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2 years ago

Expain how a hurricane develops

Physics

1 answer:

Firlakuza [10]2 years ago

6 0

Answer:

A hurricane starts to form when warm air rises and then that warm air turns into cooler air. After a while of this happening large clouds and thunderstorms are made. The clouds and thunderstorms keep growing and rotating from earth's Coriolis Effect. Then a hurricane forms.

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Nicole throws a ball at 25 m/s at an angle of 60 degrees abound the horizontal. What is the range of the ball?

just olya [345]

Answer:

the range or the ball is 48.81 m

Explanation:

given;

Nicole throws a ball at 25 m/s at an angle of 60 degrees abound the horizontal.

find:

What is the range of the ball?

solution:

let Ф = 25°

Vo = 25 m/s

consider x-motion using time of fight: x = Vox * t

where x = R = range

t = 2 Voy

R = Vo² sin (2Ф)

plugin values into the formula:

R = (25)² sin (2*25)

9.81

R = 48.81 m

therefore, the range or the ball is 48.81 m

4 0

4 years ago

Read 2 more answers

Spaceship 1 and spaceship 2 have equal masses of 200 kg. Spaceship 1 has a speed of 0 m/s, and spaceship 2 has a speed of 10 m/s

m_a_m_a [10]

Answer:

2000 kg m/s

Explanation:

The momentum of an object is a vector quantity whose magnitude is given by

$p=mv$

where

m is the mass of the object

v is the velocity of the object

and its direction is the same as the velocity.

In this problem, we have:

- Spaceship 1 has

m = 200 kg (mass)

v = 0 m/s (zero velocity)

So its momentum is

$p_1 =(200)(0)=0$

- Spaceship 2 has

m = 200 kg (mass)

v = 10 m/s (velocity)

So its momentum is

$p_2=(200)(10)=2000 kg m/s$

Therefore, the combined momentum of the two spaceships is

$p=p_1+p_2=0+2000=2000 kg m/s$

3 0

3 years ago

Suppose a gliding 2-kg cart bumps into, and sticks to, a stationary 5-kg cart. If the speed of the gliding cart before the colli

Thepotemich [5.8K]

Answer:

Therefore,

Final velocity of the coupled carts after the collision is

$v_{f}=4\ m/s$

Explanation:

Given:

Mass of Glidding Cart = m₁ = 2 kg

Mass of Stationary Cart = m₂ = 5 kg

Initial velocity of Glidding Cart = u₁ = 14 m/s

Initial velocity of Stationary Cart = u₂ = 0 m/s

To Find:

Final velocity of the coupled carts after the collision = $v_{f}=?$

Solution:

Law of Conservation of Momentum:

For a collision occurring between two objects in an isolated system, the total momentum of the two objects before the collision is equal to the total momentum of the two objects after the collision.

It is denoted by "p" and given by

Momentum = p = mass × velocity

Hence by law of Conservation of Momentum we hame

Momentum before collision = Momentum after collision

Here after collision both are stuck together so both will have same final velocity,

$m_{1}\times u_{1}+m_{2}\times u_{2}=(m_{1}+m_{2})\times v_{f}$

Substituting the values we get

$2\times 14 + 5\times 0 =(2+5)\times v_{f}$

$v_{f}=\dfrac{28}{7}=4\ m/s$

Therefore,

Final velocity of the coupled carts after the collision is

$v_{f}=4\ m/s$

8 0

3 years ago

The highest speed attained by a cyclist on level ground is 105 km/h. To attain this speed, the cyclist used a streamlined recumb

omeli [17]

Answer:

The acceleration is 0.133 m/s²

The time taken to build the final speed is 219.32 s

Explanation:

Given;

final speed of the cyclist, v = 105 km/h = 29.17 m/s²

distance traveled by the cyclist, d = 3.2 km = 3,200 m

initial speed, u = 0

Apply the following kinematic equation to determine the acceleration of the cyclist;

v² = u² + 2as

(29.17 )² = 0 + (2 x 3200)a

850.89 = 6400a

a = (850.89 ) / (6400)

a = 0.133 m/s²

The time taken to build up the final speed is given by;

v = u + at

29.17 = 0 + (0.133)t

t = 29.17 / 0.133

t = 219.32 s

4 0

3 years ago

Water runs out of a horizontal drainpipe at the rate of 135 kg/min. It falls 3.1 m to the ground. Assuming the water doesn't spl

Arlecino [84]

Answer:

The average force exerted by the water on the ground is 17.53 N.

Explanation:

Given;

mass flow rate of the water, m' = 135 kg/min

height of fall of the water, h = 3.1 m

the time taken for the water to fall to the ground;

$h = ut + \frac{1}{2}gt^2\\\\h = 0 + \frac{1}{2}gt^2\\\\t = \sqrt{\frac{2\times 3.1}{9.8} } \\\\t = 0.795 \ s$

mass of the water;

$m = m't\\\\m = 135 \ \frac{kg}{min} \ \times \ 0.795 \ s \ \times \ \frac{1 \ \min}{60 \ s} \ = 1.789 \ kg$

the average force exerted by the water on the ground;

F = mg

F = 1.789 x 9.8

F = 17.53 N

Therefore, the average force exerted by the water on the ground is 17.53 N.

7 0

3 years ago