Answer:
Total Work done =0.65 joule
Explanation:
Work done is given Mathematically as
W=F *d
Where w=work done in joules
F=applied force
d= distance moved
The work done to move the toy accros the first meter is
W1=0.5*1
W1=0.5joule
The work done to move the toy across the next 2m at an angle of 30° is
.W2=0.5*2cos30
W2=0.5*2*0.154
W2=0.154joule
Hence total work done is
W1+W2=0.5+0.154
Total Work done =0.65 joule
Velocity of submarine A is vs = 11.0m/s
frequency emitted by submarine A. F = 55.273 × 10∧3HZ
Velocity of submarine B = vO = 3.00m/s
The given equation is
f' = ((V + vO) ((v - vS)) × f
The observer on submarine detects the frequency f'.
The sign of vO should be positive as the observer of submarine B is moving away from the source of submarine A.
The speed of the sound used in seawater is 1533m/s
The frequency which is detected by submarine B is
fo = fs (V -vO/ v +vs)
= 53.273 × 10∧3hz) ((1533 m/s - 4.5 m/s)/ (1533 m/s +11 m/s)
fo = 5408 HZ
Answer:
Explanation:
Electric field talks about a region around a charged particle or object within which a force would be exerted on other charged particles or objects. to find the electric field inside the bulb we will apply the electric filed formula.
Please kindly check attachment for step by step explaination.
Answer :
The number of vacancies (per meter cube) = 5.778 × 10^22/m^3.
Explanation:
Given,
Atomic mass of silver = 107.87 g/mol
Density of silver = 10.35 g/cm^3
Converting to g/m^3,
= 10.35 g/cm^3 × 10^6cm^3/m^3
= 10.35 × 10^6 g/m^3
Avogadro's number = 6.022 × 10^23 atoms/mol
Fraction of lattice sites that are vacant in silver = 1 × 10^-6
Nag = (Na * Da)/Aag
Where,
Nag = Total number of lattice sites in Ag
Na = Avogadro's number
Da = Density of silver
Aag = Atomic weight of silver
= (6.022 × 10^23 × (10.35 × 10^6)/107.87
= 5.778 × 10^28 atoms/m^3
The number of vacancies (per meter cube) = 5.778 × 10^28 × 1 × 10^-6
= 5.778 × 10^22/m^3.
Answer:
2.7x10⁻⁸ N/m²
Explanation:
Since the piece of cardboard absorbs totally the light, the radiation pressure can be found using the following equation:
<u>Where:</u>
: is the radiation pressure
I: is the intensity of the light = 8.1 W/m²
c: is the speed of light = 3.00x10⁸ m/s
Hence, the radiation pressure is:
Therefore, the radiation pressure that is produced on the cardboard by the light is 2.7x10⁻⁸ N/m².
I hope it helps you!
