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3 years ago

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A student is trying to decide what to wear.His bedroom is at 20.0 °C.His skin temperature is 35.0 °C.The area of his exposed ski

n is 1.50 m².People all over the world have skin that is dark in the infrared,with emissivity about 0.900.Find the net energy transfer from his body by radiation in 10.0 min.

1 answer:

Law Incorporation [45]3 years ago

8 0

Answer:

vgghcgkxcjpfiffj,ncfzfzujfzxxxoifkc xuzrusdoxTXcxgifdhh

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A spring has a relaxed length of 7 cm and a stiffness of 200 N/m. How much work must you do to change its length from 10 cm to 1

s344n2d4d5 [400]

The work W needed to stretch/compress a spring from rest by a distance x is

W = 1/2 kx²

where k is the spring constant.

This means the work needed to change the length of this spring by 10 cm = 0.01 m is

W = 1/2 (200 N/m) (0.01 m)² = 0.01 J

and by 15 cm = 0.015 m is

W' = 1/2 (200 N/m) (0.015 m)² = 0.0225 J

Then the total work performed on the spring by stretching from 10 cm to 15 cm is

∆W = W' - W = 0.0225 J - 0.01 J = 0.0125 J

5 0

3 years ago

Ruby is in miami an texts her cousin, Xavier, in Seattel. her clock says 1:00 am

mihalych1998 [28]

Answer:

so its 10pm in seattle

Explanation:

5 0

3 years ago

Calculate the pressure on the bottom of a swimming pool 3. 5 m deep. How does the pressure compare with atmospheric pressure, 10

Nostrana [21]

Answer:

1.343 atm

Explanation:

The mass of water above 1 square meter of swimming pool bottom is ...

M = (3.5 m)·(1000 kg/m^3) = 3500 kg/m^2

Then the force exerted by the water on the pool bottom is ...

F = Mg = (3500 kg/m^2)(9.8 m/s^2) = 34300 N/m^2 = 34300 Pa

Compared with atmospheric pressure, this is ...

34,300/10^5 = 0.343 . . . . atmospheres

Added to the atmospheric pressure on the water's surface, the total pressure on the pool bottom is 1.343 atmospheres.

5 0

4 years ago

A 0.015 kg marble moving to the right at 0.225 m/s makes an elastic head-on collision with a 0.030 kg shooter marble moving to t

bonufazy [111]

Answer:

0.09 m/s to the right

Explanation:

The principle of conservation of linear momentum states that in a system of colliding objects, the total momentum before collision is equal to the total momentum after collision, provided there is no external force.

The total momentum, in this question, is the sum of the momentum of the the two marbles. We assume velocity to the right is positive while to the left is negative. Thus, total initial momentum is

$P_i = 0.015\times0.225+ 0.030 \times(-0.180) = 0.002025$

After collision, the first marble goes left. Let the velocity of the other marble be v. Then we have

$P_f = 0.015\times(-0.315) + 0.030v = 0.030v - 0.004725$

Equating both momenta,

$P_i = P_f$

$0.002025 = 0.030v - 0.004725$

$v = 0.09 \text{ m/s}$

Hence, the larger goes right with a velocity of 0.09 m/s (since it has a positive sign).

4 0

3 years ago

A 12 kg block is released from the top of an incline that is 5.0 m long and makes an angle of 40.0º to the horizontal. A force o

Allushta [10]

Answer:

do u have a photo that comes w/ this? so i could help more :) ?

Explanation:

8 0

3 years ago