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3 years ago

What term is used for a weather condition characterized by lack of precipitation over a long period of time?

Physics

2 answers:

sleet_krkn [62]3 years ago

5 0

The answer is Drought.

Marrrta [24]3 years ago

4 0

Answer:

Explanation:

<h2><u><em>Drought</em></u></h2>

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A merry-go-round with a rotational inertia of 600 kg m2 and a radius of 3. 0 m is initially at rest. A 20 kg boy approaches the

Margaret [11]

Hi there!

$\boxed{\omega = 0.38 rad/sec}$

We can use the conservation of angular momentum to solve.

$\large\boxed{L_i = L_f}$

Recall the equation for angular momentum:

$L = I\omega$

We can begin by writing out the scenario as a conservation of angular momentum:

$I_m\omega_m + I_b\omega_b = \omega_f(I_m + I_b)$

$I_m$ = moment of inertia of the merry-go-round (kgm²)

$\omega_m$ = angular velocity of merry go round (rad/sec)

$\omega_f$ = final angular velocity of COMBINED objects (rad/sec)

$I_b$ = moment of inertia of boy (kgm²)

$\omega_b$ = angular velocity of the boy (rad/sec)

The only value not explicitly given is the moment of inertia of the boy.

Since he stands along the edge of the merry go round:

$I = MR^2$

We are given that he jumps on the merry-go-round at a speed of 5 m/s. Use the following relation:

$\omega = \frac{v}{r}$

$L_b = MR^2(\frac{v}{R}) = MRv$

Plug in the given values:

$L_b = (20)(3)(5) = 300 kgm^2/s$

Now, we must solve for the boy's moment of inertia:

$I = MR^2\\I = 20(3^2) = 180 kgm^2$

Use the above equation for conservation of momentum:

$600(0) + 300 = \omega_f(180 + 600)\\\\300 = 780\omega_f\\\\\omega = \boxed{0.38 rad/sec}$

8 0

3 years ago

A snail travels 10m in 3000 seconds. what is the snail's average speed?

Salsk061 [2.6K]

he makes 1 m im 300 seconds btw that snails slow lol

4 0

3 years ago

Read 2 more answers

what effect would you expect the termperature of a material to have on the ease with which electrons can be ejected from it via

Ilya [14]

Answer:

The photoelectric effect is a phenomenon in which the photoelectrons are emitted from the metal when an incident electromagnetic wave hits the metal. The incident light should have a threshold frequency to meet the work function of the metal

8 0

2 years ago

20 POINTS HELP FAST PLEASE!!

s2008m [1.1K]

What two forces are required for generators and electric motors to work? “A” Electricity and magnetism.

In an electric motor, the energy transformation begins as “A” Electrical And becomes mechanical.

8 0

3 years ago

An 80 g, 40 cm long rod hangs vertically on a frictionless, horizontal axle passing through its center. A 15 g ball of clay trav

fiasKO [112]

Answer:

θ = 12.95º

Explanation:

For this exercise it is best to separate the process into two parts, one where they collide and another where the system moves altar the maximum height

Let's start by finding the speed of the bar plus clay ball system, using amount of momentum

The mass of the bar (M = 0.080 kg) and the mass of the clay ball (m = 0.015 kg) with speed (v₀ = 2.0 m / s)

Initial before the crash

p₀ = m v₀

Final after the crash before starting the movement

$p_{f}$ = (m + M) v

p₀ = $p_{f}$

m v₀ = (m + M) v

v = v₀ m / (m + M)

v = 2.0 0.015 / (0.015 +0.080)

v = 0.316 m / s

With this speed the clay plus bar system comes out, let's use the concept of conservation of mechanical energy

Lower

Em₀ = K = ½ (m + M) v²

Higher

$E_{mf}$ = U = (m + M) g y

Em₀ = $E_{mf}$

½ (m + M) v² = (m + M) g y

y = ½ v² / g

y = ½ 0.316² / 9.8

y = 0.00509 m

Let's look for the angle the height from the pivot point is

L = 0.40 / 2 = 0.20 cm

The distance that went up is

y = L - L cos θ

cos θ = (L-y) / L

θ = cos⁻¹ (L-y) / L

θ = cos⁻¹-1 ((0.20 - 0.00509) /0.20)

θ = 12.95º

4 0

3 years ago