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3 years ago

El movimiento de un bote que se transporta en un rio ocurre en ?

Physics

1 answer:

Assoli18 [71]3 years ago

5 0

<span>El movimiento de un bote que se transporta en un rio ocurre en cuantos ejes?</span>

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Is it possible that a 60kg boy riding a BMX bike could exert more pressure on a road surface than a 2500kg truck? Explain why.

zysi [14]

I really don’t know not smart

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4 years ago

Two sheets of polarizing material are placed such that their polarizing axes are 90° to each other and no light passes through t

nekit [7.7K]

Answer:

Explanation:

Polarization In this case angle of incidence is not equal to angle of polarization, hence reflected light is partially polarized and transmitted light is also partially polarized. by reflection is explained by Brewster's law,

According to this when unpolarized light incident on glass plate at an angle is called as angle of polarizing the reflected light is plane polarized, and transmitted light is partially polarized. The plane of vibration of polarized light is having plane of vibrations perpendicular to plane of incidence.

4 0

3 years ago

ANSWER ASAP PLEASE!!

eimsori [14]

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3 years ago

On a straight road (taken to be in the x direction) you drive for an hour at 60 km per hour, then quickly speed up to 120 km per

Luda [366]

Answer:

The average velocity is 180 km/hr

Explanation:

Given;

initial velocity, u = 60 km per hour

final velocity, v = 120 km per hour

initial time = 1 hour

final time = 2 hour

Initial position = 60 km/h x 1 hour = 60 km

final position = 120 km/h x 2 hour = 240 km

The average velocity is given by;

$V_{avg} = \frac{Final \ position\ - \ Initial \ position}{final \ time\ - \ initial \ time}\\\\V_{avg} = \frac{240km \ - \ 60km}{2hr\ - \ 1hr} \\\\V_{avg} = \frac{180 \ km}{1hr} \\\\V_{avg}= 180 \ km/hr$

Therefore, the average velocity is 180 km/hr

3 0

3 years ago

A 22 µF capacitor charged to 0.7 kV and a second 115 µF capacitor charged to 5.5 kV are connected to each other, with the positi

vesna_86 [32]

Answer:

0.099C

Explanation:

First, we need to get the common potential voltage using the formula

$V=\frac {C_2V_2-C_1V_1}{C_1+C_2}$

Where V is the common voltage, C and V represent capacitance and charge respectively. Subscripts 1 and 2 to represent the the first and second respectively. Substituting the above with the following given values then

$C_1=22\times 10^{-6} F\\ C_2=115\times 10^{-6} F\\ V_1= 0.7\times 10^{3}\\V_2=5.5\times 10^{3}$

Therefore

$V=\frac {115\times 10^{-6}\times 5.5\times 10^{3}-22\times 10^{6}\times 0.7\times 10^{3}}{22\times 10^{-6}+115\times 10^{-6}}=4504.3795620437$

Charge, Q is given by CV hence for the first capacitor charge will be $Q_1=C_1V$

Here, $Q_1=22\times 10^{-6}\times 4504.3795620437=0.0990963503649C\approx 0.099C$

8 0

3 years ago