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vesna_86 [32]
3 years ago
9

Is there more than one kind of balance used in a laboratory

Physics
1 answer:
Inessa [10]3 years ago
4 0
Maybe there is people who used
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Bob is threatening Tom’s life with a giant laser with wavelength (650 nm), a distance (D = 10 m) from the wall James is shackled
Fittoniya [83]

Answer:

He should stand from the center of laser pointed on the wall at 1.3 m.

Explanation:

Given that,

Wave length = 650 nm

Distance =10 m

Double slit separation d = 5 μm

We need to find the position of fringe

Using formula of distance

d\sin\theta=n\lambda

d\dfrac{y}{D}=n\lambda

y=\dfrac{\lambda D}{d}

Put the value into the formula

y=\dfrac{650\times10^{-9}\times10}{5\times10^{-6}}

y=1.3\ m

Hence, He should stand from the center of laser pointed on the wall at 1.3 m.

8 0
3 years ago
Which expression correctly describes energy using SI units
mote1985 [20]

Answer:

Joule

Explanation:

energy, work, quantity of heat

m2·kg·s-2

5 0
3 years ago
Why is the sun so far away
aev [14]

Answer:

If it was close we would die.

Explanation:

Probably has something to do with the earths original formation.

5 0
3 years ago
Read 2 more answers
Which describes Michael Faraday’s work with electricity and magnetism?
Viefleur [7K]
Genius right dear friend

8 0
3 years ago
A parallel-plate capacitor has a plate area of 0.2m^2 and a plate separation of 0.1mm. To obtain an electric field of 2.0 × 10^6
Oduvanchick [21]

Answer:

3.536*10^-6 C

Explanation:

The magnitude of the charge is expresses as Q = CV

C is the capacitance of the capacitor

V is the voltage across the capacitor

Get the capacitance

C = ε0A/d

ε0 is the permittivity of the dielectric = 8.84 x 10-12 F/m

A is the area = 0.2m²

d is the plate separation = 0.1mm = 0.0001m

Substitute

C = 8.84 x 10-12 * 0.2/0.0001

C = 1.768 x 10-8 F

Get the potential difference V

Using the formula for Electric field intensity

E = V/d

2.0 × 10^6  = V/0.0001

V = 2.0 × 10^6  * 0.0001

V = 2.0 × 10^2V

Get the charge on each plate.

Q = CV

Q =  1.768 x 10-8 * 2.0 × 10^2

Q = 3.536*10^-6 C

Hence the magnitude of the charge on each plate should be 3.536*10^-6 C

5 0
3 years ago
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