If a Charged particle at rest experiences no electromagnetic force. what happens?​

Answer:

[tex]E=6.68\times 10^{-26}\ J[/tex]

Explanation:

Given that,

The frequency of FM ratio station, f = 100.8 MHz = 100.8 × 10⁶ Hz

We need to find the energy of the wave. We know that,

Energy, E = hf

Put all the values,

[tex]E=6.63\times 10^{-34}\times 100.8\times 10^6\\\\=6.68\times 10^{-26}\ J[/tex]

So, the energy of the wave is equal to [tex]6.68\times 10^{-26}\ J[/tex].

Answer:

In the above diagram, we move by horizontally with a velocity to the right. But due to gravity the hand falls a distance as we move .

Therefore the time it takes to fall by a distance is given by =.

If = as →∞, then from the laws of motion

=12()22

Therefore for the whole journey from left to right our muscle has to give a beat up at an amount equal to the gravity force. So, the work supplied by the muscle for one trip would be

∫=∫012()22

A man is going to rub the chalk off a blackboard, he is going to choose a way to rub off the chalk in two ways

Starting from the upper left corner of the board and moving horizontally to the right and moving slightly down and then again moving to the right and then when it reaches the left corner and so on

The second way is given in the following: He starts at the top and rubs down and then up again and so on.

He is going to choose the way that is less tiring in his arms. i.e., less work done by his muscles!

Somehow from vague intuitive notion I am inclined to agree that it would be less tiring to rub off the chalk if he moves the duster beginning at the top and rubbing horizontally and gradually decreasing the height when one reach the corners till everything is off.

I tried to calculate the muscle work by considering the following ideas(please correct me if I'm wrong):

The nerves inside our muscle has to fire its signal continuously throughout the entire interval of the process of rubbing and its has to oppose the gravitational force keeping the hand up the air.

When the hand is at the highest point on the blackboard, the muscle has to work against gravity =, as the muscle continues to work against gravity throughout the interval from left to right (the horizontal path of the first case) I couldn't find the total work withstood by our muscle.

Answer:

D. Centripetal Force :)

Explanation:

voltage is 50v and capacitance is 30c what is the charge

q=1500c

Answer:

The acceleration of the object is -69.78 m/s²

Explanation:

Given;

postion of the particle:

[tex]x = 32t - 38\frac{t^3}{3} \\\\[/tex]

The velocity of the particle is calculated as the change in the position of the  particle with time;

[tex]v = \frac{dx}{dt} = 32 - 38t^2\\\\when \ the \ particle \ is \ at \ rest, \ v = 0\\\\32-38t^2 = 0\\\\38t^2 = 32\\\\t^2 = \frac{32}{38} \\\\t = \sqrt{\frac{32}{38} } \\\\t = 0.918 \ s[/tex]

Acceleration is the change in velocity with time;

[tex]a = \frac{dv}{dt} = -76t\\\\recall , \ t = 0.918 \ s\\\\a = -76(0.918)\\\\a = -69.78 \ m/s^2[/tex]

Answer:

I’ll try my best!

Answer:

C

Explanation:

The normal is the line which divides the angle between the incident ray (which is the ray of an object which strikes the mirror) and the reflected ray(the ray which is thrown back as the object hits the mirror surface) into two equal parts. The normal is always perpendicular to the surface. In the description agram Given , the Noa which is the line C, divides the reflected ray (line D) and the incident ray (line A) into two equal parts. The plane surface is line B and the other incident ray (line C) is perpendicular to B

Answer:

B. -20 m/s

Explanation:

Given the following data:

Initial velocity = 30 m/s

Acceleration = 5 m/s²

Deceleration = -5 m/s² (deceleration is the negative of acceleration)

Time = 10 seconds

To find the final velocity, we would use the first equation of motion;

[tex] V = U + at[/tex]

Where;

V is the final velocity.

U is the initial velocity.

a is the acceleration.

t is the time measured in seconds.

