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Ronch [10]
3 years ago
9

How do I do this problem?

Physics
1 answer:
lesya692 [45]3 years ago
3 0
Add then divide the hint clearly backs it up two so yeahh
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If the absolute temperature of a gas is 600 K, the temperature in degrees Celsius is: A. 705°C. B. 873°C. C. 273°C. D. 327°C
zlopas [31]

Answer:

D). 327 ^0 C

Explanation:

As we know that temperature scale is linear so we will have

\frac{^0C - 0}{100 - 0} = \frac{K - 273}{373 - 273}

now we have

\frac{^0 C - 0}{100} = \frac{K - 273}{100}

so the relation between two scales is given as

^0 C = K - 273

now we know that in kelvin scale the absolute temperature is 600 K

so now we have

T = 600 - 273 = 327 ^0 C

so correct answer is

D). 327 ^0 C

4 0
3 years ago
I need help with this
mel-nik [20]
It’s 4. Definitely 4. 99% sure it’s 4.
4 0
2 years ago
6. The hole on a level, elevated golf green is a horizontal distance of 150 m from the tee and at an elevation of 12.4 m above t
Georgia [21]

Answer:

u = 104.68 m/s

Explanation:

given,

horizontal distance = 150 m

elevation of  12.4 m

angle = 8.6°

horizontal motion = x = u cos θ. t .............(1)

vertical motion =

y = u sin \theta - \dfrac{1}{2}gt^2................(2)

from equation(1) and (2)

y = x tan \theta - \dfrac{gx^2}{2u^2cos^2\theta}..........{3}

12.4 = 150\times tan (8.6) - \dfrac{9.8\times 150^2}{2u^2cos^2(8.6)}

\dfrac{9.8\times 150^2}{2u^2cos^2(8.6)} = 10.29

\dfrac{9.8\times 150^2}{2\times 10.29\times cos^2(8.6)} = u^2

u = \sqrt{10959.34}

u = 104.68 m/s

The initial speed of the ball is u = 104.68 m/s

8 0
3 years ago
A 15 kg dog jumps out a stationary sled which has a mass of 40 kg. If
zzz [600]
Velocity of the sled is 3.2 m/s
5 0
2 years ago
A string is wrapped around a pulley with a radius of 2.0 cm. The pulley is initially at rest. A constant force of 50 N is applie
Ray Of Light [21]

Answer:

0.20kg-m^2

Explanation:

Let the linear velocity of the rope(=of pulley) is v m/s

Using kinematic equation

=> v = u + at

=>v = 0 + 4.9a

=>v = 4.9a ------------ eq1

By v^2 = u^2 + 2as

=>v^2 = 0 + 2 x v/4.9 x 1.2

=>4.9v^2 - 2.4v = 0

=>v(4.9v - 2.4) = 0

=>v = 2.4/4.9 = 0.49 m/s

Thus by v = r x omega

=>omega = v/r = 0.49/0.02 = 24.49 rad/sec

BY W = F x s = 50 x 1.2 = 60 J

=>KE(rotational) = W = 1/2 x I x omega^2

=>60 = 1/2 x I x (24.49)^2

=>I = 0.20 kg-m^2

5 0
3 years ago
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