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3 years ago

29. The x-axis of a velocity-time graph represents

Physics

1 answer:

Roman55 [17]3 years ago

8 0

Answer:

Time (s)

Explanation:

The x-axis in velocity-time graph represents the time in second and the y-axis represents the velocity in metres per second (m/s)

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An object of mass 800g hangs on a spring. Calculate the force exerted by the object if acceleration due to gravity is 10/s2

xxTIMURxx [149]

Answer:

8 N

Explanation:

Using the equation F=ma (F: force/ m: mass in kg/ a: acceleration),

F = (800/1000)(10)

F = 8 N

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Tornadoes form in unusually violent thunderstorms when there is sufficient (1) instability and (2) wind shear present in the low

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its D

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The line of latitude that is zero degrees latitude is the

Nuetrik [128]

Answer:

Equator

Explanation:

The equator is the line of latitude that separates the earth into the northern and southern hemisphere.Its magnitude is 0 degrees latitude. The equator and other lines of latitude are parallel to each other. The lines of longitude which is 0 degrees is known as the Greenwich meridian. The longitudinal lines however are not parallel as they converge at the poles.

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3 years ago

An electron is moving east in a uniform electric field of 1.55 N/C directed to the west. At point A, the velocity of the electro

valkas [14]

Answer:

Final velocity of electron, $v=6.45\times 10^5\ m/s$

Explanation:

It is given that,

Electric field, E = 1.55 N/C

Initial velocity at point A, $u=4.52\times 10^5\ m/s$

We need to find the speed of the electron when it reaches point B which is a distance of 0.395 m east of point A. It can be calculated using third equation of motion as :

$v^2=u^2+2as$ ........(1)

a is the acceleration, $a=\dfrac{F}{m}$

We know that electric force, F = qE

$a=\dfrac{qE}{m}$

Use above equation in equation (1) as:

$v^2=u^2+\dfrac{2qEs}{m}$

$v^2=(4.52\times 10^5\ m/s)^2+2\times \dfrac{1.6\times 10^{-19}\ C\times 1.55\ N/C}{9.1\times 10^{-31}\ kg}\times 0.395\ m$

v = 647302.09 m/s

$v=6.45\times 10^5\ m/s$

So, the final velocity of the electron when it reaches point B is $6.45\times 10^5\ m/s$ . Hence, this is the required solution.

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3 years ago

A charged paint is spread in a very thin uniform layer over the surface of a plastic sphere of diameter 13.0 cm , giving it a ch

Leokris [45]

a) Electric field inside the paint layer: zero

b) Electric field just outside the paint layer: $-3.62\cdot 10^7 N/C$

c) Electric field 8.00 cm outside the paint layer: $-7.27\cdot 10^7 N/C$

Explanation:

We can find the electric field inside the paint layer by applying Gauss Law: the total flux of the electric field through a gaussian surface is equal to the charge contained within the surface divided by the vacuum permittivity, mathematically:

$\int EdS = \frac{q}{\epsilon_0}$

where

E is the electric field

dS is the element of surface

q is the charge within the gaussian surface

$\epsilon_0 = 8.85\cdot 10^{-12}F/m$ is the vacuum permittivity

Here we want to find the electric field just inside the paint layer, so we take a sphere of radius $r$ as Gaussian surface, where

R = 6.5 cm = 0.065 m is the radius of the plastic sphere (half the diameter)

By taking the sphere of radius r, we note that the net charge inside this sphere is zero, therefore

$q=0$

So we have

$\int E dS=0$

which means that the electric field inside the paint layer is zero.

Now we want to find the electric field just outside the paint layer: therefore, we take a Gaussian sphere of radius

$r=R=0.065 m$

The area of the surface is

$A=4\pi R^2$

And since the electric field is perpendicular to the surface at any point, Gauss Law becomes

$E\cdot 4\pi R^2 = \frac{q}{\epsilon_0}$

The charge included within the sphere in this case is the charge on the paint layer, therefore

$q=-17.0\mu C=-17.0\cdot 10^{-6}C$

So, the electric field is:

$E=\frac{q}{4\pi \epsilon_0 R^2}=\frac{-17.0\cdot 10^{-6}}{4\pi(8.85\cdot 10^{-12})(0.065)^2}=-3.62\cdot 10^7 N/C$

where the negative sign means the direction of the field is inward, since the charge is negative.

Here we want to calculate the electric field 8.00 cm outside the surface of the paint layer.

Therefore, we have to take a Gaussian sphere of radius:

$r=8.00 cm + R = 8.00 + 6.50 = 14.5 cm = 0.145 m$

Gauss theorem this time becomes

$E\cdot 4\pi r^2 = \frac{q}{\epsilon_0}$

And the charge included within the sphere is again the charge on the paint layer,

$q=-17.0\mu C=-17.0\cdot 10^{-6}C$

Therefore, the electric field is

$E=\frac{q}{4\pi \epsilon_0 r^2}=\frac{-17.0\cdot 10^{-6}}{4\pi(8.85\cdot 10^{-12})(0.145)^2}=-7.27\cdot 10^7 N/C$

Learn more about electric field:

brainly.com/question/8960054

brainly.com/question/4273177

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5 0

3 years ago