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barxatty [35]
3 years ago
5

Precipitation clouds are observed at a distance of about 720 miles to the west of a city. The clouds are seen moving westwards a

t an average rate of 30 miles per hour. Which of these statements best predicts the weather in the city within 24 hours?
a. Rain is likely.
b. Rain is unlikely.
c. Snow is likely.
d. Snow is unlikely.
Physics
2 answers:
tankabanditka [31]3 years ago
3 0
This question may require more information for an accurate answer. If the clouds are observed west of the city and they are moving in a westerly direction it means that the clouds are moving away. Precipitation refers to rainfall, sleet, hail or snow. Since no reference was made to the temperature, season or location of the city it could be rain or snow clouds. The clouds are moving away from the city so the possible answers could be b. or d. Furthermore, by dividing the distance by the speed one would get an answer of twenty four hours. So even if the clouds were moving towards the city it would take a day to reach and precipitation would not be likely in 24 hrs.
nalin [4]3 years ago
3 0

Answer:

A) Rain is likely.

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            <span> Using conservation of energy

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A block weighting 400kg rests on a horizontal surface and support on top of it another block of weight 100kg placed on the top o
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The horizontal force applied to the block is approximately 1,420.84 N

The known parameters;

The mass of the block, w₁ = 400 kg

The orientation of the surface on which the block rest, w₁ = Horizontal

The mass of the block placed on top of the 400 kg block, w₂ = 100 kg

The length of the string to which the block w₂ is attached, l = 6 m

The coefficient of friction between the surface, μ = 0.25

The state of the system of blocks and applied force = Equilibrium

Strategy;

Calculate the forces acting on the blocks and string

The weight of the block, W₁ = 400 kg × 9.81 m/s² = 3,924 N

The weight of the block, W₂ = 100 kg × 9.81 m/s² = 981 N

Let <em>T</em> represent the tension in the string

The upward force from the string = T × sin(θ)

sin(θ) = √(6² - 5²)/6

Therefore;

The upward force from the string = T×√(6² - 5²)/6

The frictional force = (W₂ - The upward force from the string) × μ

The frictional force, F_{f2} = (981 - T×√(6² - 5²)/6) × 0.25

The tension in the string, T = F_{f2} × cos(θ)

∴ T = (981 - T×√(6² - 5²)/6) × 0.25 × 5/6

Solving, we get;

T = \dfrac{5886}{\sqrt{6^2 - 5^2} + 28.8} \approx 183.27

Frictional \ force, F_{f2} = \left (981 -  \dfrac{5886}{\sqrt{6^2 - 5^2} + 28.8}  \times \dfrac{\sqrt{6^2 - 5^2} }{6} \times  0.25 \right) \approx 219.92

The frictional force on the block W₂, F_{f2} ≈ 219.92 N

Therefore;

The force acting the block w₁, due to w₂ F_{w2} = 219.92/0.25 ≈ 879.68

The total normal force acting on the ground, N = W₁ + \mathbf{F_{w2}}

The frictional force from the ground, \mathbf{F_{f1}} = N×μ + \mathbf{F_{f2}} = P

Where;

P = The horizontal force applied to the block

P = (W₁ + \mathbf{F_{w2}}) × μ + \mathbf{F_{f2}}

Therefore;

P = (3,924 + 879.68) × 0.25 + 219.92 ≈ 1,420.84

The horizontal force applied to the block, P ≈ 1,420.84 N

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Answer:

The car that receives more kinetic energy is the small economy car.

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M is the mass of an object

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If equal forces were applied to the two cars, then the velocity of each car will be calculated as follows.

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Also assuming equal distance, equal force and assuming equal time for both cars.

The above equation will reduce to, K.E = k/m

Where k = 0.5*F²t², which is equal in both cars.

Thus, Kinetic energy will depend only on the mass of each car.

From the above expression, Kinetic Energy received by each car is inversely proportional to the mass of the car.

A small economy car (low mass)  will receive more kinetic energy while a limousine (high mass) car will receive less kinetic energy.

Therefore, the car that receives more kinetic energy is the small economy car.

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