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3 years ago

Evaporation of sweat cools the skin because

Physics

1 answer:

Law Incorporation [45]3 years ago

6 0

Answer: c. the molecules with the highest energy evaporate first, lowering the temperature of the sample

Explanation:

The process by which liquid starts to change into vapor phase at any temperature is known as evaporation.

During evaporation , the molecules which possess higher energies escape from the upper layer into vapor phase. the molecules which escape draw energy from surroundings and thus decrease the energy of the surroundings and hence lead to decrease in temperature.

As temperature of the system is directly proportional to the energy of the system , thus decrease in energy leads to decrease in temperature.

$K.E=\frac{3RT}{2}$

K.E. = Kinetic energy

T = temperature

R= gas constant

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A hockey puck of mass m1=165 g slides from left to right with an initial velocity of 15.5 m/s. It collides head on with a second

Fynjy0 [20]

Answer:

it may be -6 m/s

Explanation:

3 0

4 years ago

A water pipe is inclined 40.0° below the horizontal. The radius of the pipe at the upper end is 2.00 cm. If the gauge pressure a

aliya0001 [1]

Answer:

$P_2 = -1.9 \times 10^8 Pa$

Explanation:

As it is given that flow rate in the pipe is 20 cm^3/s

so we have

$Q = A_1v_1 = A_2v_2$

at the upper end the area is given as

$A_1 = \pi r_1^2$

$A_1 = \pi(0.02)^2 = 1.26 \times 10^{-3} cm^2$

Also at the other end

$A_2 = \pi r_2^2$

$A_2 = \pi(0.01)^2 = 0.314 \times 10^{-3} cm^2$

now the speed at two ends is given as

$v_1 = \frac{20}{1.26 \times 10^{-3}}$

$v_1 = 159.15 m/s$

$v_2 = \frac{20}{0.314 \times 10^{-3}}$

$v_2 = 637 m/s$

now by Bernoulli's theorem we have

$P_1 + \frac{1}{2}\rho v_1^2 + \rho g h_1 = P_2 + \frac{1}{2}\rho v_2^2 + \rho g h_2$

now we have

$0.112(1.013 \times 10^5) + \frac{1}{2}1000(159.15)^2 + 1000(9.81)(2.65sin40) = P_2 + \frac{1}{2}(1000)(637)^2 + 0$

Now we have

$P_2 = -1.9 \times 10^8 Pa$

4 0

3 years ago

A tube has a length of 0.025 m and a cross-sectional area of 6.5 x 10-4 m2. The tube is filled with a solution of sucrose in wat

Pepsi [2]

Answer:

The time required for sucrose transportation through the tube is 8.4319 sec.

Explanation:

Given:

L = 0.025 m

A = 6.5×10^-4 m^2

D = 5×10^-10 m^2/s

ΔC = 5.2 x 10^-3 kg/m^3

m = 5.7×10^-13 kg

Solution:

t = m×L / D×A×ΔC

t = (5.7×10^-13) × (0.025) / (5×10^-10)×(6.5×10^-4)×(5.2 x 10^-3)

t = 8.4319 sec.

5 0

3 years ago

You are in a contest with your friends to see who can drive a golf ball the farthest. should you hit a "line drive" (low to the

WITCHER [35]

Line drive would be more farthest from the very high angle shot because when we increase the angle of flight the range started to increase and at some point it becomes maximum and that angle is 45° after that as you go on increasing the angle it won't be covering more distance as compared to the max at (45°) .

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3 years ago

If this speed is based on what would be safe in wet weather, estimate the radius of curvature for a curve marked 70 km/h . The c

olganol [36]

as it is given that curved marked the speed as v = 70 km/h

so we will first convert the speed into m/s

$v = 70 km/h = 19.44 m/s$