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3 years ago

Calculate the number of electrons passing a point in the wire in 1 min when the current is 1 A

Physics

1 answer:

Vera_Pavlovna [14]3 years ago

7 0

Explanation:

When one coulomb charge passes through any cross section of the wire per second,the current passing is one ampere. Charge of electron ,e=1.6X10^-19C. n=1/(1.6X10^-19)=6.25X10^18.Sep 17, 2017

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Someone please help??

stira [4]

The efficancy is .375

5 0

3 years ago

A group of students prepare for a robotic competition and build a robot that can launch large spheres of mass M in the horizonta

jeyben [28]

Answer:

$V_0$

Explanation:

Given that, the range covered by the sphere, $M$ , when released by the robot from the height, $H$ , with the horizontal speed $V_0$ is $D$ as shown in the figure.

The initial velocity in the vertical direction is $0$ .

Let $g$ be the acceleration due to gravity, which always acts vertically downwards, so, it will not change the horizontal direction of the speed, i.e. $V_0$ will remain constant throughout the projectile motion.

So, if the time of flight is $t$ , then

$D=V_0t\; \cdots (i)$

Now, from the equation of motion

$s=ut+\frac 1 2 at^2\;\cdots (ii)$

Where $s$ is the displacement is the direction of force, $u$ is the initial velocity, $a$ is the constant acceleration and $t$ is time.

Here, $s= -H, u=0,$ and $a=-g$ (negative sign is for taking the sigh convention positive in $+y$ direction as shown in the figure.)

So, from equation (ii),

$-H=0\times t +\frac 1 2 (-g)t^2$

$\Rightarrow H=\frac 1 2 gt^2$

$\Rightarrow t=\sqrt {\frac {2H}{g}}\;\cdots (iii)$

Similarly, for the launched height $2H$ , the new time of flight, $t'$ , is

$t'=\sqrt {\frac {4H}{g}}$

From equation (iii), we have

$\Rightarrow t'=\sqrt 2 t\;\cdots (iv)$

Now, the spheres may be launched at speed $V_0$ or $2V_0$ .

Let, the distance covered in the $x-$ direction be $D_1$ for $V_0$ and $D_2$ for $2V_0$ , we have

$D_1=V_0t'$

$D_1=V_0\times \sqrt 2 t$ [from equation (iv)]

$\Rightarrow D_1=\sqrt 2 D$ [from equation (i)]

$\Rightarrow D_1=1.41 D$ (approximately)

This is in the $3$ points range as given in the figure.

Similarly, $D_2=2V_0t'$

$D_2=2V_0\times \sqrt 2 t$ [from equation (iv)]

$\Rightarrow D_2=2\sqrt 2 D$ [from equation (i)]

$\Rightarrow D_2=2.82 D$ (approximately)

This is out of range, so there is no point for $2V_0$ .

Hence, students must choose the speed $V_0$ to launch the sphere to get the maximum number of points.

7 0

3 years ago

The resistance of resistor is greater for:

Monica [59]

Answer:

c: long and thin resistor.

Explanation:

The resistance of a resistor is given by:

R = ρ*L/A

where:

R = resistance

ρ = resistivity (depends on the material)

L = length of the material

A = cross-sectional area of the material

We can see that the length is on the numerator, which means that if we increase the length, then the resistance is increased.

We also can see that the cross-sectional area is on the denominator, then if we increase the area (for example, with a ticker resistor) the resistance decreases.

Then if we want to maximize the resistance, we need to have a long and thin resistor, so the correct answer is c.

8 0

3 years ago

Please help me I have been struggling

disa [49]

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Heartz measures the number of waves per second, not per hour.
Sounds do not travel in space because it's empty
Sounds make our eardrum vibrate. The diaphragm is the muscle we use to breathe.
Soft, pale materials are best for sound proofing because they absorb the waves.
In the next paragraph, ultrasound and infrasound are inverted (ultra means "over, above" and infra means "below").

4 0

3 years ago

Need help with ASAP please

Oksanka [162]

Answer:

first blank is chemical second blank is kinetic energy

8 0

3 years ago