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3 years ago

Electric fields are vector quantities whose magnitudes are measured in units of bolts/meter(v/m) find the resultant electric fie

ld when there are two fields, E_1 and E_2, where E_1 is directed vertically upward and has magnitude 100V/m and E_2 is directed 45 to the left of E_1 and has magnitude 150V/m.

Physics

1 answer:

Grace [21]3 years ago

8 0

$E_{x} =-150 cos45 = -106.066 V/m

 E_{y} =100+150 sin45= 206.066 V/m$

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hram777 [196]

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On a windy day, the wind turbine transfers 78 W of power. Calculate the amount of energy the turbine transfers in 10s.

tangare [24]

Answer:

Energy = 780 Joules

Explanation:

Given the following data;

Power = 78 Watts

Time = 10 seconds

To find the energy transferred;

Energy refers to the amount or quantity of power which is being consumed by an individual, group of people or organization over a specific period of time.

Mathematically, energy is given by the formula;

Energy = power * time

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3 years ago

En un balde de 25 cm de altura se hace un orificio a 15 cm del suelo.calcula la velocidad de salida del agua

Contact [7]

Explanation:

En un balde de 25 cm de altura se hace un orificio a 15 cm del suelo.calcula la velocidad de salida del agua

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2 years ago

A catapult launches a test rocket vertically upward from a well, giving the rocket an initial speed of 80.6 m/s at ground level.

kow [346]

Before the engines fail, the rocket's altitude at time t is given by

$y_1(t)=\left(80.6\dfrac{\rm m}{\rm s}\right)t+\dfrac12\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t^2$

and its velocity is

$v_1(t)=80.6\dfrac{\rm m}{\rm s}+\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t$

The rocket then reaches an altitude of 1150 m at time t such that

$1150\,\mathrm m=\left(80.6\dfrac{\rm m}{\rm s}\right)t+\dfrac12\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t^2$

Solve for t to find this time to be

$t=11.2\,\mathrm s$

At this time, the rocket attains a velocity of

$v_1(11.2\,\mathrm s)=124\dfrac{\rm m}{\rm s}$

When it's in freefall, the rocket's altitude is given by

$y_2(t)=1150\,\mathrm m+\left(124\dfrac{\rm m}{\rm s}\right)t-\dfrac g2t^2$

where $g=9.80\frac{\rm m}{\mathrm s^2}$ is the acceleration due to gravity, and its velocity is

$v_2(t)=124\dfrac{\rm m}{\rm s}-gt$

(a) After the first 11.2 s of flight, the rocket is in the air for as long as it takes for $y_2(t)$ to reach 0:

$1150\,\mathrm m+\left(124\dfrac{\rm m}{\rm s}\right)t-\dfrac g2t^2=0\implies t=32.6\,\mathrm s$

So the rocket is in motion for a total of 11.2 s + 32.6 s = 43.4 s.

(b) Recall that

${v_f}^2-{v_i}^2=2a\Delta y$

where $v_f$ and $v_i$ denote final and initial velocities, respecitively, $a$ denotes acceleration, and $\Delta y$ the difference in altitudes over some time interval. At its maximum height, the rocket has zero velocity. After the engines fail, the rocket will keep moving upward for a little while before it starts to fall to the ground, which means $y_2$ will contain the information we need to find the maximum height.

$-\left(124\dfrac{\rm m}{\rm s}\right)^2=-2g(y_{\rm max}-1150\,\mathrm m)$

Solve for $y_{\rm max}$ and we find that the rocket reaches a maximum altitude of about 1930 m.

(c) In part (a), we found the time it takes for the rocket to hit the ground (relative to $y_2(t)$ ) to be about 32.6 s. Plug this into $v_2(t)$ to find the velocity before it crashes:

$v_2(32.6\,\mathrm s)=-196\frac{\rm m}{\rm s}$

That is, the rocket has a velocity of 196 m/s in the downward direction as it hits the ground.

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Luba_88 [7]

It is called a medium

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