Substituting into the formula, we have;

[tex] V = 30 + (-5)*10[/tex]

[tex] V = 30 - 50[/tex]

Final velocity, V = -20 m/s

Answer:

A) 3.367 × 10^(-6)

B) 2.97 × 10^(7) N/C

C) Upwards

Explanation:

We are given;

Mass of bee; m = 100 mg = 100 × 10^(-6) kg

Charge on bee;q=33 pC = 33 × 10^(-12)C

Electric field strength; E = 100 N/C

A) Formula for weight of bee; W = mg = 100 × 10^(-6) × 9.8 = 9.8 × 10^(-4) N

Electric force on Bee; F = qE = 33 × 10^(-12) × 100 = 33 × 10^(-10) N

ratio of the electric force on the bee to the bee's weight; F/W = (33 × 10^(-10))/(9.8 × 10^(-4)) = 3.367 × 10^(-6)

B) For the bee to be suspended in the air, it means the weight of the bee must be equal to the electric force. Thus;

mg = qE

100 × 10^(-6) × 9.8 = 33 × 10^(-12) × E

E = (100 × 10^(-6) × 9.8)/(33 × 10^(-12))

E = 2.97 × 10^(7) N/C

C) From Newton's law, sum of forces = 0.

Thus;

F_n + F + W = 0

Where F is the normal force.

Thus;

F_n = -(F + W)

F_n = - ((33 × 10^(-10)) + (9.8 × 10^(-4)))

F_n = -9.8 × 10^(-4) N

Thus, applied electric field is;

E_a = F_n/q = (-9.8 × 10^(-4))/(33 × 10^(-12)) = -2.97 × 10^(7) N/C

This is negative and so it means the direction will be opposite the Earth's electric filed which is upwards.

Answer:

The output electric power is 1338876 W.

Explanation:

Volume, V = 690 cubic meter

height, h = 220 mm = 0.22 m

efficiency = 90 %

time , t = 1 s

Let the mass is m.

m = volume x density  

m = 690 x 1000 = 690000 kg

The input power is

P = m g h = 690000 x 9.8 x 0.22 = 1497640 W

The electric power out put is

[tex]P' = 90 % of 1487640\\\\\\P' = 1338876 W[/tex]

Answer:

    v₂ = 63.62 m / s

Explanation:

For this exercise in fluid mechanics we will use Bernoulli's equation

         P₁ + ρ g v₁² +  ρ g y₁ = P₂ +  ρ g v₂² +  ρ g y₂

where the subscript 1 refers to the inside of the wing and the subscript 2 to the top of the wing.

We will assume that the distance between the two parts is small, so y₁ = y₂

        P₁-P₂ =  ρ g (v₂² - v₁²)

pressure is defined by

        P = F / A

we substitute

        ΔF / A =  ρ g (v₂² - v₁²)

         v₂² = [tex]\frac{\Delta F}{A \ \rho \ g} + v_1^2[/tex]

suppose that the area of ​​the wing is A = 1 m²

we substitute

         v₂² = [tex]\frac{1000}{1 \ 1.29 \ 9.8} + 63^2[/tex]

         v₂² = 79.10 + 3969

         v₂ = √4048.1

         v₂ = 63.62 m / s

Answer:

By measure the effective length and the time period of the pendulum.

Explanation:

Let the student take the oscillating pendulum at the planet.

He measure the time period of the pendulum  by using the stop watch or the ordinary watch.

Then measure the effective length of the pendulum which is the distance between the center of gravity of the bob and the point of suspension of the pendulum.

Now, use the formula of the time period of the pendulum,

[tex]T =2\pi\sqrt\frac{L}{g}[/tex]

Here, L is the effective length of the pendulum, g is the acceleration due to gravity at the planet and T is time period of the pendulum.  

By rearranging the terms, we get

[tex]T =2\pi\sqrt\frac{L}{g}\\\\T^{2}=4\pi^2\times\frac{L}{g}\\\\g =\frac{4\pi^2L}{T^2}[/tex]

Here, by substituting the values of L and T, the student get the value of acceleration due to gravity at that planet.

Answer:

Explanation:

Kinetic energy is at an absolute max when the potential energy is 0. At the ball's highest point, at its most absolute highest point, the velocity of the ball is 0, making KE = 0 and PE the only energy the ball has. So if this isn't a trick question, the wording is off.

Answer:

θ = 12.5 rotations

Explanation:

The number of rotations can be found by using the second equation of motion:

[tex]\theta = \omega_i t + \frac{1}{2}\alpha t^2\\\\[/tex]

where,

[tex]\theta[/tex] = angular displacement = ?

