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2 years ago

Please help me the formula is E = P x t

Physics

1 answer:

olasank [31]2 years ago

3 0

Answer:

300J

Explanation:

300w every second so every 1 second.

300 × 1 = 300J

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An airplane is flying in a horizontal circle at a speed of 480 km/h (). If its wings are tilted at angle =40° to the horizontal

german

Answer:

$R = 2162 m$

Explanation:

When wings of the airplane makes an angle of 40 degree with the horizontal so here we can say that force due to air is having two components

$F_y = mg$

$F_x = \frac{mv^2}{R}$

now we know that

$F_y = F cos40$

$F_x = F sin40$

also we know that

$v = 480 km/h$

$v = 133.3 m/s$

now plug in all data in above equations

$tan 40 = \frac{v^2}{Rg}$

$R = \frac{v^2}{g tan40}$

$R = \frac{133.3^2}{9.8 tan40}$

$R = 2162 m$

6 0

2 years ago

List out any 4 objectives of education

e-lub [12.9K]

- vocational aim
- cultural aim
- spiritual aim
- intellectual aim

6 0

2 years ago

What is the tensile strength of hair ?

Olegator [25]

Human hair

Yield strength (MPa)
140-160

Ultimate tensile strength (MPa)
200-250

Hope that helps.

6 0

2 years ago

Read 2 more answers

There are two parallel conductive plates separated by a distance d and zero potential. Calculate the potential and electric fiel

taurus [48]

Answer:

The total electric potential at mid way due to 'q' is $\frac{q}{4\pi\epsilon_{o}d}$

The net Electric field at midway due to 'q' is 0.

Solution:

According to the question, the separation between two parallel plates, plate A and plate B (say) = d

The electric potential at a distance d due to 'Q' is:

$V = \frac{1}{4\pi\epsilon_{o}}.\frac{Q}{d}$

Now, for the Electric potential for the two plates A and B at midway between the plates due to 'q':

For plate A,

$V_{A} = \frac{1}{4\pi\epsilon_{o}}.\frac{q}{\frac{d}{2}}$

Similar is the case with plate B:

$V_{B} = \frac{1}{4\pi\epsilon_{o}}.\frac{q}{\frac{d}{2}}$

Since the electric potential is a scalar quantity, the net or total potential is given as the sum of the potential for the two plates:

$V_{total} = V_{A} + V_{B} = \frac{1}{4\pi\epsilon_{o}}.q(\frac{1}{\frac{d}{2}} + \frac{1}{\frac{d}{2}}$

$V_{total} = \frac{q}{4\pi\epsilon_{o}d}$

Now,

The Electric field due to charge Q at a distance is given by:

$\vec{E} = \frac{1}{4\pi\epsilon_{o}}.\frac{Q}{d^{2}}$

Now, if the charge q is mid way between the field, then distance is $\frac{d}{2}$ .

Electric Field at plate A, $\vec{E_{A}}$ at midway due to charge q:

$\vec{E_{A}} = \frac{1}{4\pi\epsilon_{o}}.\frac{q}{(\frac{d}{2})^{2}}$

Similarly, for plate B:

$\vec{E_{B}} = \frac{1}{4\pi\epsilon_{o}}.\frac{q}{(\frac{d}{2})^{2}}$

Both the fields for plate A and B are due to charge 'q' and as such will be equal in magnitude with direction of fields opposite to each other and hence cancels out making net Electric field zero.

3 0

2 years ago

Consider a spherical volume of space that is large enough to be considered homogeneous. Also consider a particle on the surface

LenKa [72]

Answer:

Option A applies.

A. Greater than its escape speed from the mass within the volume

Explanation:

Here it is mentioned that the spherical volume is large enough for the space to be considered as homogeneous. Also, the pressure within the volume is negligible, so that will not result into the re collapse of the Universe. Now as per our knowing, Hubble's Law relates the average speed of the particle to the distance R between the Earth and the particle. So, if the particle's speed is greater than it's escape speed from the mass within the volume, then the Universe is bound to re collapse back again. Option A applies.

3 0

3 years ago