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3 years ago

If a cup of coffee is at 90°C and a person with a body temperature of 36°C touches it, how will heat flow between them?

Physics

2 answers:

Alex73 [517]3 years ago

3 0

B: from the cup to the hand.

Katena32 [7]3 years ago

3 0

Answer:

From the cup to the hand.

Explanation:

Note that heat flows from the region of higher temperature to a region of lower temperature.

If a cup of coffee is at 90°C and a person with a body temperature of 36°C touches it, therefore the heat flow will be from the coffee(higher temperature) to the body at 36°C. This is due to the temperature gradient i.e difference in temperature of the two bodies.

The heat from the coffee will be transmitted to the hand.

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A snowboarder travels 150 m down a mountain slope that is 65 degrees above horizontal. What is his vertical displacement?

choli [55]

This can be answered using trigonometric analysis. This sloped path that is 150 m long is the hypotenuse of the triangle. The adjacent angle would then be 65 degrees. Given these:

sin 65 = h / 150

Where: h = vertical displacement = 150 (sin 65)
h = 135.95 meters

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3 years ago

Two wooden crates rest on top of one another. The smaller top crate has a mass of m1 = 25 kg and the larger bottom crate has a m

faust18 [17]

Answer:

Explanation:

Let the bigger crate be in touch with the ground which is friction less. In the first case both m₁ and m₂ will move with common acceleration because m₁ is not sliding over m₂.

1 ) Common acceleration a = force / total mass

= 234 / ( 25 +91 )

= 2.017 m s⁻².

2 ) Force on m₁ accelerating it , which is nothing but friction force on it by m₂

= mass x acceleration

= 25 x 2.017

= 50.425 N

The same force will be applied by m₁ on m₂ as friction force which will act in opposite direction.

3 ) Maximum friction force that is possible between m₁ and m₂

= μ_s m₁g

= .79 x 25 x 9.8

= 193.55 N

Acceleration of m₁

= 193 .55 / 25

= 7.742 m s⁻²

This is the common acceleration in case of maximum tension required

So tension in rope

= ( 25 +91 ) x 7.742

= 898 N

4 ) In case of upper crate sliding on m₂ , maximum friction force on m₁

= μ_k m₁g

= .62 x 25 x 9.8

= 151.9 N

Acceleration of m₁

= 151.9 / 25

= 6.076 m s⁻².

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3 years ago

The maximum gauge pressure in a hydraulic lift is 18.0 atm what is the largest size vehicle

victus00 [196]

There is a missing part in the question. Found the complete text on internet:
"What is the largest size vehicle (kg) it can lift if the diameter of the output line is 28.0 cm? "

Solution
The diameter of the piston is 28.0 cm, this means its radius is 14.0 cm (half the diameter), so the area of the piston is
 $A=\pi r^2 = \pi (0.14 m)^2 =6.15 \cdot 10^{-2} m^2$

The maximum pressure of the lift is
 $p=18.0 atm = 1.82 \cdot 10^6 Pa$

Therefore the maximum force the piston can lift is
 $F=pA=(1.82 \cdot 10^6 Pa)(6.15 \cdot 10^{-2} m^2)=1.12 \cdot 10^5 N$

And the size (the mass) of the vehicle is
 $m= \frac{F}{g}= \frac{1.12 \cdot 10^5 Pa}{9.81 m/s^2} =1.14 \cdot 10^4 kg$

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