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2 years ago

What is the first stage of a thunderstorm?

Physics

2 answers:

lianna [129]2 years ago

8 0

The first stage is cumulus stage.

mariarad [96]2 years ago

7 0

1. B

2. A

3. B

4. A

5. C

6. B

7. B

8. A

You might be interested in

If the column of mercury in a barometer stands at 72.6 cm, what is the atmospheric pressure?

lakkis [162]

Answer:pressure = density * acceleration due to gravity * height

H=72.6cm= 0.726m

P=0.726*13.6*10^3*9.8

P=96761.28Pa

Explanation:

6 0

2 years ago

. Since the man is walking at a constant velocity, he has what acceleration

docker41 [41]

If he’s walking at a constant velocity there is no acceleration.

6 0

2 years ago

Two uniform solid cylinders, each rotating about its central (longitudinal) axis, have the same mass of 3.56 kg and rotate with

Natali5045456 [20]

Answer:

(a) 3107.98 J

(b) 14530.6 J

Explanation:

mass, m = 3.56 kg

angular speed, ω = 179 rad/s

Moment of inertia of solid cylinder, I = 1/2 mr^2

where, m is the mass and r be the radius of the cylinder.

(a) radius, r = 0.330 m

I = 0.5 x 3.56 x 0.330 x 0.330 = 0.194 kgm^2

The formula for the rotational kinetic energy is given by

$K = \frac{1}{2}I\omega ^{2}$

K = 0.5 x 0.194 x 179 x 179 = 3107.98 J

(b) radius, r = 0.714 m

I = 0.5 x 3.56 x 0.714 x 0.714 = 0.907 kgm^2

The formula for the rotational kinetic energy is given by

$K = \frac{1}{2}I\omega ^{2}$

K = 0.5 x 0.907 x 179 x 179 = 14530.6 J

3 0

2 years ago

In Anchorage, collisions of a vehicle with a moose are so common that they are referred to with the abbreviationMVC. Suppose a 1

lara31 [8.8K]

Answer:

Part a)

$f = \frac{8}{9}$

Part b)

$f = \frac{120}{169}$

Part c)

So from above discussion we have the result that energy loss will be more if the collision occurs with animal with more mass

Explanation:

Part a)

Let say the collision between Moose and the car is elastic collision

So here we can use momentum conservation

$m_1v_{1i} = m_1v_{1f} + m_2v_{2f}$

$1000 v_o = 1000 v_{1f} + 500 v_{2f}$

also by elastic collision condition we know that

$v_{2f} - v_{1f} = v_o$

now we have

$2v_o = 2v_{1f} + v_o + v_{1f}$

now we have

$v_{1f} = \frac{v_o}{3}$

Now loss in kinetic energy of the car is given as

$\Delta K = \frac{1}{2}m(v_o^2 - v_{1f}^2)$

$\Delta K = \frac{1}{2}m(v_o^2 - \frac{v_o^2}{9})$

so fractional loss in energy is given as

$f = \frac{\Delta K}{K}$

$f = \frac{\frac{4}{9}mv_o^2}{\frac{1}{2}mv_o^2}$

$f = \frac{8}{9}$

Part b)

Let say the collision between Camel and the car is elastic collision

So here we can use momentum conservation

$m_1v_{1i} = m_1v_{1f} + m_2v_{2f}$

$1000 v_o = 1000 v_{1f} + 300 v_{2f}$

also by elastic collision condition we know that

$v_{2f} - v_{1f} = v_o$

now we have

$10v_o = 10v_{1f} + 3(v_o + v_{1f})$

now we have

$v_{1f} = \frac{7v_o}{13}$

Now loss in kinetic energy of the car is given as

$\Delta K = \frac{1}{2}m(v_o^2 - v_{1f}^2)$

$\Delta K = \frac{1}{2}m(v_o^2 - \frac{49v_o^2}{169})$

so fractional loss in energy is given as

$f = \frac{\Delta K}{K}$

$f = \frac{\frac{60}{169}mv_o^2}{\frac{1}{2}mv_o^2}$

$f = \frac{120}{169}$

Part c)

So from above discussion we have the result that energy loss will be more if the collision occurs with animal with more mass

8 0

2 years ago

A merry-go-round on a playground consists of a horizontal solid disk with a weight

bezimeni [28]

Answer:α = 1.00 rad/s², τ = 90.0 N•m, KEr = 1.12 kJ

Explanation:

m = 795/9.81 = 81.0 kg

ω₁ = 47.79 rev/min(2π rad/rev) / (60 s/min) = 5.00 rad/s

α = (ω₁ - ω₀)/τ = (5.00 - 0.00)/5.00 = 1.00 rad/s²

I = ½mR² = ½(81.0)(1.49²) = 90.0 kg•m²

τ = Iα = 90.0(1.00) = 90.0 N•m

KEr = ½Iω² = ½(90.0)5.00² = 1,124.477 ≈ 1.12 kJ

5 0

2 years ago