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yarga [219]
2 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]2 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]2 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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If the horse lifts a 50 kilogram mass a vertical distance of 3 meters in 2 seconds, what is the horse’s minimum power output?
yaroslaw [1]

Answer:

735watts

Explanation:

Power = Force*distance/time

Given

Force = mg = 50 * 9.8'

Force = 490N

distance = 3m

time = 2s

Substitute

Power = 490*3/2

Power = 1470/2

Power = 735Watts

Hence the minimum power output is 735watts

7 0
3 years ago
Best answer gets brainliest!!!!
qaws [65]
1 well.. the light when in a far distance seems to fade 
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4 0
3 years ago
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A bionic man running at 6.5 m/sec , east is acceleration at a uniform rate of 1.5 m/sec^2 east over a displacement of 100.0 m ea
tangare [24]
Given:
u = 6.5 m/s, initial velocity 
a = 1.5 m/s², acceleration
s = 100.0 m, displacement

Let v =  the velocity attained after the 100 m displacement.
Use the formula
v² = u² + 2as
v² = (6.5 m/s)² + 2*(1.5 m/s²)*(100 m) = 342.25 (m/s)²
v = 18.5 m/s

Answer: 18.5 m/s
5 0
3 years ago
A student sits on a pivoted stool while holding a pair of weights. The stool is free to rotate about a vertical axis with neglig
Blababa [14]

Answer:

<u>Please Mark As Brainliest!!</u>

a) 4.99 rad/sec b) 6.24 rad/sec c) 7.03 J

Explanation:

a)  If the student completes one turn in 1.26 sec, this is called the period of the movement.

If we take into account that the angle rotated during one turn is 2π rads, by definition of angular velocity, we can get this value as follows:

ω = Δθ / Δt = 2*π rad / 1.26 seg = 4.99 rad/sec.

b) As no external torques are acting on the system, the total angular momentum must be conserved, so we can write the following equation:

Li = Lf   ⇒  I₁ * ω₁  = I₂* ω₂

So, we can solve for ω₂, as follows:

ω₂ = (I₁ * ω₁) / I₂ = 6.24 rad/sec

c) Appying the work-energy theorem, we know that the work done by the student, must be equal to the change in the kinetic energy, which in this case is only rotational, so we can write:

W = 1/2 I₂* ω₂² - 1/2 I₁ ω₁²

W =1/2 ((2.25 kg.m² * (6.24)²) (rad/sec)² - (1.8 kg.m²* (4.99)²) (rad/sec)²)  

W = 7.03 J

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