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3 years ago

13

The passenger-side rear view mirror on a car says, “Objects in the mirror may be closer than they appear”. Assuming the images a

re not inverted, this mirror must be:

1 answer:

Ipatiy [6.2K]3 years ago

4 0

Answer:

the mirror is a convex mirror

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A light ray traveling through medium 1, index of refraction n1n1n_1 = 1.75, reaches the interface between medium 1 and medium 2,

horrorfan [7]

Answer:

θ₁ = 35.32°

Explanation:

given,

refractive index of medium 1 = n₁ = 1.75

refractive index of medium 2 = n₂ = 1.24

condition to describe the refracted angle

$\theta_{refracted} + \theta_{reflected}=90^0$

$\theta_2 = 90^0-\theta_1$ ...(1)

Using Snell's Law

n₁ sin θ₁ = n₂ sin θ₂

θ₁ , θ₂ is the angle of incidence and refractive index

n₁. n₂ is the refractive index medium 1 and medium 2

1.75 x sin θ₁ = 1.24 x sin θ₂

From equation (1)

1.75 x sin θ₁ = 1.24 x sin (90-θ₁)

1.75 sin θ₁ = 1.24 cos θ₁

tan θ₁ = 0.708

θ₁ = 35.32°

Hence, angle of incidence is equal to θ₁ = 35.32°

7 0

3 years ago

PLZ HELP I DONT GET IT

GaryK [48]

Answer:

0.1 L

Explanation:

From the question given above, we obtained the following data:

Initial volume (V₁) = 0.05 L

Initial Pressure (P₁) = 207 KPa

Final pressure (P₂) = 101 KPa

Final volume (V₂) =?

We can obtain the new volume (i.e the final volume) of the gas by using the Boyle's law equation as illustrated below:

P₁V₁ = P₂V₂

207 × 0.05 = 101 × V₂

10.35 = 101 × V₂

Divide both side by 101

V₂ = 10.35 / 101

V₂ = 0.1 L

Thus, the new volume of the gas is 0.1 L

6 0

2 years ago

Ocean waves pass through two small openings, 20.0 m apart, in a breakwater. You're in a boat 70.0 m from the breakwater and init

Klio2033 [76]

Answer:

λ = 5.65m

Explanation:

The Path Difference Condition is given as:

δ= $(m+\frac{1}{2})\frac{lamda}{n}$ ;

where lamda is represent by the symbol (λ) and is the wavelength we are meant to calculate.

m = no of openings which is 2

∴δ= $\frac{3*lamda}{2}$

n is the index of refraction of the medium in which the wave is traveling

To find δ we have;

δ= $\sqrt{70^2+(33+\frac{20}{2})^2 }-\sqrt{70^2+(33-\frac{20}{2})^2 }$

δ= $\sqrt{4900+(\frac{66+20}{2})^2}-\sqrt{4900+(\frac{66-20}{2})^2}$

δ= $\sqrt{4900+(\frac{86}{2})^2 }-\sqrt{4900+(\frac{46}{2})^2 }$

δ= $\sqrt{4900+43^2}-\sqrt{4900+23^2}$

δ= $\sqrt{4900+1849}-\sqrt{4900+529}$

δ= $\sqrt{6749}-\sqrt{5429}$

δ= 82.15 -73.68

δ= 8.47

Again remember; to calculate the wavelength of the ocean waves; we have:

δ= $\frac{3*lamda}{2}$

δ= 8.47

8.47 = $\frac{3*lamda}{2}$

λ = $\frac{8.47*2}{3}$

λ = 5.65m

3 0

3 years ago

A plate carries a charge of 3.8 UC, while a rod carries a charge of 1.9 C. How many electrons must be transferred from the plate

frosja888 [35]

Answer:

$N_{electrons}=Q_{transfered}/q_{electron}=5.94*10^{18}electrons$

Explanation:

The total charge is distributed over the two objects:

$Q_{total}/2=(3.8*10^{-6}C+1.9C)/2=0.9500019C\\$

The plate and the rod must have $Q_{total}/2\\$ . So the charge transferred from the plate to the rod is:

$Q_{transfered}=3.8*10^{-6}C-Q_{total}/2=3.8*10^{-6}C-0.9500019C=-0.9499981C\\$

Number of electrons:

$N_{electrons}=Q_{transfered}/q_{electron}=-0.9499981C/(-1.6*10^{-19}C)=5.94*10^{18}electrons$

4 0

2 years ago

A tube 1.20 m long is closed at one end. A stretched wire is placed near the open end. The wire is 0.350 m long and has a mass o

Ksju [112]

Answer:

71.4583 Hz

67.9064 N

Explanation:

L = Length of tube = 1.2 m

l = Length of wire = 0.35 m

m = Mass of wire = 9.5 g

v = Speed of sound in air = 343 m/s

The fundamental frequency of the tube (closed at one end) is given by

$f=\frac{v}{4L}\\\Rightarrow f=\frac{343}{4\times 1.2}\\\Rightarrow f=71.4583\ Hz$

The fundamental frequency of the wire and tube is equal so he fundamental frequency of the wire is 71.4583 Hz

The linear density of the wire is

$\mu=\frac{m}{l}\\\Rightarrow \mu=\frac{9.5\times 10^{-3}}{0.35}\\\Rightarrow \mu=0.02714\ kg/m$

The fundamental frequency of the wire is given by

$f=\frac{1}{2l}\sqrt{\frac{T}{\mu}}\\\Rightarrow f^2=\frac{1}{4l^2}\frac{T}{\mu}\\\Rightarrow T=f^2\mu 4l^2\\\Rightarrow T=71.4583^2\times 0.02714\times 4\times 0.35^2\\\Rightarrow T=67.9064\ N$

The tension in the wire is 67.9064 N

7 0

2 years ago