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NISA [10]
3 years ago
13

Weather changes occur primarily because of

Physics
2 answers:
OleMash [197]3 years ago
6 0
<span>Weather occurs primarily due to air pressure, temperature and moisture differences between one place to another. Therefore, the correct answer that would best complete the given statement above would be second option. </span>Weather changes occur primarily because of <span>heat exchange between the sun and atmosphere. Hope this helps.</span>
NikAS [45]3 years ago
4 0

Answer:

Incoming solar radiations

Explanation:

Weather is defined as the condition of atmosphere at any given time. Weather depends upon the temperature, precipitation, air pressure, and cloud covering.

Daily changes in weather comes because of storms and winds while seasonal changes in weather comes because earth revolves around the sun.

As earth is not flat so, when sun shines near the equator region of earth it becomes warm but the polar region at that time is cold because less solar radiations reach that region. Hence, Incoming solar radiations will be the right answer.

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What is the volume of water in 150ml of the 35% w/w of sucrose solution with a specific gravity of 1.115?
pentagon [3]

Answer:108.71 mL

Explanation:

Given

Volume of sample V=150 mL

concentration of sucrose solution 35 % w/w i.e. In  100 gm of sample 35 gm is sucrose

specific gravity =1.115

Density of solution \rho _s=1.115\times density\ of\ water

Thus

\rho _s=1.15\times 1 gm/mL=1.115\ gm/mL

mass of sample M=1.115\times 150=167.25\ gm

mass of sucrose m_s=0.35\times 167.25=58.53\ gm

mass of Water m_w=108.71 gm

Volume of water =108.71\times 1=108.781 mL

7 0
3 years ago
In a double-slit experiment, if the central diffraction peak contains 13 interference fringes, how many fringes are contained wi
Ronch [10]

Answer:

Explanation:

Width of central diffraction peak is given by the following expression

Width of central diffraction peak= 2 λ D/ d₁

where d₁ is width of slit and D is screen distance and λ is wave length.

Width of other fringes become half , that is each of  secondary diffraction fringe is equal to

λ D/ d₁

Width of central interference  peak is given by the following expression

Width of  each of  bright fringe =  λ D/ d₂

where d₂ is width of slit and D is screen distance and λ is wave length.

Now given that the central diffraction peak contains 13 interference fringes

so ( 2 λ D/ d₁)  /  λ  D/ d₂ = 13

then (  λ D/ d₁)  /  λ  D/ d₂ = 13 / 2

= 6.5

no of fringes  contained within each secondary diffraction peak = 6.5

6 0
3 years ago
10. How does surface area affect the amount of air resistance an object experiences?
ikadub [295]

Answer:cross-sectional area, and thus surface area, increases the amount of air resistance an object experiences

Explanation:

6 0
3 years ago
The flywheel of an engine has moment of inertia 2.50 kg m2 about its rotation axis. What constant torque is required to bring it
MrRissso [65]

Answer:

Explanation:

From the question we are told that

   The moment of inertia is  I = 2.50 \ kg \cdot m^2

    The final  angular speed is w_f =  400 rev/min  =  \frac{400 * 2\pi}{60}  = 41.89 \ rad/s

     The time taken is  t =  8.0 s

      The initial angular speed is  w_i  =  0\ rad/s

Generally the average angular acceleration is mathematically represented as

        \alpha  =  \frac{w_f - w_i }{t}

=>     \alpha  =  \frac{41.89}{8}

=>      \alpha  = 5.24 \ rad/s^2

Generally the torque is mathematically represented as

   \tau  =  I  *  \alpha

=>    \tau   =  5.24 *  2.50

=>     \tau   =  13.09 \  N \cdot m

5 0
3 years ago
I need help with 1-10
e-lub [12.9K]

Answer:

1) C. Energy

2) A. Joule

3) D. Joule

4) B. Potential

5) A. Greater

6) C. Largest

7) A. Speed and mass

8) A. Kinetic

9) A. Kinetic

10) D. Height and mass

3 0
3 years ago
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