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dolphi86 [110]
3 years ago
5

Eliza went for a swim. When she came out of the water, she felt cold. At the moment, a strong wind blew, and she felt even colde

r. Explain why did she felt even colder when a string wind blew.
Physics
1 answer:
yaroslaw [1]3 years ago
4 0
She did because when you are in water for a long period of time it might feel warm but when you come out it is going to feel cold again just like how you felt going in the water
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A girl at an airport rolls a ball north on a moving walking that moves east. If the ball’s speed with respect to the walkway is
sweet-ann [11.9K]

We have to add two vectors.

Vector #1: 0.15 m/s north

Vector #2: 1.50 m/s east

Their sum:

Magnitude: √(0.15² + 1.50²)

Magnitude = √(0.0225+2.25)

Magnitude = √2.2725

Magnitude = <em>1.5075 m/s</em>

Direction = arctan(0.15/1.50) north of east

Direction = <em>5.71° north of east</em>

4 0
3 years ago
We use electrical devices leopard deuce motion, light, and sound. Which aspect of energy explains why these devices are possible
poizon [28]

Answer:

B) Energy can change from one form to another

Explanation:

It goes from electrical energy from the mains into motion, light and sound

3 0
3 years ago
How much force is applied if a 130kg mass is accelerated at 5 m/s^2?​
STALIN [3.7K]

Answer:

650N

Explanation:

f= ma

130kg x 5 m/s^2 = 650N

5 0
2 years ago
Consider the Uniform Circular Motion Gizmo configured as shown. Notice that, under the current settings, |a|=0.50m/s2. What chan
Eddi Din [679]

To increase the centripetal acceleration to 2.00 m/s^2, you can double the speed or decrease the radius by 1/4

Explanation:

An object is said to be in uniform circular motion when it is moving at a constant speed in a circular path.

The acceleration of an object in uniform circular motion is called centripetal acceleration, and it is given by

a=\frac{v^2}{r}

where

v is the speed of the object

r is the radius of the circular path

In the problem, the original centripetal acceleration is

a=0.50 m/s^2

We want to increase it by a factor of 4, i.e. to

a'=2.00 m/s^2

We notice that the centripetal acceleration is proportional to the square of the speed and inversely proportional to the radius, so we can do as follows:

- We can double the speed:

v' = 2v

This way, the new acceleration is

a'=\frac{(2v)^2}{r}=4(\frac{v^2}{r})=4a

so, 4 times the original acceleration

- We can decrease the radius to 1/4 of its original value:

r'=\frac{1}{4}r

So the new acceleration is

a'=\frac{v^2}{(r/4)}=4(\frac{v^2}{r})=4a

so, the acceleration has increased by a factor 4 again.

Learn more about centripetal acceleration:

brainly.com/question/2562955

brainly.com/question/6372960

#LearnwithBrainly

5 0
2 years ago
A tennis ball is dropped from 1.13 m above the
Alex73 [517]

Answer:

-4.71 m/s

Explanation:

Given:

y₀ = 1.13 m

y = 0 m

v₀ = 0 m/s

a = -9.8 m/s²

Find: v

v² = v₀² + 2a (y − y₀)

v² = (0 m/s)² + 2(-9.8 m/s²) (0 m − 1.13 m)

v = -4.71 m/s

7 0
2 years ago
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