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castortr0y [4]
3 years ago
8

Question 1

Physics
1 answer:
Yuki888 [10]3 years ago
4 0

Answer:

2.Weather patterns are similar along the coast of a continent but different when moving from the coast to the center of a continent.

You might be interested in
What is the magnitude of the speed of a person at the equator due to rotational speed of the Earth?
notka56 [123]
The circumference of the Earth at the equator is listed as 24,901 miles.
So his speed is

                     24,901 miles per day.

Convert it to units that we have a better feel for:

                   (24,901 mi/da) x (1 da / 24 hrs)

             =    (24,901 / 24)  (miles/hour)

             =  about  1,038 miles per hour.

You'll find a huge number of people on the internet these days,
telling you that you could not be moving at that speed and not
feel it, so therefore the Earth is not spinning, and it's not a globe.

I have a lot of feelings and comments about those people, their
lines of reasoning, and their levels of education and intelligence,
so don't get me started.

I just want to guarantee you that everything you're learning about
the Earth and the solar system in school is well founded, and it's
all based on the life's work of some of the smartest people of the
past 300 years of human history.  Everything you're taught about
the Earth has good reasons behind it, whereas those other people
have nothing.

A person on Earth's equator is moving from west to east at roughly
1,038 miles per hour, relative to any point on the Earth's rotation axis.
4 0
3 years ago
In a fast-pitch softball game the pitcher is impressive to watch, as she delivers a pitch by rapidly whirling her arm around so
svetlana [45]

Answer:

(a). The magnitude of the total acceleration of the ball is 239.97 m/s².

(b). The angle of the total acceleration relative to the radial direction is 11.0°

Explanation:

Given that,

Radius of the circle = 0.681 m

Angular acceleration = 67.7 rad/s²

Angular speed =18.6 rad/s

We need to calculate the centripetal acceleration of the ball

Using formula of centripetal acceleration

a_{c}=\omega^2\times r

Put the value into the formula

a_{c}=(18.6)^2\times0.681

a_{c}=235.5\ m/s^2

We need to calculate the tangential acceleration of the ball

Using formula of tangential acceleration

a_{t}=r\alpha

Put the value into the formula

a_{t}=0.681\times67.7

a_{t}=46.104\ m/s^2

(a). We need to calculate the magnitude of the total acceleration of the ball

Using formula of total acceleration

a=\sqrt{a_{c}^2+a_{t}^2}

Put the value into the formula

a=\sqrt{(235.5)^2+(46.104)^2}

a=239.97\ m/s^2

(b). We need to calculate the angle of the total acceleration relative to the radial direction

Using formula of the direction

\theat=\tan^{-1}(\dfrac{a_{t}}{a_{c}})

Put the value into the formula

\theta=\tan^{-1}(\dfrac{46.104}{235.5})

\theta=11.0^{\circ}

Hence, (a). The magnitude of the total acceleration of the ball is 239.97 m/s².

(b). The angle of the total acceleration relative to the radial direction is 11.0°

5 0
4 years ago
A meter stick is held vertically above your hand, with the lower end between your thumb and first finger. On seeing the meter st
inessss [21]

Answer:

t=0.193s

Explanation:

What is said is that the meter fell d=18.3cm=0.183m under the action of gravity. We can use the formula for accelerated motion:

d=v_0t+\frac{at^2}{2}

Since it departed from rest it will mean that:

d=\frac{at^2}{2}

So our time will be:

t=\sqrt{\frac{2d}{a}}

Which for our values is:

t=\sqrt{\frac{2(0.183m)}{(9.81m/s^2)}}=0.193s

7 0
3 years ago
A violin string E vibrates faster then another violin string G when bowed. Which string produced a sound with higher frequency?​
Lisa [10]

Violin string E

!!!!!!!!!!!!!!

5 0
3 years ago
Calculate the amount of force required to produce motion in a car of 1000kg with an acceleration of 5m/s square.​
atroni [7]

Answer:

5000 N

Explanation:

F=ma

=1000*5

=5000N

4 0
3 years ago
Read 2 more answers
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