If your heart is not strong enough or efficient enough, it is difficult to

A physicist drives through a stop light. When he is pulled over, he tells the police officer that the Doppler shift made the red

Orlov [11]

The physicist traveling, according to his own testimony at -6.6 × 10⁷ m/s.

<h3>How fast was the physicist traveling, according to his own testimony?</h3>

Using the formula for doppler shift for light,

λ' = λ√[(1 + v/c)/(1 - v/c)] where

λ = wavelength of source,
λ' = wavelength of observer,
v = speed of source and
c = speed of light

Given that the driver is moving away from the stop light, we take the driver as the source. Since, the Doppler shift made the red light of wavelength 650nm appear green to him, with a wavelength of 520nm. we have

λ' = wavelength of source = 650 nm,
λ' = wavelength of observer = 520 nm

So, substituting the values of the variables into the equation, we have

λ' = λ√[(1 + v/c)/(1 - v/c)]

520 nm = 650 nm√[(1 + v/c)/(1 - v/c)]

520/650 = √[(1 + v/c)/(1 - v/c)]

0.8 = √[(1 + v/c)/(1 - v/c)]

Squaring both sides, we have

0.8² = (1 + v/c)/(1 - v/c)

0.64 = (1 + v/c)/(1 - v/c)

0.64(1 - v/c) = (1 + v/c)

0.64 - 0.64v/c = 1 + v/c

0.64 - 1 = v/c + 0.64v/c

-0.36 = 1.64v/c

-0.2195 = v/c

v = -0.22c

v = -0.22 × 3 × 10⁸ m/s

v = -0.66 × 10⁸ m/s

v = -6.6 × 10⁷ m/s

So, the physicist traveling, according to his own testimony at -6.6 × 10⁷ m/s.

Learn more about doppler shift for light here:

brainly.com/question/28499579

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5 0

2 years ago

How long does it take to raise the temperature of the air in a good-sized living room (3.00m×5.00m×8.00m) by 10.0∘C? Note that t

tekilochka [14]

Answer : The time required is, 16.1 minutes.

Explanation :

First we have to calculate the amount of heat required to increase the temperature is:

$Q=mC\Delta T\\\\Q=\rho VC\Delta T$

$(m=\rho V)$

where,

Q = amount of heat required = ?

m = mass

$\rho$ = density of air = $1.20kg/m^3$

V = volume of air

C = specific heat of air = $1006J/kg^oC$

$\Delta T$ = change in temperature = $10.0^oC$

Now put all the given values in above formula, we get:

$Q=\rho VC\Delta T$

$Q=(1.20kg/m^3)\times (3.00m\times 5.00m\times 8.00m)\times (1006J/kg^oC)\times (10.0^oC)$

$Q=1.449\times 10^6J$

Now we have to calculate the time required.

Formula used :

$t=\frac{Q}{P}$

where,

t = time required = ?

Q = amount of heat required = $1.449\times 10^6J$

P = power = 1500 W

Now put all the given values in above formula, we get:

$t=\frac{1.449\times 10^6J}{1500W}$

$t=966s\times \frac{1min}{60s}=16.1min$

Thus, the time required is, 16.1 minutes.

5 0

3 years ago

A billion years ago, Earth and its moon were just 200000 kilometers apart. Express this distance in meters.

Luda [366]

Answer:

The value is $x = 200000000 \ m$

Explanation:

From the question we are told that

The distance between the earth and the moon is $d = 200000 \ km$

Generally ,

$1 \ km = 1000 \ m$

$200000 \ km \to x \ m$

=> $x = \frac{200000 * 1000}{1 }$

=> $x = 200000000 \ m$

4 0

3 years ago

Two forces act on an object. The first force has a magnitude of 17.0 N and is oriented 48.0° counterclockwise from the x ‑axis,

Ivanshal [37]

Answer:

Fn: magnitude of the net force.

Fn=30.11N , oriented 75.3 ° clockwise from the -x axis

Explanation:

Components on the x-y axes of the 17 N force(F₁)

F₁x=17*cos48°= 11.38N

F₁y=17*sin48° = 12.63 N

Components on the x-y axes of the the second force(F₂)

F₂x= −19.0 N

F₂y= 16.5 N

Components on the x-y axes of the net force (Fn)

Fnx= F₁x +F₂x= 11.38N−19.0 N= -7.62 N

Fny= F₁y +F₂y= 12.63 N +16.5 N = 29.13 N

Magnitude of the net force.

$F_{n} =\sqrt{(F_{nx})^{2} +(F_{ny}) ^{2} }$

$F_{n} =\sqrt{(-7.62)^{2} +(29.13) ^{2} }$

$F_{n} = 30.11N$

Direction of the net force (β)

$\beta =tan^{-1} (\frac{29.13}{7.62} )$

β=75.3°

Magnitude and direction of the net force

Fn= 30.11N , oriented 75.3 ° clockwise from the -x axis

In the attached graph we can observe the magnitude and direction of the net force

6 0

3 years ago

A vector has a magnitude of 46.0 m and points in a direction 20.0° below the positive x-axis. A second vector, , has a magnitude

irina1246 [14]

Answer with Step-by -step explanation:

We are given that

b. $\mid A\mid=46 m$

$\theta=20^{\circ}$ below the positive x-axis

Therefore, the angle made by vector A in counter clockwise direction when measure from positive x-axis= $x=360-20=340^{\circ}$

x-component of vector A= $A_x=\mid A\mid cosx=46cos 340=46\times 0.94=43.24$

y-Component of vector A= $A_y=\mid A\mid sinx=46sin340=46(-0.34)=-15.64$

Magnitude of vector B=86 m

The vector B makes angle with positive x- axis= $x'=42^{\circ}$

x-component of vector B= $B_x=86cos42=63.64$

y-Component of vector B= $B_y=86sin42=57.62$

Vector A= $A_xi+A_yj=43.24i-15.64j$

Vector B= $B_xi+B_yj=63.64i+57.62j$

Vector C=A+B

Substitute the values

$C=43.24i-15.64j+63.64i+57.62j$

$C=106.88i+41.98j$

c.Direction= $\theta=tan^{-1}(\frac{y}{x})=tan^{-1}(\frac{41.98}{106.88})=21.5^{\circ}$

The direction of the vector C=21.5 degree

6 0

3 years ago