All categories

3 years ago

Conventionally, the field strength around a charged object is the direction of the force acting on a .

Physics

2 answers:

damaskus [11]3 years ago

8 0

Answer:

Field Strength:

Any charged body has the capacity to effect any test charge that comes inside its field or region. It can be also defined as,

"The total amount or magnitude of force,F or intensity felt by any unit test charge when it enters an electromagnetic or electric field."

For an electric field its unit will be, volt per meter or simply V/m.

Explanation:

A unit test charge inside an electric field:

When a unit test charge enters a given parameters or area set by the charged particle then it will surely experience a force,F equal to the magnitude of that charge body and it will be different as it continues to move closer or far inside the field.

As, we have, "E=F/q",(where "F" is the field strength and "q" is the unit test charge placed inside the field).

ExtremeBDS [4]3 years ago

7 0

Unit positive test charge

You might be interested in

Effciency of a lever is never 100% or more. why?Give reason

Troyanec [42]

Answer:

Ideally, the work output of a lever should match the work input. However, because of resistance, the output power is nearly always be less than the input power. As a result, the efficiency would go below $100\%$ .

Explanation:

In an ideal lever, the size of the input and output are inversely proportional to the distances between these two forces and the fulcrum. Let $D_\text{in}$ and $D_\text{out}$ denote these two distances, and let $F_\text{in}$ and $F_\text{out}$ denote the input and the output forces. If the lever is indeed idea, then:

$F_\text{in} \cdot D_\text{in} = F_\text{out} \cdot D_\text{out}$ .

Rearrange to obtain:

$\displaystyle F_\text{in} = F_\text{out} \cdot \frac{D_\text{out}}{D_\text{in}}$

Class two levers are levers where the perpendicular distance between the fulcrum and the input is greater than that between the fulcrum and the output. For this ideal lever, that means $D_\text{in} > D_\text{out}$ , such that $F_\text{in} < F_\text{out}$ .

Despite $F_\text{in} < F_\text{out}$ , the amount of work required will stay the same. Let $s_\text{out}$ denote the required linear displacement for the output force. At a distance of $D_\text{out}$ from the fulcrum, the angular displacement of the output force would be $\displaystyle \frac{s_\text{out}}{D_\text{out}}$ . Let $s_\text{in}$ denote the corresponding linear displacement required for the input force. Similarly, the angular displacement of the input force would be $\displaystyle \frac{s_\text{in}}{D_\text{in}}$ . Because both the input and the output are on the same lever, their angular displacement should be the same:

$\displaystyle \frac{s_\text{in}}{D_\text{in}} =\frac{s_\text{out}}{D_\text{out}}$ .

Rearrange to obtain:

$\displaystyle s_\text{in}=s_\text{out} \cdot \frac{D_\text{in}}{D_\text{out}}$ .

While increasing $D_\text{in}$ reduce the size of the input force $F_\text{in}$ , doing so would also increase the linear distance of the input force $s_\text{in}$ . In other words, $F_\text{in}$ will have to move across a longer linear distance in order to move $F_\text{out}$ by the same $s_\text{out}$ .

The amount of work required depends on both the size of the force and the distance traveled. Let $W_\text{in}$ and $W_\text{out}$ denote the input and output work. For this ideal lever:

$\begin{aligned}W_\text{in} &= F_\text{in} \cdot s_\text{in} \\ &= \left(F_\text{out} \cdot \frac{D_\text{out}}{D_\text{in}}\right) \cdot \left(s_\text{out} \cdot \frac{D_\text{in}}{D_\text{out}}\right) \\ &= F_\text{out} \cdot s_\text{out} = W_\text{out}\end{aligned}$ .

In other words, the work input of the ideal lever is equal to the work output.

The efficiency of a machine can be measured as the percentage of work input that is converted to useful output. For this ideal lever, that ratio would be $100\%$ - not anything higher than that.

On the other hand, non-ideal levers take in more work than they give out. The reason is that because of resistance, $F_\text{in}$ would be larger than ideal:

$\displaystyle F_\text{in} = F_\text{out} \cdot \frac{D_\text{out}}{D_\text{in}} + F(\text{resistance})$ .

As a result, in real (i.e., non-ideal) levers, the work input will exceed the useful work output. The efficiency will go below $100\%$ ,

4 0

3 years ago

If the car passes point A with a speed of 20 m/s and begins to increase its speed at a constant rate of at = 0.5 m/s2 , determin

IceJOKER [234]

Answer:

$1.68 \frac{m}{s^2}$

Explanation:

Please find the image for the question as attached file.

Solution -

Given -

First of all we will calculate the velocity at point C,

As per newton's third law of motion-

$V_C^2 = V_A^2 + 2 a_t (S_C - S_A)\\$

Substituting the given values in above equation, we get -

$V_C^2 = 20^2 + 2*0.5*(100-0)\\V_C = 22.361 \frac{m}{s}$

Now we will determine the radius of curvature for the curve shown in the attached image

$Y = 16 - \frac{1}{625} X^2\\$

Differentiating on both the sides, we get -

$\frac{dy}{dx} = -3.2 (10^-3) X\\\frac{d^2y}{d^2x} = -3.2 (10^-3)\\Curve = \frac{[1+(\frac{dy}{dx})^2]^{\frac{3}{2}} }{\frac{d^2y}{d^2x}} \\Curve = 312.5$ meter

Acceleration on curved path

$a = \frac{V_C^2}{Curve} \\a = \frac{22.361^2}{312.5} \\a= 1.60 \frac{m}{s^2}$

Final acceleration

$a_f = \sqrt{0.5^2 + 1.6^2} \\a_f = 1.68\frac{m}{s^2}$

5 0

3 years ago

What gas makes up most of our atmosphere in the troposphere?

Agata [3.3K]

78 nitrogen I think

4 0

3 years ago

Read 2 more answers

One evolutionary strategy that has maximized gas diffusion is

Akimi4 [234]

Answer:

Increasing its surface area

Explanation:

In Science, increasing the surface area of a gas increases its collision rate and is one evolutionary strategy that has maximized gas diffusion by catalysing the rate of gaseous reaction.

7 0

3 years ago

A canon is tilled back 30.0 degrees and shoots a cannon ball at 155 m/s. What is the

Umnica [9.8K]

Answer:

$= \frac{(115^2)(\sin 30)^2}{2\times 9,8} \\\\= 168.7m$

Therefore, highest point that the cannon ball reaches is 168.7m

Explanation:

the cannon is fired at an angle 30 o to the horizonatal with a speed of 155 m/s

highest point that the cannon ball reaches?

$H_{max}=\frac{V^2\sin ^2 \theta}{2g}$

g = 9.8m/s2

$= \frac{(115^2)(\sin 30)^2}{2\times 9,8} \\\\= 168.7m$

Therefore, highest point that the cannon ball reaches is 168.7m

6 0

3 years ago