravity acting on an object is the object's ______. A. acceleration B. mass C. weight D. matter
ryzh [129]

Answer:

d

Explanation:

3 0
3 years ago
Read 2 more answers
A block of mass m=1.2 kg is held at rest against a spring with a force constant k=730N/m. Initially the spring is compressed a d
igor_vitrenko [27]

Answer:

Explanation:

potential energy of compressed spring

= 1/2 k d²

= 1/2 x 730 d²

= 365 d²

This energy will be given to block of mass of 1.2 kg in the form of kinetic energy .

Kinetic energy after crossing the rough patch

= 1/2 x 1.2 x 2.3²

= 3.174 J

Loss of energy

= 365 d² - 3.174  

This loss is due to negative work done by frictional force

work done by friction = friction force x width of patch

= μmg d ,   μ = coefficient of friction , m is mass of block , d is width of patch

= .44 x 1.2 x 9.8 x .05

= .2587 J

365 d² - 3.174   = .2587

365 d² = 3.4327

d² = 3.4327 / 365

= .0094

d = .097 m

= 9.7 cm

If friction increases , loss of energy increases . so to achieve same kinetic energy , d will have to be increased so that initial energy increases so compensate increased loss .

5 0
3 years ago
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