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3 years ago

I WILL GIVE YOU BRAINLIEST! PLEASE HELP

Physics

1 answer:

Mice21 [21]3 years ago

5 0

Answer:

0.00547s

Explanation:

Step one:

given data

mass= 0.4kg

force= 1060.9N

velocity = 14.5m/s

Frome

Ft= mv

substitute

t= mv/F

t=0.4*14.5/1060.9

t=5.8/1060.9

t=0.00547s

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What is the frequency of radiation whose wavelength is 2.40 x 10-5 cm?

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<span>To begin, the formula for finding frequency when wavelength is known is "f = c / w" when c is the constant velocity (3 * 10^8 m/s). To convert the wavelength into a common form (m/s), it will have to be multiplied by 10^-2. This leaves the equation as "f = 3.0 * 10^8 / (2.4 * 10^-5 * 10^-2), or 2.4 * 10^-7. This gives 1.25 * 10^15 m/s as the frequency.</span>

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3 years ago

A particle (charge = +0.8 mC) moving in a region where only electric forces act on it has a kinetic energy of 6.7 J at point A.

Maksim231197 [3]

Answer:

The kinetic energy of the particle as it moves through point B is 7.9 J.

Explanation:

The kinetic energy of the particle is:

$\Delta K = \Delta E_{p} = q\Delta V$

<u>Where</u>:

K: is the kinetic energy

$E_{p}$ : is the potential energy

q: is the particle's charge = 0.8 mC

ΔV: is the electric potential = 1.5 kV

$\Delta K = q \Delta V= 0.8 \cdot 10^{-3} C*1.5 \cdot 10^{3} V = 1.2 J$

Now, the kinetic energy of the particle as it moves through point B is:

$\Delta K = K_{f} - K_{i}$

$K_{f} = \Delta K + K_{i} = 1.2 J + 6.7 J = 7.9 J$

Therefore, the kinetic energy of the particle as it moves through point B is 7.9 J.

I hope it helps you!

8 0

3 years ago

A straight wire 20 cm long, carrying a current of 4 A, is in a uniform magnetic field of 0.6 T. What is the force on the wire wh

zheka24 [161]

Answer:

Magnetic force, F = 0.24 N

Explanation:

It is given that,

Current flowing in the wire, I = 4 A

Length of the wire, L = 20 cm = 0.2 m

Magnetic field, B = 0.6 T

Angle between force and the magnetic field, θ = 30°. The magnetic force is given by :

$F=ILB\ sin\theta$

$F=4\ A\times 0.2\ m\times 0.6\ T\ sin(30)$

F = 0.24 N

So, the force on the wire at an angle of 30° with respect to the field is 0.24 N. Hence, this is the required solution.

7 0

3 years ago

Read 2 more answers

The world's most powerful laser is the LFEX laser in Japan. It can produce a 2 petawatt(2×10^15W) laser pulse that last for 1 ps

kogti [31]

Answer:

$2.82942\times 10^{24}\ W\m^2$

Explanation:

d = Diameter of spot = 30 μm

r = Radius of spot = $\frac{d}{2}=\frac{30}{2}=15\ mu m$

P = Power of the laser = $2\times 10^{15}\ W$

A = Area = $\pi r^2$

Intensity is given by

$I=\frac{P}{A}\\\Rightarrow I=\frac{2\times 10^{15}}{\pi\times (15\times 10^{-6})^2}\\\Rightarrow I=2.82942\times 10^{24}\ W/m^2$

The light intensity within this spot is $2.82942\times 10^{24}\ W/m^2$

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3 years ago

A spherical ball of charged particles has a uniform charge density. In terms of the ball's radius R, at what radial distances in

eimsori [14]

Answer:

Electric field at radius r inside the solid sphere is

Electric field at radius r between inner radius and outer radius of the shell is

E=0 N/C

Explanation:Given

The radius of the solid sphere is

The charge on the solid sphere is

The inner radius of the shell is

The outer radius of the shell is

The total charge on the shell is

PART(A)

The magnitude of electric field at radius r where \\The volumetric charge density of the solid sphere will be

The charge enclosed by the radius r inside the solid sphere is

rho=

According to gauss law

PART(B)

The electric field at radius r where

The shell is conducting so due to induction of charge

The charge induced on the inner surface of the shell is equal in magnitude of the total charge on the solid sphere but polarity will be changed because a conducting shell has no net electric field inside the shell

So,

The charge on the inner surface of the shell is

Due to conservation of the charge on the shell

The charge accumulated on the outer surface of the shell is

The charge enclosed by the radius r where

According to gauss law

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2 years ago