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2 years ago

Science- look at the picture and i will mark you as brainlyliset if you answer correctly.

Physics

2 answers:

exis [7]2 years ago

6 0

Answer:

It would be C

Explanation:

Because adding HCI would cause a poor conductor of electricity

hram777 [196]2 years ago

4 0

It’s either C or D sorry hope this helps •~•

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A 121 turn 121 turn circular coil of radius 2.85 cm 2.85 cm is immersed in a uniform magnetic field that is perpendicular to the

sashaice [31]

Answer:

0.074 V

Explanation:

Parameters given:

Number of turns, N = 121

Radius of coil, r = 2.85 cm = 0.0285 m

Time interval, dt = 0.179 s

Initial magnetic field strength, Bin = 55.1 mT = 0.0551 T

Final magnetic field strength, Bfin = 97.9 mT = 0.0979 T

Change in magnetic field strength,

dB = Bfin - Bin

= 0.0979 - 0.0551

dB = 0.0428 T

The magnitude of the average induced EMF in the coil is given as:

|Eavg| = |-N * A * dB/dt|

Where A is the area of the coil = pi * r² = 3.142 * 0.0285² = 0.00255 m²

Therefore:

|Eavg| = |-121 * 0.00255 * (0.0428/0.179)|

|Eavg| = |-0.074| V

|Eavg| = 0.074 V

5 0

2 years ago

Read 2 more answers

A crate resting on a rough horizontal floor is to be moved horizontally. The coefficient of static friction is 0.36. To start th

luda_lava [24]

Answer:

Explanation:

The direction of force will be in upward direction making an angle of θ with the vertical .

Reaction force R = mg - F cosθ

Friction force = μR

= .36 (mg - F cosθ )

Horizontal component of applied force

= F sinθ

For equilibrium

F sinθ = .36 (mg - F cosθ)

F sinθ + .36 F cosθ =.36 mg

F (sinθ + .36 cosθ) = .36 mg

F R( cosδsinθ +sinδ cosθ) = .36 mg ( Rcosδ = 1 . Rsinδ= .36 )

F R sin( θ+δ ) = . 36 mg

F = .36 mg / Rsin( θ+δ )

For minimum F , sin( θ+δ ) should be maximum

sin( θ+ δ ) = sin 90

θ+ δ = 90

Rsinδ / Rcosδ = .36

δ = 20⁰

θ = 70⁰ Ans

5 0

3 years ago

Coherent light with wavelength 598 nm passes through two very narrow slits, and the interference pattern is observed on a screen

Stella [2.4K]

Answer:

1.196 μm

Explanation:

D = Screen distance = 3 m

$\lambda$ = Wavelength = 598 m

y = Distance of first-order bright fringe from the center of the central bright fringe = 4.84 mm

d = Slit distance

$tan\theta=\frac{y}{D}\\\Rightarrow \theta=tan^{-1}{\frac{y}{D}}\\\Rightarrow \theta=tan^{-1}{\frac{4.84\times 10^{-3}}{3}}\\\Rightarrow \theta=0.09243\ ^{\circ}$

$sin\theta=\frac{\lambda}{d}\\\Rightarrow d=\frac{\lambda}{sin\theta}\\\Rightarrow d=\frac{598\times 10^{-9}}{sin0.09243}\\\Rightarrow d=0.00037066\ m$

For first dark fringe

$dsin\theta=\frac{\lambda'}{2}\\\Rightarrow \lambda'=2dsin\theta\\\Rightarrow \lambda'=2\times 0.00037066\times sin0.09243\\\Rightarrow \lambda'=1.196\times 10^{-6}\\\Rightarrow \lambda'=1.196\ \mu m$