A simple pendulum makes 540 complete oscillations in a minute, so its frequency is ........... HZ

Calculate the electric field at the center of a square

pantera1 [17]

Answer:

$E_y=1175510.2\ N.C^{-1}$

The Magnitude of electric field is in the upward direction as shown directly towards the charge $q_1$ .

Explanation:

Given:

side of a square, $a=52.5\ cm$

charge on one corner of the square, $q_1=+45\times 10^{-6}\ C$

charge on the remaining 3 corners of the square, $q_2=q_3=q_4=-27\times 10^{-6}\ C$

Distance of the center from each corners $=\frac{1}{2} \times diagonals$

$diagonal=\sqrt{52.5^2+52.5^2}$

$diagonal=74.25\cm=0.7425\ m$

∴Distance of center from corners, $b=0.3712\ m$

Now, electric field due to charges is given as:

$E=\frac{1}{4\pi\epsilon_0}\times \frac{q}{b^2}$

For charge $q_1$ we have the field lines emerging out of the charge since it is positively charged:

$E_1=9\times 10^9\times \frac{45\times 10^{-6}}{0.3712^2}$

$E_1=2938775.5\ N.C^{-1}$

Force by each of the charges at the remaining corners:

$E_2=E_3=E_4=9\times 10^9\times \frac{27\times 10^{-6}}{0.3712^2}$

$E_2=E_3=E_4=1763265.3\ N.C^{-1}$

Now, net electric field in the vertical direction:

$E_y=E_1-E_4$

$E_y=1175510.2\ N.C^{-1}$

Now, net electric field in the horizontal direction:

$E_y=E_2-E_3$

$E_y=0\ N.C^{-1}$

So the Magnitude of electric field is in the upward direction as shown directly towards the charge $q_1$ .

8 0

3 years ago

You exert 375 J to move a 45-kg object 5 m. How much force is required?

Serjik [45]

The amount of force required is 69J

8 0

3 years ago

Approximately, What is the value of the Hubble Constant, as measured by scientists? Hypothetically, if the value of the Hubble C

Serhud [2]

Answer:

The current value of the Hubble's constant = 73 km/sec/Mpc.

t = 71.9 trillion years will be the new age of universe if the Hubble constant = 700 km/s/Mpc

Explanation:

The current value of the Hubble's constant = 73 km/sec/Mpc. However, recent discoveries in the cosmology contradicts the idea of Hubble constant as being fixed. Some scientists are not agreeing on this value and the debate is going on.

Hubble law states that how fast universe is expanding or in other words, galaxies are expanding separating with a speed directly proportional to the distance of galaxies to the earth.

Hence,

v is directly proportional to d

where, v = apparent velocity

d = distance

if we equate velocity and distance then there comes Hubble constant.

v = $H_{0}$ x d

$H_{0}$ = 73 km/sec/Mpc

where, Mpc = Mega Parsec = 1 Mpc = 3.086 x $10^{19}$ km

We can use Hubble constant to tell the age of universe.

t = d/v

t = d/( $H_{0}$ xd)

t = 1/ $H_{0}$

Scientist calculated the age of universe by using Hubble constant, which is 13.4 billion years.

Now, if we hypothetically change the value of Hubble constant,

from $H_{0}$ = 73 km/sec/Mpc to $H_{0}$ = 700 km/sec/Mpc

then the age of universe will be:

t = 1/ $H_{0}$

first convert the units of new $H_{0}$ into 1/s

$H_{0}$ = (700) x (/3.08 x $10^{19}$ )

$H_{0}$ = 227.27 x $10^{-19}$ = 2.27 x $10^{-21}$ 1/s

So,

Age of universe will be:

t = 1/ $H_{0}$ = 1/2.27x $10^{-21}$ 1/s

t = 2.27 x $10^{21}$ s

t = 71.9 trillion years

t = 71.9 trillion years will be the new age of universe if the Hubble constant = 700 km/s/Mpc

6 0

3 years ago

How much force is needed to keep the bowling ball moving towards the pins once it has

vazorg [7]

Answer:

Explanation:

The amount of force needed needs to be greater than all the forces acting in the opposite direction that the bowling ball was thrown. This includes air resistance, floor friction, gravity, and any other force involved. As long as the force acting on the bowling ball that is causing it to go in the direction of the pins is slightly greater than the opposite acting forces then it will continue in that direction. Since no values are provided we cannot calculate the actual precise value of force needed.

6 0

3 years ago

When an object is at rest on an inclined plane, what are the different forces acting on it?

elena-s [515]

a) Gravitational force only

b) Normal force only

c) Gravitational force, normal force, and frictional force

d) Gravitational force and normal force

4 0

3 years ago

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