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3 years ago

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Suppose the Opera station broadcasts at 90.5MHz and the Rock and Roll station broadcasts at 107.1MHz. 1. Which station's signal

has waves with longer waveslength? 2. Which station's signal has waves with higher energy?

1 answer:

Allisa [31]3 years ago

3 0

1). The "Opera" at 90.5 MHz has lower frequency,
so its wavelength is longer.

2). Whichever station is transmitting with higher power
has waves with higher energy.

Each photon of the transmission at 107.1 MHz carries more energy,
because the energy of a photon is proportional to its frequency.
But it's a mistake to think that the station at higher frequency has
waves with higher energy.

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Answer:

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Explanation:

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on the surface of planet x a body with a mass of 10 kilograms weighs 40 newtons. The magnitude of the acceleration due to gravit

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Based on the Newton's second law of motion, the value of the net force acting on the object is equal to the product of the mass and the acceleration due to gravity. If we let a be the acceleration due to gravity, the equation that would allow us to calculate it's value is,
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Answer:

$v_{2.6b}=11.18\ m.s^{-1}$

$v_{7.2b}=14.19\ m.s^{-1}$

$s_{bb}=226.3305\ m$

$a_{db}=-3.7386\ m.s^{-2}$ negative sign denotes deceleration.

$t_b=21.3956\ s$

$a_y=1.1065\ m.s^{-2}$

Explanation:

Given:

initial speed of blue car, $u_b=0\ m.s^{-1}$

initial speed of yellow car, $u_y=0\ m.s^{-1}$

acceleration rate of blue car, $a_b=4.3\ m.s^{-2}$

time for which the blue car accelerates, $t_{ab}=3.3\ s$

time for which the blue car moves with uniform speed before decelerating, $t_{ub}=14.3\ s$

total distance covered by the blue car before coming to rest, $s_b=253.26 \ m$

distance at which the the yellow car intercepts the blue car just as the blue car come to rest, $s_y=253.26 \ m$

1)

<u>Speed of blue car after 2.6 seconds of starting the motion:</u>

Applying the equation of motion:

$v_{2.6b}=u_b+a_b.t$

$v_{2.6b}=0+4.3\times 2.6$

$v_{2.6b}=11.18\ m.s^{-1}$

<u>Speed of blue car after 7.2 seconds of starting the motion:</u>

∵The car accelerates uniformly for 3.3 seconds after which its speed becomes uniform for the next 14.3 second before it applies the brake.

so,

$v_{7.2b}=u+a_b\times t_{ab}$

$v_{7.2b}=0+4.3\times 3.3$

$v_{7.2b}=14.19\ m.s^{-1}$

<u>Distance travelled by the blue car before application of brakes:</u>

This distance will be $s_{bb}=$ (distance travelled during the accelerated motion) + (distance travelled at uniform motion)

<em>Now the distance travelled during the accelerated motion:</em>

$s_{ab}=u_b.t_{ab}+\frac{1}{2} a_{b}.t_{ab}^2$

$s_{ab}=0\times 3.3+0.5\times 4.3\times 3.3^2$

$s_{ab}=23.4135\ m$

<em>Now the distance travelled at uniform motion:</em>

$s_{ub}=14.19\times 14.3$

$s_{ub}=202.917\ m$

Finally:

$s_{bb}=s_{ab}+s_{ub}$

$s_{bb}=23.4135+202.917$

$s_{bb}=226.3305\ m$

<u>Acceleration of the blue car once the brakes are applied</u>

Here we have:

initial velocity, $u=14.19\ m.s^{-1}$

final velocity, $v=0\ m.s^{-1}$

distance covered while deceleration, $s_{db}=s_b-s_{bb}$

$\Rightarrow s_{db}=253.26 -226.3305=26.9295\ m$

Using the equation of motion:

$v^2=u^2+2a_{db}.s_{db}$

$0^2=14.19^2+2\times a_{db}\times 26.9295$

$a_{db}=-3.7386\ m.s^{-2}$ negative sign denotes deceleration.

<u>The total time for which the blue car moves:</u>

$t_b=t_a+t_{ub}+t_{db}$ ........................(1)

<em>Now the time taken to stop the blue car after application of brakes:</em>

Using the eq. of motion:

$v=u+a_{db}.t_{db}$

$0=14.19-3.7386\times t_{db}$

$t_{db}=3.7956\ s$

Putting respective values in eq. (1)

$t_b=3.3+14.3+3.7956$

$t_b=21.3956\ s$

<u>For the acceleration of the yellow car:</u>

We apply the law of motion:

$s_y=u_y.t_y+\frac{1}{2} a_y.t_y^2$

<em>Here the time taken by the yellow car is same for the same distance as it intercepts just before the stopping of blue car.</em>

Now,

$253.26=0\times 21.3956+0.5\times a_y\times 21.3956^2$

$a_y=1.1065\ m.s^{-2}$

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AveGali [126]

Answer:

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Explanation:

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