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Varvara68 [4.7K]
3 years ago
15

The abbreviation for the SI unit of temperature is _____. m s kg K

Physics
1 answer:
zysi [14]3 years ago
4 0
The answer would be Kelvin which is written as K

In short, Your Answer would be Option D

Hope this helps!
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swimmers at a water park have a choice of two frictionless water slides as shown in the figure. although both slides drop over t
LuckyWell [14K]
If swimmers had a choice of the water slides shown in this figure,
they would all go home dry, since there is no figure.  I'll have to try to
answer this question based on only the words in the text, augmented
only by my training, education, life experience, and human logic.

-- Both slides are frictionless.  So no energy is lost as a swimsuit
scrapes along the track, and the swimmer's kinetic energy at the
bottom is equal to the potential energy he had at the top.

-- Both slides start from the same height.  So the same swimmer
has the same potential energy at the top of either one, and therefore
the same kinetic energy at the bottom of either one.

-- So the difference in the speeds of two different swimmers
on the slides depends only on the difference in the swimmers'
mass, and is not influenced by the shape or length of the slides
(as long as the slides remain frictionless).

If both swimmers have the same mass, then  v₁ = v₂ .
4 0
3 years ago
What is the impulse
vichka [17]
Impulse is the integral of a force, F.
Hope this helps.
(Please mark this brainliest, I would really appreciate it) Thanks!
5 0
3 years ago
If an object was accelerating at 10 m/s2, and a mass of 1 kg, what was size of the force acting on the object?
ryzh [129]

Answer:

10 N

Explanation:

f = ma

= 10m/s^2 * 1 kg

=10N

7 0
3 years ago
An engine absorbs 1.69 kJ from a hot reservoir at 277°C and expels 1.25 kJ to a cold reservoir at 27°C in each cycle.
Anna35 [415]

Answer

Given,

Energy absorbed, Q_H = 1.69\ kJ

Energy expels,Q_C =  1.25\ kJ

Temperature of cold reservoir, T = 27°C

a) Efficiency of engine

 \eta = \dfrac{Q_H - Q_C}{Q_H}\times 100

 \eta = \dfrac{1.69 - 1.25}{1.69}\times 100

\eta =26.03 %

b) Work done by the engine

 W = Q_H- Q_C

 W =1.69 - 1.25

 W = 0.44\ kJ

c) Power output

     t = 0.296 s

   P = \dfrac{W}{t}

   P = \dfrac{0.44}{0.296}

   P = 1.486\ kW

8 0
3 years ago
The pressure on a volume of liquid V = 1.0 mº at the surface is approximately equal to the atmospheric pressure Patm = 1.00 x 10
Alexus [3.1K]

Answer:-2.86*10⁻⁴

Explanation: Use the equation change in volume = (change in pressure * original volume) / Bulks Modulus. ΔV = (-Δp*V₀) / B

Plugging in your numbers, you should get ΔV = (-2.29*10⁷*1) / (8*10¹⁰) = -2.86*10⁻⁴

ΔP = P₂-P₁  ----> ΔP = 2.30*10⁷ - 1.00*10⁵ = 2.29*10⁷

3 0
3 years ago
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