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yarga [219]
3 years ago
15

1. How does an area’s weather differ from the area’s climate? (1 point). A.Weather involves temperature and precipitation and cl

imate involves only temperature.. B. An area’s weather depends on where it is located on Earth and the area’s climate does not.. C.An area’s weather does not change very much and an area’s climate changes many times.. D.Weather is the area’s day-to-day conditions and climate is the area’s average conditions.. 2. All of the following factors contribute to Earth’s climate EXCEPT (1 point) . latitude.. longitude.. transport of heat by winds.. shape and elevation of landmasse
Physics
2 answers:
Deffense [45]3 years ago
5 0
An area's weather differs from an area's climate because weather is the area's daily conditions. The climate is the average conditions of the area.

Longitude does not contribute to the earth's climate. Longitude are imaginary lines that have no impact.
makkiz [27]3 years ago
4 0
1. "Weather is the area’s day-to-day conditions and climate is the area’s average conditions" is how <span>an area’s weather differ from the area’s climate. The correct option among all the options given in the question is option "D".

2. </span>All of the following factors contribute to Earth’s climate except longitude. <span>The correct option among all the options given in the question is the second option.</span>
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Long as there is a sufficient source of energy within the Sun.
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4 years ago
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You throw a ball with a speed of 25.0 m/s at an angle of 40.0â—¦ above the horizontal directly toward a wall. The wall is 22.0 m
Alina [70]

The ball takes about 1.15 seconds to reach the wall

<h3>Further explanation</h3>

Acceleration is rate of change of velocity.

\large {\boxed {a = \frac{v - u}{t} } }

\large {\boxed {d = \frac{v + u}{2}~t } }

<em>a = acceleration ( m/s² )</em>

<em>v = final velocity ( m/s )</em>

<em>u = initial velocity ( m/s )</em>

<em>t = time taken ( s )</em>

<em>d = distance ( m )</em>

Let us now tackle the problem!

This problem is about Projectile Motion

<u>Given:</u>

initial speed = u = 25 m/s

angle of speed = θ = 40.0°

horizontal distance = x = 22.0 m

<u>Unknown:</u>

time taken by the ball = t = ?

<u>Solution:</u>

<em>We will use this following formula to find the time taken by the ball to reach the wall:</em>

x = u_x t

x = u \cos \theta ~t

22 = 25 \cos 40^o ~t

t = 22 \div ( 25 \cos 40^o )

t \approx 1.15 \texttt { seconds}

\texttt{ }

<h2>Conclusion :</h2>

The ball takes about 1.15 seconds to reach the wall

\texttt{ }

<h3>Learn more</h3>
  • Velocity of Runner : brainly.com/question/3813437
  • Kinetic Energy : brainly.com/question/692781
  • Acceleration : brainly.com/question/2283922
  • The Speed of Car : brainly.com/question/568302

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Kinematics

Keywords: Velocity , Driver , Car , Deceleration , Acceleration , Obstacle , Projectile , Motion , Horizontal , Vertical , Release , Point , Ball , Wall

7 0
3 years ago
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A ball with 100 J of PE is released from a height of 10 m. What will be the KE of the ball at 5
harkovskaia [24]

Answer:

The kinetic energy is: 50[J]

Explanation:

The ball is having a potential energy of 100 [J], therefore

PE = [J]

The elevation is 10 [m], and at this point the ball is having only potential energy, the kinetic energy is zero.

E_{p} =m*g*h\\where:\\g= gravity[m/s^{2} ]\\m = mass [kg]\\m= \frac{E_{p} }{g*h}\\ m= \frac{100}{9.81*10}\\\\m= 1.01[kg]\\\\

In the moment when the ball starts to fall, it will lose potential energy and the potential energy will be transforme in kinetic energy.

When the elevation is 5 [m], we have a potential energy of

P_{e} =m*g*h\\P_{e} =1.01*9.81*5\\\\P_{e} = 50 [J]\\

This energy is equal to the kinetic energy, therefore

Ke= 50 [J]

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If you were to stir salt in water until the salt disappeared, which is the solute?
emmainna [20.7K]
Salt is the solute in that situation

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How long does it take for the Earth to make a complete revolution around the sun?
timurjin [86]

365 days so a year basically

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