Should i beat up my sibling cause they mean to me?

A school bus takes 0.530 h to reach the school from your house. If the average velocity of the bus is 19.0 km/h to the east, wha

kifflom [539]

Answer:

$x_{distance}=10.07km$

Explanation:

Given data

time=0.530 h

Average velocity Vavg=19.0 km/s

To find

Displacement Δx

Solution

The Formula for average velocity is given as

$V_{avg}=(x_{distance} )/(t_{time} )\\ x_{distance}=V_{avg}*(t_{time} )\\x_{distance}=(19.0km/h)*(0.530h)\\x_{distance}=10.07km$

3 0

3 years ago

By using electroscope, if the body is negatively charged then due to electrostatic induction, then on the leaves there will be a

Kryger [21]

(b) negative charge

This is the answer

6 0

2 years ago

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If the electron just misses the upper plate as it emerges from the field, find the magnitude of the electric field.

Damm [24]

Answer:

The magnitude of the electric field be 171.76 N/C so that the electron misses the plate.

Explanation:

As data is incomplete here, so by seeing the complete question from the search the data is

vx_0=1.1 x 10^6

ax=0 As acceleration is zero in the horizontal axis so

Equation of motion in horizontal direction is given as

$s_x=v_x_0 t$

$t=\frac{s_x}{v_x}\\t=\frac{2 \times 10^{-2}}{1.1 \times 6}\\t=1.82 \times 10^{-8} s$

Now for the vertical distance

vy_o=0

than the equation of motion becomes

$s_y=v_y_0 t+\frac{1}{2} at^2\\s_y=\frac{1}{2} at^2\\0.5 \times 10^{-2}=\frac{1}{2} a(1.82 \times 10^{-8})^2\\a=3.02 \times 10^{13} m/s^2$

Now using this acceleration the value of electric field is calculated as

$E=\frac{F}{q}\\E=\frac{ma}{q}\\E=\frac{m_ea}{q_e}\\$

Here a is calculated above, m is the mass of electron while q is the charge of electron, substituting values in the equation

$E=\frac{9.1\times 10^{-31} \times 3.02 \times 10^{13} }{1.6 \times 10^{-19}}\\E=171.76 N/C$

So the magnitude of the electric field be 171.76 N/C so that the electron misses the plate.

5 0

3 years ago

Why the center of Newton rings is dark?

RideAnS [48]

The point of contact the path difference is zero but one of the interfering ray is reflected so the effective path difference becomes λ/2 thus the condition of minimum intensity is created in the center.

3 0

3 years ago

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The zone of earthquakes and volcanoes surrounding the pacific ocean is called

gulaghasi [49]

Answer:

The Ring of Fire

Explanation:

The ring of fire is also called the Circum-Pacific Belt, it is a path along the pacific ocean consisting of active volcanoes and frequent earthquakes.

It has a length of approximately 40,000 kilometers. It lies on the edge of tectonic plates where the in-earth vibrations and geothermal energies are prone to erupt out.

Ring of fire inhibits about 75% o the earth's volcanoes and 95% of earthquakes occur in this region.

3 0

2 years ago