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3 years ago

3.9 divided by 15 gcuu

Physics

2 answers:

Vsevolod [243]3 years ago

8 0

0.26 there you go buddy

Gnoma [55]3 years ago

8 0

0.26 smennwnwnwsne e

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Which type of error is a false positive

maks197457 [2]

Answer: Type I Error

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2 years ago

An electron is accelerated from rest by an electric potential difference of 437 V, and then enters a region of uniform magnetic

Lelu [443]

The radius of the electron's path is once it enters the magnetic field 3.16*10^-4 m.

The electric capacity difference, additionally called voltage, is the outside work needed to bring a rate from one region to some other area in an electric discipline. The electric powered capacity distinction is the change of potential electricity skilled by way of a test rate that has a fee of +1.

The distinction in ability among two points represents the paintings worried or the electricity released in the switch of a unit amount of power from one factor to the other.

Electric ability is the work executed in step with unit fee in bringing the fee from infinity to that factor against electrostatic pressure. In a conductor, electrons float simplest when there is a difference in electric stress at its ends. This is additionally known as the capacity distinction.

Learn more about the electric potential difference here brainly.com/question/14306881

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2 years ago

a football is thrown upward at a 31° angle to the horizontal.the acceleration of gravity is 9.8m/s^2. To throw the ball a distan

mr Goodwill [35]

The ball's vertical position $y$ in the air at time $t$ is

$y=v_0\sin31^\circ\,t-\dfrac g2t^2$

The ball is at its original height when $y=0$ , which happens at

$v_0\sin31^\circ\,t-\dfrac g2t^2=\dfrac t2\left(2v_0\sin31^\circ-gt)=0$

$\implies t=0\text{ and }t=\dfrac{2v_0\sin31^\circ}g$

Meanwhile, the ball's horizontal position $x$ at time $t$ is

$x=v_0\cos31^circ\,t$

So when the ball reaches its original height a second time, the ball will have traveled a horizontal distance of

$x=\dfrac{2{v_0}^2\sin31^\circ\cos31^\circ}g=\dfrac{{v_0}^2\sin(2\cdot31^\circ)}g$

(which you might recognize as the formula for the range of a projectile)

To reach a distance of $x=77\,\rm m$ , the initial speed $v_0$ would be

$77\,\mathrm m=\dfrac{{v_0}^2\sin62^\circ}{9.8\,\frac{\rm m}{\mathrm s^2}}\implies v_0=29\dfrac{\rm m}{\rm s}$

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3 years ago

Which of these statements about resistance is true? Choose the best answer.

bonufazy [111]

'C' and 'D' are the same statement, and none of the 3 choices
is good.

Electrical energy is produced in a generator, a solar panel, or
a battery, not in things with resistance.

From the generator or battery, current flows through a circuit of
one or more components.

The greater the resistance of a component, the more energy is
LOST as the current flows through it. The component dissipates
the energy in the form of heat.

8 0

3 years ago

Read 2 more answers

A car traveling in a straight line has a velocity of 5.0 m/s. After an acceleration of 0.75 m/s/s, the cars velocity is 8.0. In

Bogdan [553]

Vs - velocity on beginning
ve - velocity on ending. You've got:
$v_s = 5 \frac{m}{s} \\ v_e=8 \frac{m}{s} \\ \hbox{Then:} \\ \Delta v=v_e - v_s = 8 \frac{m}{s} - 5\frac{m}{s}=3 \frac{m}{s} \\ a=0,75 \frac{m}{s^2} \\ \hbox{And from formula:} \\ a=\frac{\Delta v}{\Delta t} \qquad \Rightarrow \qquad \Delta t= \frac{\Delta v}{a} \\ \hbox{Substitute:} \\ \Delta t=\frac{3\frac{m}{s}}{0,75 \frac{m}{s^2}}= \frac{3}{\frac{3}{4}} s= 3 \cdot \frac{4}{3} s= 4 s$
So he needed 4 second.

3 0

3 years ago