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SVEN [57.7K]
3 years ago
8

If pizza is round will it roll off a cliff?

Physics
1 answer:
Rina8888 [55]3 years ago
6 0
If a pizza is round like a ball it will roll off a cliff.If it’s just flat and round it won’t.
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Can work done=mass*acceleration*displacement(work=m*a*s)
Airida [17]

no, work is = force * distance or displacement


5 0
3 years ago
1. How is power defined? *
ANEK [815]

Answer:

D. the rate at which work is accomplished

Explanation:

8 0
2 years ago
A 4.0 cm × 4.2 cm rectangle lies in the xy-plane. You may want to review (Pages 664 - 668) . Part A What is the electric flux th
padilas [110]

Answer:

(A). The flux is 0.336 N.m²/C

(B). The flux is zero.

Explanation:

Given that,

Length = 4.2 cm

Width = 4.0 cm

Electric field E=(150 i-200 k)\ N/C

Area vector is perpendicular to xy plane

(A). We need to calculate the flux

Using formula of flux

\phi=E\cdot A

Where, E = electric field

A = area

Put the value into the formula

\phi=(150 i-200 k)\times(4.2\times10^{-2}\times4.0\times10^{-2})k

\phi=-200\times4.2\times10^{-2}\times4.0\times10^{-2}

\phi=-0.336\ N.m^2/C

(B). Given electric field

E=(150i-200j)\ N/C

We need to calculate the flux

Using formula of flux

\phi=E\cdot A

Put the value into the formula

\phi=(150 i-200 j)\times(4.2\times10^{-2}\times4.0\times10^{-2})k

Here, The component of k is not given

So, the flux is

\phi=0

Hence, (A). The flux is -0.336 N.m²/C

(B). The flux is zero.

7 0
3 years ago
Read 2 more answers
What causes the bright lines seen in the emission spectrum?
Snezhnost [94]
Base in your questions that ask what cause the bright lines seen in the emission spectrum and i think the best answer to that is the H2 gas is used when protons was heated so the electron absorb all the photons and get exited and resulted by given of a light.
7 0
3 years ago
A light source radiates 60.0 W of single-wavelength sinusoidal light uniformly in all directions. What is the amplitude of the m
Anna35 [415]

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The intensity depending on the power is defined as

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Replacing the values that we have,

I = \frac{60}{(4*\pi (0.7)^2)}

I = 9.75 Watt/m^2

The definition of intensity tells us that,

I = \frac{1}{2}\frac{B_o^2 c}{\mu}

Where,

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Then replacing with our values we have,

9.75 = \frac{Bo^2 (3*10^8)}{(4\pi*10^{-7})}

Re-arrange to find the magnetic Field B_0

B_o = 2.86*10^{-7} T

Therefore the amplitude of the magnetic field of this light is B_o = 2.86*10^{-7} T

6 0
3 years ago
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