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sweet [91]
3 years ago
15

18°C = _____ -255 K 0 K 18 K 291 K

Physics
2 answers:
dimulka [17.4K]3 years ago
7 0
I think it is 291 

hope this helps
Travka [436]3 years ago
6 0
291 K, to convert from Celsius to Kelvins just add 273 
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What is the squirrels escape velocity in mph if the squirrel accelerates at a constant 1.5 m/s squared from rest for 2.5s
Zina [86]

Answer:

3.75 m/s

Explanation:

From the question given above, the following data were obtained:

Acceleration (a) = 1.5 m/s²

Initial velocity (u) = 0 m/s

Time (t) = 2.5 s

Final velocity (v) =?

a = (v – u) / t

1.5 = (v – 0) / 2.5

1.5 = v / 2.5

Cross multiply

v = 1.5 × 2.5

v = 3.75 m/s

Hence, the escape velocity of the squirrel is 3.75 m/s

7 0
3 years ago
The psychologist known for latent learning and cognitive maps is _________. A. Robert Rescorla B. Edward Tolman C. William James
Lubov Fominskaja [6]

Answer:

B

Explanation:

B. Edward Tolman

4 0
3 years ago
Read 2 more answers
Given that water at standard pressure freezes at 0∘c, which corresponds to 32∘f, and that it boils at 100∘c, which corresponds t
Zolol [24]
The temperature difference of 1 K is equivalent to the temperature difference of 1 °C. Therefore, we find the relationship between the change in °F and °C.
A change of 212 - 32 °F is the same as a change of 100 - 0 °C. Thus:
(212 - 32) °F = (100 - 0) °C
1 °C = 1.8 °F
1 K = 1.8 °F
6 0
3 years ago
A piece of warm concrete is placed in a cold-water tank, and energy flows between the concrete and the water. Which way does the
sammy [17]

Answer:

Heat flows from hot to cold objects. When a hot and a cold body are in thermal contact, they exchange heat energy until they reach thermal equilibrium, with the hot body cooling down and the cold body warming up. This is a natural phenomenon we experience all the time.

Explanation:

3 0
2 years ago
A sinusoidal wave of angular frequency 1,203 rad/s and amplitude 3.1 mm is sent along a cord with linear density 3.9 g/m and ten
kobusy [5.1K]

Answer:

18.7842493212 W

Explanation:

T = Tension = 1871 N

\mu = Linear density = 3.9 g/m

y = Amplitude = 3.1 mm

\omega = Angular frequency = 1203 rad/s

Average rate of energy transfer is given by

P=\dfrac{1}{2}\sqrt{T\mu}\omega^2y^2\\\Rightarrow P=\dfrac{1}{2}\sqrt{1871\times 3.9\times 10^{-3}}\times 1203^2\times (3.1\times 10^{-3})^2\\\Rightarrow P=18.7842493212\ W

The average rate at which energy is transported by the wave to the opposite end of the cord is 18.7842493212 W

7 0
3 years ago
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