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2 years ago

What happens to potential energy when a skier goes down the hill? A. Decrease B. Increase

Physics

1 answer:

luda_lava [24]2 years ago

4 0

Question: What happens to potential energy when a skier goes down the hill?

Answer: Decrease

Explanation: As the skier begins the descent down the hill potential energy is lost and kinetic energy is going to start acting

question answered by

(jacemorris04)

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A length change 0.08 m will occur for an object that is L= 56 m long. If the coefficient of thermal expansion is5.3 x 10 /C and

ArbitrLikvidat [17]

Answer:

Increase in temperature = 269.54 °C

Explanation:

We have equation for thermal expansion

ΔL = LαΔT

Change in length, ΔL = 0.08 m

Length, L = 56 m

Coefficient of thermal expansion, α = 5.3 x 10⁻⁶ °C⁻1

Change in temperature, ΔT = T - 253

Substituting

0.08 = 56 x 5.3 x 10⁻⁶ x (T - 253)

(T - 253) = 269.54

T = 522.54 °C

Increase in temperature = 269.54 °C

4 0

2 years ago

In the compound MgS, the sulfide ion has 1. lost one electrons. 2. lost two electrons. 3. gained one electron. 4. gained two ele

never [62]

Answer:

4.has gained two electrons

Explanation:

There exist electrovalent bonding the compound MgS . In electrovalent bonding, there is a transfer of electrons from the metal to non-metal.

Magnesium atom has an atomic number 12 and its electron configuration is 2,8,2

Sulfur atom , a non-metal has atomic number of 16 and its electron configuration = 2,8,6

This means that magnesium as a metal needs to loose two electrons from its valence shell to attain its stable structure.Also sulfur requires two more electron to achieve its octet structure.

Hence a transfer of electrons will take place from magnesium atom to sulfur atom, sulfur gaining two electrons.

6 0

2 years ago

Read 2 more answers

In a double-slit experiment, light from two monochromatic light sources passes through the same double slit. The light from the

MatroZZZ [7]

Answer:

The wavelength is $\lambda_2 = 534 *10^{-9} \ m$

Explanation:

From the question we are told that

The wavelength of the first light is $\lambda _ 1 = 587 \ nm$

The order of the first light that is being considered is $m_1 = 10$

The order of the second light that is being considered is $m_2 = 11$

Generally the distance between the fringes for the first light is mathematically represented as

$y_1 = \frac{ m_1 * \lambda_1 * D}{d}$

Here D is the distance from the screen

and d is the distance of separation of the slit.

For the second light the distance between the fringes is mathematically represented as

$y_2 = \frac{ m_2 * \lambda_2 * D}{d}$

Now given that both of the light are passed through the same double slit

$\frac{y_1}{y_2} = \frac{\frac{m_1 * \lambda_1 * D}{d} }{\frac{m_2 * \lambda_2 * D}{d} } = 1$

=> $\frac{ m_1 * \lambda _1 }{ m_2 * \lambda_2} = 1$

=> $\lambda_2 = \frac{m_1 * \lambda_1}{m_2}$

=> $\lambda_2 = \frac{10 * 587 *10^{-9}}{11}$

=> $\lambda_2 = 534 *10^{-9} \ m$

4 0

2 years ago

How can you make the potential energy as high as possible in a magnetic field between one electromagnet and one piece of iron?

harkovskaia [24]

In step 1, to increase the potential energy, the iron will move towards the electromagnet.

In step 2, to increase the potential energy, the iron will move towards the electromagnet.

<h3>Potential energy of a system of magnetic dipole</h3>

The potential energy of a system of dipole depends on the orientation of the dipole in the magnetic field.

$U = \mu B$

where;

$\mu$ is the dipole moment
B is the magnetic field

$B = \frac{\mu_0 I}{2\pi r}$

$U = \mu\times (\frac{\mu_0 I}{2\pi r} )$

Increase in the distance (r) reduces the potential energy. Thus, we can conclude the following;

In step 1, to increase the potential energy, the iron will move towards the electromagnet.
In step 2, when the iron is rotated 180, it will still maintain the original position, to increase the potential energy, the iron will move towards the electromagnet.

Learn more about potential energy in magnetic field here: brainly.com/question/14383738

7 0

1 year ago

What is the minimum frequency with which a 200-turn, flat coil of cross sectional area 300 cm2 can be rotated in a uniform 30-mT

sergiy2304 [10]

Answer:

The minimum frequency of the coil is 7.1 Hz

Explanation:

Given;

number of turns, N = 200 turns

cross sectional area, A = 300 cm² = 300 x 10⁻⁴ m²

magnitude of magnetic field strength, B = 30 x 10⁻³ T

maximum value of the induced emf, E = 8 V

Maximum induced emf is given as;

E = NBAω

where

ω is angular velocity (ω = 2πf)

E = NBA2πf

where;

f is the minimum frequency, measured in hertz (Hz)

f = E / (NBA2π)

f = 8 / (200 x 30 x 10⁻³ x 300 x 10⁻⁴ x 2 x 3.142)

f = 7.073 Hz

f = 7.1 Hz

Therefore, the minimum frequency of the coil is 7.1 Hz

8 0

2 years ago