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Pavlova-9 [17]
2 years ago
11

I need help! Pleeeeeeeeaaaaaaaaaaaaasssssssssssseeeeeeeeeeee?

Physics
1 answer:
serious [3.7K]2 years ago
4 0

Answer:

np 500000 eeeeeeeeeeee

Explanation:

346 763 999 eeeeeeeeeeeee

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Calculate the angle of refraction of 30.0° light shines from water into ice. The indices of refraction for water and ice are 1.3
WITCHER [35]

Answer:

The angle of refraction Ø2 equals 62.95° ≈ 63°

Explanation:

The relationship between the angles of incidence and

refraction , with respect to light or other waves passing through two different substances or media, such as glass, water or air is given by Snell's Law.

Snell's Law states that the when light travels from one medium to another, it generally refracts.

It is given by the mathematical expression;

[SinØ1°/SinØ2°] = [n2/n1]

Cross multiplying, we have;

n1 × SinØ1° = n2 × SinØ2°

where, n is the indices of refraction of each substance

Ø is the angle between the ray and the line normal to the surface.

Given the following values;

n1 = 1.36 n2 = 1.31 Ø1 = 30° Ø = ?

n1 × SinØ1° = n1 × SinØ2°

SinØ2° = [n1 × SinØ1°]/n2

substituting the values respectively;

SinØ2° = [1.36 × Sin30°]/1.31

SinØ2° = [1.36 × 0.5]/1.31

SinØ2° = 0.68 × 1.31

SinØ2° = 0.8906

Ø2° = Sin–¹(0.8906)

Ø2° = 62.95° ≈ 63°

4 0
3 years ago
In which parts of a plant would u expect phototropism to occur?
Lubov Fominskaja [6]

Answer:

chloroplasts

Explanation:

Most plant shoots exhibit positive phototropism, and rearrange their chloroplasts in the leaves to maximize photosynthetic energy and promote growth.

3 0
3 years ago
Read 2 more answers
Enzyme reaction and deactivation Lipase is being investigated as an additive to laundry detergent for removal of stains from fab
natita [175]

Answer:

t= 27.38 mins [this the time taken by the enzyme to hydrolyse 80% of the fat present]

Explanation:given values

Half life of lipase t_1/2 = 8 min x 60s/min = 480 s

Rate constant for first order reaction

k_d = 0.6932/480 = 1.44 x 10^-3 s-1

Initial fat concentration S_0 = 45 mol/m3 = 45 mmol/L

rate of hydrolysis Vm0 = 0.07 mmol/L/s

Conversion X = 0.80

Final concentration S = S_0(1-X) = 45 (1-0.80) = 9 mol/m3

K_m = 5mmol/L

time take is given by

t= -\frac{1}{K_d}ln[1-\frac{K_d}{V_m_0}(k_mln\frac{s_0}{s}+(s_0-s))]

all values are given and putting these value we get

t=1642.83 secs

which is equal to

t= 27.38 mins [this the time taken by the enzyme to hydrolyse 80% of the fat present]

5 0
3 years ago
A bicyclist travels 4.5 km west, then travels 6.7 km at an angle 27.0 degrees South of West. What is the magnitude of the bicycl
Dimas [21]

Answer:

<em>10.90km</em>

Explanation:

Magnitude of the total displacement is expressed using the equation

d = √dx²+dy²

dx is the horizontal component of the displacement

dy is the vertical component of the displacement

dy = -6.7sin27°

dy = -6.7(0.4539)

dy = -3.042

For the  horizontal component of the displacement

dx = -4.5 - 6.7cos27

dx = -4.5 -5.9697

dx = -10.4697

Get the magnitude of the bicyclist's total displacement

Recall that: d = √dx²+dy²

d = √(-3.042)²+(-10.4697)²

d = √9.2538+109.6146

d = √118.8684

<em>d = 10.90km</em>

<em>Hence the magnitude of the bicyclist's total displacement is 10.90km</em>

<em></em>

6 0
3 years ago
Estimate the electric field at a point 2.40 cm perpendicular to the midpoint of a uniformly charged 2.00-m-long thin wire carryi
nadya68 [22]

Answer:

E = 1.85*10^{12}\frac{N}{C}

Explanation:

Hi!

The perpendicular distance 2.4cm, is much less than the distance to both endpoints of the wire, which is aprox 1m. Then the edge effect is negligible at this field point, and we can aproximate the wire as infinitely long.

The electric filed of an infinitely long wire is easy to calculate. Let's call z the axis along the wire. Because of its simmetry (translational and rotational), the electric field E must point in the radial direction,  and it cannot depende on coordinate z. To calculate the field Gauss law is used, as seen in the image, with a cylindrical gaussian surface. The result is:

E = \frac{\lambda}{2\pi \epsilon_0 r}\\\lambda=\text{charge per unit length}=\frac{4.95 \mu C}{2 m} = 2.475 \frac{C}{m}\\r=\text{perpendicular distance to wire}\\\epsilon_0=8.85*10^{-12}\frac{C^2}{Nm^2}

Then the electric field at the point of interest is estimated as:

E = \frac{\22.475}{2\pi*( 8.85*10^{-12})*(2.4*10^{-2})}\frac{N}{C}=1.85*10^{12}\frac{N}{C}

6 0
4 years ago
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