How would taping a coin to the balloon affect the overall motion of the balloon?

Given a temperature of 300 Kelvin, what is the approximate temperature in degrees Celsius?

S_A_V [24]

Jun 16, 2012 - Given a temperature of 300 Kelvin, what is the approximate temperature in degrees Celsius? –73°C 27°C 327°C 673°C.

3 0

4 years ago

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Many web sites describe how to add wires to your clothing to keep you warm while riding your motorcycle. The wires are added to

Tomtit [17]

Answer:

$P=42.075W$

Explanation:

The power provided by a resistor (wire in this case) is given by:

$P=\frac{V^2}{R}$ .

The resistance of a wire is given by:

$R=\frac{\rho L}{A}$

Where for the resistivity the one of the copper should be used: $\rho=1.68\times10^{-8}\Omega m$ .

The area A is that of a circle, which written in terms of its diameter is:

$A=\pi r^2=\pi (d/2)^2=\frac{\pi d^2}{4}$

Putting all together:

$P=\frac{AV^2}{\rho L}=\frac{\pi d^2V^2}{4\rho L}$

Which for our values is:

$P=\frac{\pi (0.00025m)^2(12V)^2}{4(1.68\times10^{-8}\Omega m)(10m)}=42.075W$

7 0

4 years ago

A camera with a 50.0-mm focal length lens is being used to photograph a person standing 3.00 m away. (a) How far from the lens m

kirill [66]

a) 50.8 mm

b) The whole image (1:1)

c) It seems reasonable

Explanation:

a)

To project the image on the film, the distance of the film from the lens must be equal to the distance of the image from the lens. This can be found by using the lens equation:

$\frac{1}{f}=\frac{1}{p}+\frac{1}{q}$

where

f is the focal length of the lens

p is the distance of the object from the lens

q is the distance of the image from the lens

In this problem:

f = 50.0 mm = 0.050 m is the focal length (positive for a convex lens)

p = 3.00 m is the distance of the person from the lens

Therefore, we can find q:

$\frac{1}{q}=\frac{1}{f}-\frac{1}{p}=\frac{1}{0.050}-\frac{1}{3.00}=19.667m^{-1}\\q=\frac{1}{19.667}=0.051 m=50.8 mm$

b)

Here we need to find the height of the image first.

This can be done by using the magnification equation:

$\frac{y'}{y}=-\frac{q}{p}$

where:

y' is the height of the image

y = 1.75 m is the height of the real person

q = 50.8 mm = 0.0508 m is the distance of the image from the lens

p = 3.00 m is the distance of the person from the lens

Solving for y', we find:

$y'=-\frac{qy}{p}=-\frac{(0.0508)(1.75)}{3.00}=-0.0296 m=-29.6mm$

(the negative sign means the image is inverted)

Therefore, the size of the image (29.6 mm) is smaller than the size of the film (36.0 mm), so the whole image can fit into the film.

c)

This seems reasonable: in fact, with a 50.0 mm focal length, if we try to take the picture of a person at a distance of 3.00 m, we are able to capture the whole image of the person in the photo.

3 0

3 years ago

The kicker now kicks the ball with the same speed as in the number of 4,but at 60.0°from the horizontal or 30.0° from the vertic

stealth61 [152]

Answer:

Explanation:

1) The time of flight equation for projectile motion can be used here to find total time in air.

t = 2vsin∅ / g

where v is speed, Ф is launch angle

t = 2×4×sin 60 / 9.8

t = 0.71 seconds

2) Distance where it hit the ground is called as range and has the following standard equation

D = v² sin2Ф/g

D = 4²sin 2×60 / 9.8

D = 1.41m

3) Maximum elevation is maximum time reached

h = v² sin²Ф / 2g

h = 4²sin² 60 / 2*9.8

h = 0.61 m

3 0

3 years ago

Why don’t trade winds blow straight toward the equator Apex

rosijanka [135]

Trade winds near equator blows in curve path instead of straight path. This is because of earth rotation. This effect of earth rotation that cause wind to move in curve motion is called Coriolis effect. These kind of wind blows at the northeast of the North hemisphere and southeast of the South hemisphere. The trade wind are warm and it blows due to rising of hot air from equator.

6 0

3 years ago