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3 years ago

Where are some places that cyber bullying occurs? Check all that apply

Physics

1 answer:

horsena [70]3 years ago

5 0

Answer:

B, E, F

Explanation:

Because cyberbullying happens online, text messages, social media networks, and chat rooms can all have cyberbullying, because they are all ways to chat online. D does not apply, because on most websites there are not ways to chat.

You might be interested in

How much time will elapse between seeing and hearing an event?

julia-pushkina [17]

Depends on how far away the event is and what the temperature is as this affects the speed of sound.

For example, let's say you're 600 meters away and the temperature has no affect.

The speed of sound would be roughly 340 m/s so the time it would take to hear the sound would be 600/340 = 1.76 seconds

The speed of light (c) is 3.0 X 10^8 m/s so the time it would take to see the event would be 600/3 X 10^8 = 2 X 10^-7

Subtract: 1.76 - (2 X 10^-7) = approx. 1.76

3 0

3 years ago

Use Eq. cosϕ=R/Z to show that the average power delivered by the source in an L−R−C series circuit is given by Pav = I^2rmsR .

Evgen [1.6K]

Answer:

Explanation:

In a L C R circuit, the average power is given by

$P_{av}=V_{rms}I_{rms}Cos\phi$

As given in the question

CosФ = R / Z

And we know that

$V_{rms}=I_{rms}\times Z$

$P_{av}=I_{rms}\times Z\times I_{rms}\times \frac{R}{Z}$

$P_{av}=I_{rms}^{2}\times R$

6 0

3 years ago

A body which has surface area 5cm² and temperature of 727°C radiates 300J of energy in one minute. Calculate it's emissivity giv

cestrela7 [59]

<h2>Answer: 0.17</h2>

Explanation:

The Stefan-Boltzmann law establishes that a black body (an ideal body that absorbs or emits all the radiation that incides on it) "emits thermal radiation with a total hemispheric emissive power proportional to the fourth power of its temperature":

$P=\sigma A T^{4}$ (1)

Where:

$P=300J/min=5J/s=5W$ is the energy radiated by a blackbody radiator per second, per unit area (in Watts). Knowing $1W=\frac{1Joule}{second}=1\frac{J}{s}$

$\sigma=5.6703(10)^{-8}\frac{W}{m^{2} K^{4}}$ is the Stefan-Boltzmann's constant.

$A=5cm^{2}=0.0005m^{2}$ is the Surface area of the body

$T=727\°C=1000.15K$ is the effective temperature of the body (its surface absolute temperature) in Kelvin.

However, there is no ideal black body (ideal radiator) although the radiation of stars like our Sun is quite close. So, in the case of this body, we will use the Stefan-Boltzmann law for real radiator bodies:

$P=\sigma A \epsilon T^{4}$ (2)

Where $\epsilon$ is the body's emissivity

(the value we want to find)

Isolating $\epsilon$ from (2):

$\epsilon=\frac{P}{\sigma A T^{4}}$ (3)

Solving:

$\epsilon=\frac{5W}{(5.6703(10)^{-8}\frac{W}{m^{2} K^{4}})(0.0005m^{2})(1000.15K)^{4}}$ (4)

Finally:

$\epsilon=0.17$ (5) This is the body's emissivity

3 0

3 years ago

If an object has a mass of 38 kg, what is its approximate weight on earth?

klasskru [66]

38*10=380 N
To be more exact, 38 should be multiplied by 9.8 instead of 10.

3 0

3 years ago

Block B (of mass m) is initially at rest. Block A (of mass 3m) travels toward B with an initial speed v0 (vee nought) and collid

NeTakaya

Answer:

The maximum height reached by the two blocks is approximately 0.1147959 × v₀²

Explanation:

The mass of block B = m

The mass of block A = 3·m

The initial velocity of block B, v₂ = 0 m/s

The initial velocity of block A, v₁ = v₀

The amount of friction between the blocks and the surface = Negligible friction

By the Law of conservation of linear momentum, we have;

Total initial momentum = Total final momentum

3·m·v₁ + m·v₂ = (3·m + m)·v₃ = 4·m·v₃

Plugging in the values for the velocities gives;

3·m × v₀ + m × 0 = (3·m + m)·v₃ = 4·m·v₃

∴ 3·m × v₀ = 4·m·v₃

$\therefore v_3 = \dfrac{3}{4} \times v_0 = 0.75 \times v_0$

The kinetic energy, K.E. of the combined blocks after the collision is given as follows;

K.E. = 1/2 × mass × v²

$\therefore K.E. = \dfrac{1}{2} \times 4\cdot m \times \left (\dfrac{3}{4} \cdot v_0 \right )^2 = \dfrac{9}{8} \cdot m\cdot v_0^2$

The potential energy, P.E., gained by the two blocks at maximum height = The kinetic energy, K.E., of the two blocks before moving vertically upwards

The potential energy, P.E. = m·g·h

Where;

m = The mass of the object at the given height

g = The acceleration due to gravity

h = The height at which the object of mass, 'm', is located

Therefore, for h = The maximum height reached by the two blocks, we have;

P.E. = K.E.

$m \cdot g \cdot h = \dfrac{9}{8} \cdot m\cdot v_0^2$

$h = \dfrac{\dfrac{9}{8} \cdot m\cdot v_0^2}{m \cdot g } = \dfrac{9}{8} \cdot \dfrac{ v_0^2}{ g } = \dfrac{9}{8} \cdot \dfrac{ v_0^2}{ 9.8} = \dfrac{42}{392} \cdot v_0^2 \approx 0.1147959 \cdot v_0^2$

The maximum height reached by the two blocks, h ≈ 0.1147959·v₀².

7 0

2 years ago