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3 years ago

What type of clouds usually accompany cold fronts?

Physics

2 answers:

ASHA 777 [7]3 years ago

6 0

Answer: Cumulus

Explanation: Most large cloud fronts are made up of cumulus clouds, large storm clouds are cumulonimbus clouds.

alina1380 [7]3 years ago

4 0

The correct answer is A

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A wheel turns through 5.5 revolutions while being accelerated from rest at 20rpm/s.(a) What is the final angular speed ? (b) How

alina1380 [7]

Answer:

(a) The final angular speed is 12.05 rad/s

(b) The time taken to turn 5.5 revolutions is 5.74 s

Explanation:

Given;

number of revolutions, θ = 5.5 revolutions

acceleration of the wheel, α = 20 rpm/s

number of revolutions in radian is given as;

θ = 5.5 x 2π = 34.562 rad

angular acceleration in rad/s² is given as;

$\alpha = \frac{20 \ rev}{min} *\frac{1}{s} *(\frac{2\pi \ rad}{1 \ rev } *\frac{1 \ min}{60 \ s}) \\\\\alpha = 2.1 \ rad/s^2$

(a)

The final angular speed is given as;

$\omega _f^2 = \omega_i ^2 + 2\alpha \theta\\\\\omega _f^2 = 0 + 2\alpha \theta\\\\\omega _f^2 = 2\alpha \theta\\\\\omega _f = \sqrt{2\alpha \theta}\\\\ \omega _f = \sqrt{2(2.1) (34.562)}\\\\ \omega _f = 12.05 \ rad/s$

(b) the time taken to turn 5.5 revolutions is given as

$\omega _f = \omega _i + \alpha t\\\\12.05 = 0 + 2.1t\\\\t = \frac{12.05}{2.1} \\\\t = 5.74 \ s$

3 0

2 years ago

A 1.50-kg iron horseshoe initially at 550°C is dropped into a bucket containing 25.0 kg of water at 20.0°C. What is the final te

Ber [7]

Answer:

Te = 23.4 °C

Explanation:

Given:-

- The mass of iron horseshoe, m = 1.50 kg

- The initial temperature of horseshoe, Ti_h = 550°C

- The specific heat capacity of iron, ci = 448 J/kgC

- The mass of water, M = 25 kg

- The initial temperature of water, Ti_w = 20°C

- The specific heat capacity of water, cw = 4186 J/kgC

Find:-

What is the final temperature of the water–horseshoe system?

Solution:-

- The interaction of horseshoe and water at their respective initial temperatures will obey the Zeroth and First Law of thermodynamics. The horseshoe at higher temperature comes in thermal equilibrium with the water at lower temperature. We denote the equilibrium temperature as (Te) and apply the First Law of thermodynamics on the system:

m*ci*( Ti_h - Te) = M*cw*( Te - Ti_w )

- Solve for (Te):

m*ci*( Ti_h ) + M*cw*( Ti_w ) = Te* (m*ci + M*cw )

Te = [ m*ci*( Ti_h ) + M*cw*( Ti_w ) ] / [ m*ci + M*cw ]

- Plug in the values and evaluate (Te):

Te = [1.5*448*550 + 25*4186*20 ] / [ 1.5*448 + 25*4186 ]

Te = 2462600 / 105322

Te = 23.4 °C

7 0

3 years ago

Read 2 more answers

While seeds are forming, another plant part forms around them. Animals eat this part and carry the seeds to a new spot.

Eddi Din [679]

Answer:

<em>The correct option is D) fruit</em>

Explanation:

Both the gymnosperm and the angiosperms plants produce seed. Seeds are an essential part of reproduction in plants.

After fertilization has occurred, the ripened ovule is the part which turns into a seed. The size of the seeds depends on the type if plant.

After fertilization, the ovary forms the fruit. As the ovule is present in the ovule hence after fertilization, the seed which was made from the ovule gets enclosed in the fruit which was made from the ovary.

6 0

3 years ago

Volleyball was invented in 1905.

Lady_Fox [76]

Answer:

REALLY??

Explanation:

8 0

2 years ago

Read 2 more answers

The tires on your truck have 0.35 m radius. In a straight line, you drive 2600 m. What is the angular displacement of the tire,

Travka [436]

1) In a circular motion, the angular displacement $\theta$ is given by
$\theta = \frac{S}{r}$
where S is the arc length and r is the radius. The problem says that the truck drove for 2600 m, so this corresponds to the total arc length covered by the tire: $S=2600 m$ . Using the information about the radius, $r=0.35 m$ , we find the total angular displacement:
$\theta = \frac{2600 m}{0.35 m} =7428 rad$

2) If we put larger tires, with radius $r=0.60 m$ , the angular displacement will be smaller. We can see this by using the same formula. In fact, this time we have:
$\theta = \frac{2600 m}{0.60 m}=4333 rad$

8 0

3 years ago