Can a body possess velocity at the same time in horizontal and vertical directions?

The Hall effect can be used to calculate the charge-carrier number density in a conductor. If a conductor carrying a current of

kicyunya [3294]

Answer:

$6.6*10^{27}e/m^3$

Explanation:

When calculating Hall voltage, it is crucial to have the current, magnetic field strength, length, area, and number of charge carriers available. The Hall voltage can be expressed using the equation:

$V_h = \frac{iB}{neL}$

Where:

i= the current

B= the magnetic field strength

L = the length

n = the number of charge carriers

e= charge of an electron

We need to replace values and solve for n:

$n= \frac{iB}{V_h e L}$

$n= \frac{2*1.2}{4.5*10^{-6}*5^10^{-3}*1.6*10^{-19}}$

$n= 6.6*10^{27}electron.m^{-3}$

As a result, the charge carrier density is $6.6*10^{27}e/m^3$

5 0

3 months ago

A solid sphere is released from the top of a ramp that is at a height h1 = 2.30 m. It rolls down the ramp without slipping. The

Yuliya22 [3333]

Answer:

The horizontal distance d that the ball covers before it lands is 1.72 m.

Explanation:

Given,

Height of ramp $h_{1}=2.30\ m$

Height of bottom of ramp $h_{2}=1.69\ m$

Diameter = 0.17 m

We need to determine the horizontal distance d the ball travels before landing.

We need to calculate the time

Using the equation of motion

$h_{2}=ut+\dfrac{1}{2}gt^2$

$t=\sqrt{\dfrac{2h_{2}}{g}}$

$t=\sqrt{\dfrac{2\times1.69}{9.8}}$

$t=0.587\ sec$

Next, we can find the ball's velocity

Using the kinetic energy formula

$K.E=\dfrac{1}{2}mv^2+\dfrac{1}{2}I\omega^2$

$K.E=\dfrac{1}{2}mv^2+\dfrac{1}{2}\times(\dfrac{2}{5}mr^2)\times(\dfrac{v}{r})^2$

$K.E=\dfrac{7}{10}mv^2$

By applying the conservation of energy

$K.E=mg(h_{1}-h_{2})$

$\dfrac{7}{10}mv^2=mg(h_{1}-h_{2})$

$v^2=\dfrac{10}{7}\times g(h_{1}-h_{2})$

We substitute the values into the equation

$v=\sqrt{\dfrac{10\times9.8\times(2.30-1.69)}{7}}$

$v=2.922\ m/s$

Next, we determine the horizontal distance d the ball travels before landing

Using the distance formula

$d =vt$

Where. d = distance

t = time

v = velocity

We substitute the values into the formula

$d=2.922\times 0.587$

$d=1.72\ m$

Thus, the horizontal distance d that the ball travels before it lands is 1.72 m.

8 0

2 months ago

A bowling pin is thrown vertically upward such that it rotates as it moves through the air, as shown in the figure. Initially, t

Ostrovityanka [3204]

Answer:

Explanation:

The equation used to determine the maximum height of the bowling pin during its trajectory is given by;

H = u²/2g

where u, the initial speed/velocity, equals 10m/s

g stands for gravitational acceleration = 9.81m/s²

Substituting in the values gives us

H = 10²/2(9.81)

H = 100/19.62

Consequently, the highest point of the bowling pin's center of mass is approximately 5.0m.

3 0

4 months ago

A person weighing 55 kg walks by applying 160 N of force on the ground, while pushing a 10-kg object. If the person accelerates

Sav [3153]

Answer:

I'm uncertain

Explanation:

since I didn't provide a correct answer, continue with my inquiries and you can use 'I'm uncertain' for 100 points.

6 0

3 months ago

Can a body possess velocity at the same time in horizontal and vertical directions?​

Can a body possess velocity at the same time in horizontal and vertical directions?