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4 years ago

While calculating electric field should we include the sign of the charge in the formula?

1 answer:

3 0

I think that the sign of the charge in the formula would be not needed. The sign is not important is if you want just the magnitude (the absolute value) of the electric field in a given point in space. Hope this answers the question. Have a nice day.

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Image caught on Screen is called

bixtya [17]

Answer:

<h2>Virtual image</h2>

Explanation:

<h3>Virtual image can be caught on a screen</h3>

hope this helps you.

will give the brainliest!

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4 0

3 years ago

Read 2 more answers

A student (m = 68 kg) falls freely from rest and strikes the ground. During the collision with the ground, he comes to rest in a

Gnesinka [82]

Answer:

5.7141 m

Explanation:

Here the potential and kinetic energy will balance each other

$mgh=\frac{1}{2}mv^2\\\Rightarrow v=\sqrt{2gh}$

This is the initial velocity of the system and the final velocity is 0

t = Time taken = 0.04 seconds

F = Force = 18000 N

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s²

Equation of motion

$v=u+at\\\Rightarrow a=\frac{v-u}{t}$

From Newton's second law

$F=ma\\\Rightarrow F=m\frac{v-u}{t}\\\Rightarrow 18000=68\frac{0-\sqrt{2gh}}{0.04}\\\Rightarrow \frac{18000}{68}\times 0.04=-\sqrt{2\times 9.81\times h}\\\Rightarrow 10.58823=-\sqrt{2\times 9.81\times h}$

Squarring both sides

$112.11061=2\times 9.81\times h\\\Rightarrow h=\frac{112.11061}{2\times 9.81}\\\Rightarrow h=5.7141\ m$

The height from which the student fell is 5.7141 m

5 0

3 years ago

At this output power, how long would it take an 89 kg cyclist (including the mass of his bike) to climb the famed 1100-m-high al

Artemon [7]

In general the route that follows the 1100-m-high alpe d'huez mountain stage takes anywhere from four to six hours in athletes from around the world. A beginner would probably take 8 hours or so. World records were recently broken with a time just under three hours. The most friction for cyclist is the competition itself.

7 0

3 years ago

A wooden block of mass m = 9 kg starts from rest on an inclined plane sloped at an angle θ = 30° from the horizontal. The block

jenyasd209 [6]

Answer:

Fk = 21.645N

Explanation:

Let Fb be Force of block and thus;

Fb= mg where m is mass of block and g is acceleration due to gravity

Thus Fb= 9kg x 9.81N/kg = 88.29 N

Now, the question says this force Fb rests at an inclined plane [email protected] 30° angle

Thus;

Force parallel to inclined plane = 88.29 x sin30° = 44.145 N.

Force perpendicular to the inclined plane = 88.29 x cos30 = 76.46 N

Now, when an object is falling freely, we know that

h = (1/2)at^(2)

From the question, the height is 5m and t= 2 seconds

Thus;

5 = (1/2)a(2)^(2)

2a = 5 and thus,

a = 5/2 = 2.5 m/s^(2)

Now, in inclined planes, perpendicular force - kinetic friction force = Resultant force

Thus let perpendicular be Fp and kinetic friction force be Fk and so;

Fp - Fk = F

F= ma = 9 x 2.5 = 22.5N

Thus, 44.145 - Fk = 22.5

Thus, Fk = 44.145 - 22.5 = 21.645N

5 0

3 years ago

A soccer ball is kicked from the ground with an initial speed v at an upward angle θ. A player a distance d away in the directio

Nady [450]

Answer:

The speed of player is given by

$V=\frac{2v^{2} 2sin\alpha.cos\alpha-gd }{2vsin\alpha}$

Explanation:

The time of flight for a projectile motion is given by

$T=\frac{2vsin\alpha }{g}$ (i)

where t is the time of flight, v is the initial speed, and α is the angle.

Now the person must also reach the impact point of ball in the same time as above.

Now the total distance D the player needs to cover is basically R horizontal range of projectile minus the distance d, range R is given by,

$R=\frac{2v^{2} sin2\alpha }{g}$

Now the distance the player must cover is given by

D= R-d

D= $\frac{2v^{2} sin2\alpha }{g}$ - d

$D=\frac{2v^{2}sin2\alpha-gd}{g}$ (ii)

Now the average speed of player is given by

$V=\frac{D}{T}$ (iii)

Replacing the values of D and T from eq. (i) and (ii) in eq. (iii).

$V=\frac{\frac{2v^{2} sin2\alpha-gd }{g}}{\frac{2vsin\alpha }{g} }$

$V=\frac{2v^{2} 2sin\alpha.cos\alpha-gd }{2vsin\alpha}$

3 0

3 years ago