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1 year ago

Q. 1 MWH is equal to ------- joules a.3.610^10 b.3.610^6 c.3.6*10^9 d.3.6

1 answer:

5 0

Correct option is B
1 kWh is equal to 3.6 × 10
6

1 Watt-sec is a Joule a watt expended for a second.
Because there’s 3600 seconds in an hour,
then 3600 Ws = 1 Wh = 3600 Joules = 3.6 kJ
so 1000 Wh = 1 kWh = 1000 × 3.6 kJ = 3.6 MJ = 3.6 × 10
6

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2 years ago

In the photoelectric effect, a photon with an energy of 5.3 × 10–19 J strikes an electron in a metal. Of this energy, 3.6 × 10–1

valentina_108 [34]

Answer:

The velocity of the photo electron is $6.11\times 10^5\ m/s$ .

Explanation:

Given that,

Supplied energy, $E_s=5.3\times 10^{-19}\ J$

Minimum energy of the electron to escape from the metal, $E_e=3.6\times 10^{-19}\ J$

We need to find the velocity of the photo electron. The energy supplied by the photon is equal to the sum of minimum escape energy and the kinetic energy of the escaping electron. So,

$5.3\times 10^{-19}\ J=3.6\times 10^{-19}\ J+K\\\\K=5.3\times 10^{-19}-3.6\times 10^{-19}\\\\K=1.7\times 10^{-19}\ J$

The formula of kinetic energy is given by :

$K=\dfrac{1}{2}mv^2\\\\v=\sqrt{\dfrac{2K}{m}} \\\\v=\sqrt{\dfrac{2\times 1.7\times 10^{-19}}{9.1\times 10^{-31}}} \\\\v=6.11\times 10^5\ m/s$

So, the velocity of the photo electron is $6.11\times 10^5\ m/s$ .

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2 years ago

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7 0

2 years ago

Read 2 more answers

An irregular object of mass 3 kg rotates about an axis, about which it has a radius of gyration of 0.2 m, with an angular accele

Citrus2011 [14]

Answer:

The magnitude of the applied torque is $6.0\times10^{-2}\ N-m$

(e) is correct option.

Explanation:

Given that,

Mass of object = 3 kg

Radius of gyration = 0.2 m

Angular acceleration = 0.5 rad/s²

We need to calculate the applied torque

Using formula of torque

$\tau=I\times\alpha$

Here, I = mk²

$\tau=mk^2\times\alpha$

Put the value into the formula

$\tau=3\times(0.2)^2\times0.5$

$\tau=0.06\ N-m$

$\tau=6.0\times10^{-2}\ N-m$

Hence, The magnitude of the applied torque is $6.0\times10^{-2}\ N-m$

7 0

3 years ago

Calculate curls of the following vector functions (a) AG) 4x3 - 2x2-yy + xz2 2

aleksandr82 [10.1K]

Answer:

The curl is $0 \hat x -z^2 \hat y -4xy \hat z$

Explanation:

Given the vector function

$\vec A (\vec r) =4x^3 \hat{x}-2x^2y \hat y+xz^2 \hat z$

We can calculate the curl using the definition

$\nabla \times \vec A (\vec r ) = \left|\begin{array}{ccc}\hat x&\hat y&\hat z\\\partial/\partial x&\partial/\partial y&\partial/\partial z\\A_x&X_y&A_z\end{array}\right|$

Thus for the exercise we will have

$\nabla \times \vec A (\vec r ) = \left|\begin{array}{ccc}\hat x&\hat y&\hat z\\\partial/\partial x&\partial/\partial y&\partial/\partial z\\4x^3&-2x^2y&xz^2\end{array}\right|$

So we will get

$\nabla \times \vec A (\vec r )= \left( \cfrac{\partial}{\partial y}(xz^2)-\cfrac{\partial}{\partial z}(-2x^2y)\right) \hat x - \left(\cfrac{\partial}{\partial x}(xz^2)-\cfrac{\partial}{\partial z}(4x^3) \right) \hat y + \left(\cfrac{\partial}{\partial x}(-2x^2y)-\cfrac{\partial}{\partial y}(4x^3) \right) \hat z$

Working with the partial derivatives we get the curl

$\nabla \times \vec A (\vec r )=0 \hat x -z^2 \hat y -4xy \hat z$

6 0

3 years ago

Q. 1 MWH is equal to ------- joules a.3.6*10^10 b.3.6*10^6 c.3.6*10^9 d.3.6

Q. 1 MWH is equal to ------- joules a.3.610^10 b.3.610^6 c.3.6*10^9 d.3.6