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ivanzaharov [21]
3 years ago
9

Why do you often freeze after a shower? Help mee

Physics
1 answer:
V125BC [204]3 years ago
5 0
When water evaporates of your body it cools you down, hence why after a shower when the water droplets begin to evaporate you feel cold
You might be interested in
If your front lawn is 18.0 feet wide and 20.0 feet long, and each square foot of lawn accumulates 1050 new snow flakes every min
Ivanshal [37]
<span>First, we need to determine the entire area of your front line by multiplying its length times its width.
18.0*20.0 = 360.0 square feet
We can use the rate of accumulation of snow, combined with this figure, to determine how much snow accumulates on your lawn per minute.
360.0 sq ft * 1050 flakes/min/sq ft = 378,000 flakes/min
We can then use the mass of a snowflake to calculate total snow accumulation per minute.
378,000 flakes/min * 2.00 mg/flake = 756,000 mg/min
Finally, we can use this number to determine accumulation per hour.
756,000 mg/min * 60 min/hr =
45,360,000 mg/hr</span>
8 0
3 years ago
A ray of light traveling in glass (refractive index na) reflects at the flat interface between the glass and water (refractive i
pickupchik [31]

Answer:

θr  will increase.

Explanation:

We know that when light travel from one medium to another medium the relationship between refractive index and angle given as

na. sinθa = nb . sinθr

sin \theta _r=\dfrac{n_asin\theta _a}{n_b}

If na is increased (by using a different type of glass) but θa is kept the same then sinθr will also increase.

So answer is θr  will increase.

5 0
3 years ago
A baseball player throws a ball horizontally at the same moment another baseball player jobs a second ball straight down from th
attashe74 [19]

I don't know what you mean when you say he "jobs" the other ball, and the answer to this question really depends on that word.

I'm going to say that the second player is holding the second ball, and he just opens his fingers and lets the ball <u><em>drop</em></u>, at the same time and from the same height as the first ball.

Now I'll go ahead and answer the question that I've just invented:

Strange as it may seem, <em>both</em> balls hit the ground at the <em>same time</em> ... the one that's thrown AND the one that's dropped.  The horizontal speed of the thrown ball has no effect on its vertical acceleration, so both balls experience the same vertical behavior.

And here's another example of the exact same thing:

Say you shoot a bullet straight out of a horizontal rifle barrel, AND somebody else <em>drops</em> another bullet at exactly the same time, from a point right next to the end of the rifle barrel.  I know this is hard to believe, but both of those bullets hit the ground at the same time too, just like the baseballs ... the bullet that's shot out of the rifle and the one that's dropped from the end of the barrel.

7 0
3 years ago
A 1.80- kgkg monkey wrench is pivoted 0.250 mm from its center of mass and allowed to swing as a physical pendulum. The period f
madam [21]

Answer:

A. 1.125×10^-7 kgm^2

B. 6.64875 rad/s

Explanation:

The moment of inertia is defined as a quantity expressing a body's tendency to resist angular acceleration, which is the sum of the products of the mass of each particle in the body with the square of its distance from the axis of rotation.

A. Moment of inertia = m1✖r1^2

=1.80 x (2.5x10^-4)^2

= 6.25x10^8 x 1.80

= 1.125 x 10^-7 kgm^2.

B. w is represented as Angular speed.

V is velocity, T is time in period.

Velocity= distance / time.

V= 2.5x10^-4 / 0.940

V= 2.6595 metre per seconds

w= v/r

w= 2.6595 / 0.400

w= 6.64875 rad/s.

6 0
3 years ago
In the future, people will only enjoy one sport: Electrodes. In this sport, you gain points when you cause metallic discs hoveri
pochemuha

Answer:

  • Disk C and Disk D

Explanation:

The total charge in the disks

q_1 + q_2 = q_{total}

must be conserved before and after bringing them together.

Lets equate the sum of the initial charge with the sum of the final for the disk:

q_{1_i} + q_{2_i} = q_{1_f} + q_{2_f} = 2 * (+8.5) \mu C

q_{1_i} + q_{2_i} = +17 \mu C

So, the initial charges must sum +17 μC.

Now, as there are no charges over +17 μC, this means that both charges must be positive.

As the only positive charges are q_C and q_D, this disk must be the ones we are looking for. And, as we can see, they sum 17 μC:

q_{C} + q_{D} = 5 \mu C + 12 \mu C = 17 \mu C

3 0
3 years ago
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