ωi = initial angular speed = 0 rad/s

t = time = 5 s

α = angular acceleration = 2π rad/s²

Therefore,

[tex]\theta = (0\ rad/s)(5\ s)+\frac{1}{2}(2\pi\ rad/s^2)(5\ s)^2\\\\\theta = 78.54\ rad[/tex]

converting it to no. or rotations:

[tex]\theta = (78.54\ rad)(\frac{1\ rotation}{2\pi\ rad})[/tex]

θ = 12.5 rotations

Answer: 90

Explanation: a=Fnet/M

=27/0.30

=90

the acceleration is 90

Answer:

The acceleration of the ball is 83.333ms2 [forward].

Explanation:

i hope it helps :)

Answer:

the process of finding the age of fossils is called carbon dating because they use samples of carbon and compare it's characteristics to the samples of carbon from each set time

Answer:

wool and fibers

Explanation:

Answer:

Một ô tô có khối lượng m=1000kg đang chạy với vận tốc 18km/h thì hãm phanh.​Biết lực hãm là 2000N. Tính quãng đường xe còn chạy thêm trước khi dừng ... Chiếu phương trình của định luật II Newtơn mà →F=m. ... chuyển động ta có F=m​a, suy ra gia tốc chuyển động của xe ( với F=2000N) ...

Explanation:

Answer:

the correct answer is D

Explanation:

In this exercise, the vectors are in the same west-east direction, so we can assume that the positive direction is east and perform the algebraic sum.

       R = δ + ε

where

       δ = 2.0 m

       ε = 7.0 m

the positive sign indicates that it is heading east

        R = 2.0 + 7.0

        R = 9.0 m

the direction is east

the correct answer is D

Answer:

= 0.257

Explanation:

Applied force on the safe by Bonnie and Clyde is F= 387 N + 341 N

= 728 N

Given safe slides with constant speed .So, force of friction =applied force

= 728 N

μ*normal force = 728 N

from this the safes coefficient of kinetic friction on the bank floor is

μ = 728 N / normal force

= 728 N / Mg

= 728 N / ( 289 * 9.8 )

= 0.257

Answer:

The final speed is 5.78 m/s.

Explanation:

mass, m = 375 g = 0.375 kg

initial velocity, u = 4 m/s

Distance, s = 2.5 m

Angle, A = 30 degree

Force, F = 1.5 N

let the final velocity is v.

Use the work energy theorem

Work done = change in kinetic energy

[tex]W= 0.5 m(v^2 - u^2)\\\\F s cos A= 0.5 m (v^2 - u^2)\\\\1.5\times 2.5\times cos30= 0.5\times 0.375\times (v^2 - 16)\\\\v = 5.78 m/s[/tex]

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Answer:

Explanation:

The first thing we are asked to find is the Force experienced by the box. That is found in the formula:

F - f = ma where F is the force exerted by the box, f is the friction opposing the box, m is the mass, and a is the acceleration (NOT the same as the pull of gravity). But F can be rewritten in terms of the angle of inclination also:

[tex]wsin\theta-f=ma[/tex] where w is the weight of the box. We will use this version of the formula because it will help us answer the second question, which is to solve for a. Filling in:

First we need the weight of the box. Having the mass, we find the weight:

w = mg so

w = 25.14(10) so

w = 251.4 N (I am not paying any attention at all to the sig fig's here, since I noticed no one on this site does!) Now we have the weight. Filling that in:

251.4sin(30) - f = ma Before we go on to fill in for f, let's answer the first question. F = 251.4sin(30) so

F = 125.7   And in order to answer what a is equal to, we find f:

f = μ[tex]F_n[/tex] where Fn is the weight of the object.

f = .25(251.4) so

f = 62.85. Filling everything in now altogether to solve for a, the only missing value:

125.7 - 62.85 = 25.14a and

62.85 = 25.14a so

a = 2.5 m/s/s

Now we have to move on to another set of equations to answer the last part. The last part involves the y-dimension. In this dimension, what we know is that

a = -10 m/s/s

v₀ = 0 (it starts from rest)

Δx = -13.30 m (negative because the box falls this fr below the point fro which it started). Putting all that together in the equation for displacement:

Δx = v₀t + [tex]\frac{1}{2}at^2[/tex] and we are solving for time:

[tex]-13.30=0t+\frac{1}{2}(-10)t^2[/tex] and

[tex]t=\sqrt{\frac{2(-13.30)}{-10} }[/tex] so

t = 1.6 seconds to reach the bottom of the slope from 13.30 m high.

Answer:

The center of gravity of the tile is at the center of the tile:

Examining 1/2 of the tile, each mass  point will have an equal mass point

on the other side of the tile (considering a line thru the center of the tile)

with coordinates opposite and equal of the point being considered.

