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3 years ago

What are the four factors that determine weather?

Physics

1 answer:

garik1379 [7]3 years ago

8 0

Solar Radiation, Orbital Distance, Air Pressure, and the Abundance of water.

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If a photon has frequency = 2.00 x 1014s-1 and the speed of light = 3.00 x 108ms-1, then what is its wavelength?

butalik [34]

Answer:

The photon has a wavelength of $1.5x10^{-6}m$

Explanation:

The speed of a wave can be defined as:

$v = \nu \cdot \lambda$ (1)

Where v is the speed, $\nu$ is the frequency and $\lambda$ is the wavelength.

Equation 1 can be expressed in the following way for the case of an electromagnetic wave:

$c = \nu \cdot \lambda$ (2)

Where c is the speed of light.

Therefore, $\lamba$ $\lambda$ can be isolated from equation 2 to get the wavelength of the photon.

$\lambda = \frac{c}{\nu}$ (3)

$\lambda = \frac{3.00x10^{8}m/s}{2.00x10^{14}s^{-1}}$

$\lambda = 1.5x10^{-6}m$

Hence, the photon has a wavelength of $1.5x10^{-6}m$

Summary:

Photons are the particles that constitutes light.

3 0

3 years ago

Consider a father pushing a child on a playground merry-go-round. The system has a moment of inertia of 84.4 kg.m^2. The father

Sophie [7]

Answer:

Explanation:

Given that:

the initial angular velocity $\omega_o = 0$

angular acceleration $\alpha$ = 4.44 rad/s²

Using the formula:

$\omega = \omega_o+ \alpha t$

Making t the subject of the formula:

$t= \dfrac{\omega- \omega_o}{ \alpha }$

where;

$\omega = 1.53 \ rad/s^2$

∴

$t= \dfrac{1.53-0}{4.44 }$

t = 0.345 s

Using the formula:

$\omega ^2 = \omega _o^2 + 2 \alpha \theta$

here;

$\theta$ = angular displacement

∴

$\theta = \dfrac{\omega^2 - \omega_o^2}{2 \alpha }$

$\theta = \dfrac{(1.53)^2 -0^2}{2 (4.44) }$

$\theta =0.264 \ rad$

Recall that:

2π rad = 1 revolution

Then;

0.264 rad = (x) revolution

$x = \dfrac{0.264 \times 1}{2 \pi}$

x = 0.042 revolutions

Here; force = 270 N

radius = 1.20 m

The torque = F * r

$\tau = 270 \times 1.20 \\ \\ \tau = 324 \ Nm$

However;

From the moment of inertia;

$Torque( \tau) = I \alpha \\ \\ Since( I \alpha) = 324 \ Nm. \\ \\ Then; \\ \\ \alpha= \dfrac{324}{I}$

given that;

I = 84.4 kg.m²

$\alpha= \dfrac{324}{84.4} \\ \\ \alpha=3.84 \ rad/s^2$

For re-tardation; $\alpha=-3.84 \ rad/s^2$

Using the equation

$t= \dfrac{\omega- \omega_o}{ \alpha }$

$t= \dfrac{0-1.53}{ -3.84 }$

$t= \dfrac{1.53}{ 3.84 }$

t = 0.398s

The required time it takes= 0.398s

5 0

2 years ago

How is a magnetable to pick up bits of metal without actually touching them?

lisov135 [29]

C.There is an invisible magnetic field around the magnet where it exerts magnetic force

7 0

2 years ago

Two black holes (the remains of exploded stars), separated by a distance of

jolli1 [7]

The largest mass is 4.7 x 10³⁰ kg and the smallest mass is 5 x 10²⁹ kg.

The given parameters;

distance between the two black holes, r = 10 AU = 1.5 x 10¹² m
gravitational force between the two black holes, F = 6.9 x 10²⁵ N.
combined mass of the two black holes = 5.20 x 10³⁰ kg

The product of the two masses is calculated from Newton's law of universal gravitational as follows;

$F = \frac{Gm_1m_2}{r^2} \\\\m_1m_2 = \frac{Fr^2}{G} \\\\m_1m_2 = \frac{(6.9\times 10^{25}) \times (1.5\times 10^{12})^2}{6.67\times 10^{-11}} \\\\m_1m_2 = 2.328 \times 10^{60} \ kg^2$

The sum of the two masses is given as;

m₁ + m₂ = 5.2 x 10³⁰ kg

m₂ = 5.2 x 10³⁰ kg - m₁

The first mass is calculated as follows;

m₁(5.2 x 10³⁰ - m₁) = 2.328 x 10⁶⁰

5.2 x 10³⁰m₁ - m₁² = 2.328 x 10⁶⁰

m₁² - 5.2 x 10³⁰m₁ + 2.328 x 10⁶⁰ = 0

solve the quadratic equation using formula method;

a = 1, b =- 5.2 x 10³⁰, c = 2.328 x 10⁶⁰

$m_1 = \frac{-b \ \ +/- \ \ \sqrt{b^2 - 4ac} }{2a} \\\\m_1 = \frac{-(-5.2\times 10^{20}) \ \ +/- \ \ \sqrt{(-5.2\times 10^{20})^2 - 4(1\times 2.328\times 10^{60})} }{2(1)} \\\\m_1 = 4.7 \times 10^{30} \ kg \ \ or \ \ 4.9 \times 10^{29} \ kg$

The second mass is calculated as follows;

m₂ = 5.2 x 10³⁰ kg - m₁

m₂ = 5.2 x 10³⁰ kg - 4.7 x 10³⁰ kg

m₂ = 5 x 10²⁹ kg

m₂ = 5.2 x 10³⁰ kg - 4.9 x 10²⁹ kg

m₂ = 4.7 x 10³⁰ kg

Thus, the largest mass is 4.7 x 10³⁰ kg and the smallest mass is 5 x 10²⁹ kg.

Learn more here:brainly.com/question/9373839

3 0

2 years ago

An air-filled parallel-plate capacitor is connected to a battery and allowed to charge up. Now a slab of dielectric material is

natita [175]

Answer:

The charge on the capacitor had increased

Explanation:

The expression for the capacitance of an air-filled parallel-plate capacitor is as follows as;

$C=\frac{\epsilon _{0}A}{d}$

Here, C is the capacitance, $\epsilon _{0}$ is the permittivity of free space, A is the area and d is the distance between the parallel plate capacitor.

When a slab of dielectric material is placed between the plates of the capacitor then the expression for the capacitance is as follows;

$C=\frac{K\epsilon _{0}A}{d}$

Here, K is the dielectric constant.

In the given problem, a slab is inserted between the plates of the capacitor then the capacitance of the capacitor will increase in this case. Therefore, the option (a) is true.

The expression for the charge stored in the capacitor is as;

Q= CV

Here, Q is the charge and V is the potential.

The charge will also increase in this case as the charge stored in the capacitor is directly proportional to the capacitance. Therefore, the option (d) is not true.

The expression for the energy stored in the capacitor is as follows;

$E=\frac{CV^{2}}{2}$

The voltage is constant in the given problem but the capacitance increases then the energy stored in the capacitor will increase. Therefore, the option (b) is not true.

The voltage across the capacitor will remain same as the capacitor is still connected to the battery. Therefore, the option (c) is not true.

Therefore, only option (a) is true.

3 0

3 years ago