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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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A rocket ship has several engines and thrusters. While the Solid Rocket Booster (SRB) and main engines only work together during
Katarina [22]

A rocket ship is accelerated by the SRB and the main engines for 2.0 minutes and the main engines for 8.5 minutes after the launch. The acceleration of the ship during the first 2.0 minutes is 11 m/s² (D).

A rocket ship has several engines and thrusters. We can divide its initial movement into 2 parts:

  • From t = 0 min to t = 2.0 min, the SRB and the main engines act together and the speed goes from 0 m/s (rest) to 1341 m/s.
  • From t = 2.0 min to t = 8.5 min, the main engines alone accelerate the ship form 1341 m/s to 7600 m/s.

We want to know the acceleration in the first part (first 2.0 minutes). We need to consider that:

  • The speed increases from 0 m/s to 1341 m/s.
  • The time elpased is 2.0 min.
  • 1 min = 60 s.

The acceleration of the ship during the first 2.0 minutes is:

a = \frac{\Delta v }{t} ) \frac{(1341m/s-0m/s)}{2.0min} \times \frac{1min}{60s}  = 11 m/s^{2}

A rocket ship is accelerated by the SRB and the main engines for 2.0 minutes and the main engines for 8.5 minutes after the launch. The acceleration of the ship during the first 2.0 minutes is 11 m/s² (D).

Learn more: brainly.com/question/16274121

3 0
2 years ago
2. If you want 0. 250 a (250 milliamps) to flow around a circuit with a resistance of 400 ohms, what voltage do you need?
grigory [225]

Answer:

0.000625 V

Explanation:

The formula linking current , resistance and voltage is :

V = I/R

Voltage = Current / Resistance

Now we substitute values given in question :

Voltage = 0.250 / 400

Voltage (V) = 0.000625

Our final answer is 0.000625 V

Hope this helped and have a good day

5 0
1 year ago
An object was thrown at a certain angle above the ground.It reaches a maximum height of 42.50 meters and hits back the ground 76
aksik [14]

Answer:

Explanation:  Please see my attached calculations.

7 0
3 years ago
A star with a large luminosity would have a relatively _____ absolute magnitude.
Mademuasel [1]
A star with large luminosity would have a relatively low absolute magnitude. Absolute magnitude is a number that tells how bright a star is from the Earth. However, this scale is backwards and logarithmic, so having a large absolute magnitude value means that the star is faint.
3 0
3 years ago
Students in an introductory physics lab are performing an experiment with a parallel-plate capacitor made of two circular alumin
Alex777 [14]

Answer:

0.92 μC

Explanation:

In a parallel-plate capacitor, the electric field formed is equal to the charge density divited by the vacuum permisivity e0, as there are no dielectric between the plates. e0 is equal to 8.85*10^-12 C^2/Nm^2. The charge density is the total charge of each individual plate divided by its area. Then, the maximum charge allowed will be equal to:

E = \frac{o}{e_0} = \frac{Q}{Ae_0} \\ Q = E*A*e_0 = 3*10^6 N/C * (0.25*\pi *(0.21m)^2)*8.85*10^{-12}C^2/Nm^2 = 9.196 *10^{-7} C

or 0.92 μC

7 0
3 years ago
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