Each mass point will have an identical mass point on the opposite side of the tile, with coordinates opposite and equal to the point being analyzed, while looking at only half of the tile.

A hypothetical location where the gravitational force is said to appear to act is the human body's centre of gravity (COG). That is the location where the entire body's bulk appears to be concentrated[2]. The COG need not be located inside the actual boundaries of an object or a person because it is a hypothetical point. Imagine the thing balanced on one finger to get an idea of its COG (there are objective measurements).

The COG is positioned anatomically so that it is about anterior to the second sacral vertebra. The specific placement of the COG, however, continually shifts with each new posture of the torso and limbs since humans do not remain stable in the anatomical position. The dimensions of the body

The center of the gravity is at half of the length of tile.

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Answer:

The dipole always tries to align in the direction of electric field so that the net torque is zero.

Explanation:

When a dipole placed in an electric field at any arbitrary angle, it experiences a torque.

Let the dipole moment is p, electric field is E and the angle is A.  

The torque is given by

Torque = p E sin A

where, A is the angle between the electric field and dipole moment.

So, the dipole always tries to align in the direction of electric field so that the net torque is zero.

Answer:

an unconscious needs rooted in childhood are the foundation of psychoanalytic theory.

Explanation:

Answer:

Example:Opening of a bottle cap by tool

when we hold a tool and open the bottle cap this is because , force x tool force .

The load arm is usually shorter than the effort arm in second order levers. Moving a large weight hence requires less effort. A force multiplier lever or effort multiplier lever is the name for this kind of lever. A boat's oars, for instance, can increase the force.

Second-order levers are devices with the input force farthest from the fulcrum and the output force on the same side of the fulcrum. A wheelbarrow is an excellent illustration of a second-order lever.

A second-order lever will have an output force greater than an input force, similar to first-order levers. The output journey, however, will be shorter than the input length. Both the input and output forces in this situation will move in the same direction.

Learn more about lever here:

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Two connection methods are used for brushless DC motors. One method is to connect the coils in a loop as we compared it with the rotor winding of DC motors in Fig. 2.27. This method is called a Δ (delta) connection.

CORRECT ME IF IM WRONG!!

>>> JAZMINE

(a) 2.001N

(b) 2.001N

A sketch of the scenario has been attached to this response.

Since only the force vector F₁ is required, the only force shown in the sketch is F₁.

As shown in the sketch;

The x-component of the force vector F₁ = [tex]F_{x}[/tex]

The y-component of the force vector F₁ = [tex]F_{y}[/tex]

The magnitude of F₁ as given in the question = 2.83N

The angle that the force makes with respect to the x-axis = 45.0°

Using the trigonometric ratio, we see that;

(a) cos 45.0° = [tex]\frac{F_x}{F_1}[/tex]

=>  [tex]F_{x}[/tex] =  F₁ cos 45.0°

=> [tex]F_{x}[/tex] =  2.83 cos 45.0°

=> [tex]F_{x}[/tex] =  2.83 x 0.7071

=> [tex]F_{x}[/tex] =  2.001N

(b) Also;

sin 45.0° = [tex]\frac{F_y}{F_1}[/tex]

=>   [tex]F_{y}[/tex] =  F₁ sin 45.0°

=>  [tex]F_{y}[/tex] =  2.83 sin 45.0°

=>  [tex]F_{y}[/tex] =  2.83 x 0.7071

=>  [tex]F_{y}[/tex] =  2.001N

Therefore, the x-component and y-component of the force vector F₁ is 2.001N

The x and y component of vector F1 is mathematically given as

F_x =  2.001N

F_y=  2.001N

Question Parameters:

Generally, the equation for the x-component  is mathematically given as

x=Fsin\theta

Therefore

F_x =  F₁ cos 45.0°

F_x =  2.83 x 0.7071

F_x =  2.001N

For y component

x=Fcos\theta

F_y =  F₁ sin 45.0

F_y =  2.83 x 0.7071

F_y=  2.001N

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Answer:

The downwards acceleration is 3.53 m/s2.

Explanation:

Let the true weight is m g.

The reading of the balance, R = 0.64 mg

Let the acceleration is a.

As the apparent weight is less than the true weight so the elevator goes down wards with some acceleration.

Use Newton's second law

m g - R = m a

m g - 0.64 m g = m a

0.36 g = a

a = 3.53 m/s2