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V125BC [204]
3 years ago
9

answers If visible light and radio waves are both examples of electromagnetic waves, why can’t we see radio waves?

Physics
1 answer:
Nata [24]3 years ago
4 0

Answer:

Radio Waves:

Radio waves being the lowest-energy form of light and are known to be produced by electrons spiraling around magnetic fields. These Magnetic fields are generated by stars, including our sun, and many weird celestial objects like black holes and neutron stars.

Explanation:

All electromagnetic radiation is light, but we can only see a small portion of this radiation that is the portion we call visible light. Cone-shaped cells in our eyes act as receivers tuned to the wavelengths in this narrow band of the spectrum

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What is the equation for finding the amount of inertia in forces and motion? Explain how you solve that equation.​
jolli1 [7]

Answer:

Inertia = angular momentum / angular velocity

4 0
2 years ago
Suppose a cup of coï¬ee is at 100 degrees Celsius at time t = 0, it is at 70 degrees at t = 10 minutes, and it is at 50 degrees
Alexandra [31]

Answer:

T ambient = 10 degrees

Explanation:

Using Newton's Law of Cooling:

T(t) = Tamb + (Ti - Tamb)*e^(-kt)  ..... Eq 1

Ti = 100

We have two points to evaluate the above equation as follows:

T = 70 @ t = 10 using Eq 1  

70 = Tamb + (100 - Tamb)*e^(-10k)   ... Eq 2

T = 50 @ t = 20 using Eq 1

50 = Tamb + (100 - Tamb)*e^(-20k)   ... Eq 3

Solving the above Eq 2 and Eq 3 simultaneously:

Using Eq 2:

(70 - Tamb) / (100 - Tamb) = e^(-10k)  

Squaring both sides we get:

((70 - Tamb) / (100 - Tamb))^2 = e^(-20k)   .... Eq 4

Substitute Eq 4 into Eq 3

50 = Tamb + (100 - Tamb)*((70 - Tamb) / (100 - Tamb))^2

After simplification:

50 = (Tamb (100-Tamb) + (70-Tamb)^2) / (100 - Tamb)

5000 - 50*Tamb = 4900 - 40*Tamb

Tamb = 100 / 10 = 10 degrees

6 0
3 years ago
A long string is wrapped around a 6.6-cm-diameter cylinder, initially at rest, that is free to rotate on an axle. The string is
lys-0071 [83]

Answer:

\omega_f=571.42\ rpm

Explanation:

It is given that,

Diameter of cylinder, d = 6.6 cm

Radius of cylinder, r = 3.3 cm = 0.033 m

Acceleration of the string, a=1.5\ m/s^2

Displacement, d = 1.3 m

The angular acceleration is given by :

\alpha =\dfrac{a}{r}

\alpha =\dfrac{1.5}{0.033}

\alpha =45.46\ rad/s^2

The angular displacement is given by :

\theta=\dfrac{d}{r}

\theta=\dfrac{1.3}{0.033}

\theta=39.39\ rad

Using the third equation of rotational kinematics as :

\omega_f^2-\omega_i^2=2\alpha \theta

Here, \omega_i=0

\omega_f=\sqrt{2\alpha \theta}

\omega_f=\sqrt{2\times 45.46\times 39.39}

\omega_f=59.84\ rad/s

Since, 1 rad/s = 9.54 rpm

So,

\omega_f=571.42\ rpm

So, the angular speed of the cylinder is 571.42 rpm. Hence, this is the required solution.

5 0
3 years ago
you have been called to testify as a as an expert witness in a trial involving a head-on collision Car A weighs 1515 pounds and
GrogVix [38]

Answer:

70.6 mph

Explanation:

Car A mass= 1515 lb

Car B mass=1125 lb  

Speed of car B is 46 miles/h

Distance before locking, d=19.5 ft

Coefficient of kinetic friction is 0.75

Initial momentum of car B=mv where m is mass and v is velocity in ft/s  

46 mph*1.46667=67.4666668 ft/s

Momentum_B=1125*67.4666668 ft/s

Initial momentum of car A is given by

Momentum_A=1515v_a where v_a is velocity of A

Taking East as positive and west as negative then the sum of initial momentum is

1515v_a-(1125*67.4666668 ft/s)

The common velocity is represented as v_c hence after collision, the final momentum is

Momentum_final=(m_a+m_b)v_c=(1515+1125)v_c=2640v_c

From the law of conservation of linear momentum, sum of initial and final momentum equals each other hence

1515v_a-(1125*67.4666668 ft/s)= 2640v_c

The acceleration of two cars a=-\mug=-0.75*32.17=-24.1275 ft/s^{2}

From kinematic equation

v^{2}=u^{2}+2as hence

v^{2}-u^{2}=2as

0^{2}-(v_c)^{2}=2*-24.1275*19.5

v_c=\sqrt{2*24.1275*19.5}=30.67 ft/s

Substituting the value of v_c in equation 1515v_a-(1125*67.4666668 ft/s)= 2640v_c

1515v_a-(1125*67.4666668 ft/s)= 2640*30.67

1515v_a=(1125*67.4666668 ft/s)+2640*30.67

v_a=\frac {156868.8}{1515}=103.5438 ft/s

\frac {103.5438}{1.46667}=70.59787 mph\approx 70.60 mph

3 0
3 years ago
A car is traveling at 50 mi/h when the brakes are fully applied, producing a constant deceleration of 38 ft/s2. what is the dist
e-lub [12.9K]

Convert 38 ft/s^2 to mi/h^2. Then we se the conversion factor > 1 mile = 5280 feet and 1 hour = 3600 seconds.

So now we show it > 38  \frac{ft}{s^2}  x  \frac{1mi}{5280ft} x  \frac{(3600s)^2}{(1h)^2} = 93272.27  \frac{mi}{h^2}

Then we have to use the formula of constant acceleration to determine the distance traveled by the car before it ended up stopping.

Which the formula for constant acceleration would be > v_2^2=v_1^2 + 2as

The initial velocity is 50mi/h (v_1=50)

When it stops the final velocity is (v_2=0)

Since the given is deceleration it means the number we had gotten earlier would be a negative so a = -93272.27

Then we substitute the values in....

0^2 = 50^2 + 2(-93272.27)s

0 = 2500 - 186544.54s

Isolate S next.

185644.54s= 2500

s =  2500/(185644.54)

s=0.0134


So we can say the car stopped at 0.0134 miles before it came to a stop but to express the distance traveled in feet we need to use the conversion factor of 1 mile = 5280 feet in otherwards > 0.0134 mi *  \frac{5280ft}{1mi}  = 70.8 ft
So this means that the car traveled in feet 70.8 ft before it came to a stop.

4 0
3 years ago
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