When energy moves from one form to another it is said to be?

You get pulled over for running a red light. You explain to the police officer that the light appeared green to you due to the D

Ludmilka [50]

Answer:

$ 3085713685.71

Explanation:

$\lambda_0$ = Actual wavelength = 700 nm

$\lambda$ = Changed wavelength = 500 nm

Let the wavelength of red color be 700 nm and green be 500 nm

Change in wavelength is

$\Delta \lambda=700-500\\\Rightarrow \theta\lambda=200\ nm$

We have the relation

$\dfrac{\Delta\lambda}{\lambda_0}=\dfrac{v}{c}\\\Rightarrow v=\dfrac{\Delta\lambda}{\lambda_0}c\\\Rightarrow v=\dfrac{200}{700}\times 3\times 10^8\\\Rightarrow v=85714285.7143\ m/s$

The speed of the vehicle is 85714285.7143 m/s

$\dfrac{85714285.7143\times 3600}{1000}=308571428.571\ km/h$

By how much was the car speeding

$308571428.571-60=308571368.571\ km/h$

The number of 10 km/h in the above speed

$308571368.571\times\dfrac{1}{10}=30857136.8571$

Cost of the ticket

$30857136.8571\times 100=\$ 3085713685.71$

The cost of the ticket is $ 3085713685.71

8 0

3 years ago

Read 2 more answers

Consider going around a horizontal turn to the right. If the coaster suddenly slipped off the track, what path would it follow?

Svetlanka [38]

Answer:

Explanation: It would go straight because objects in motion stay in motion and it would stay the same direction

5 0

3 years ago

Interactive Solution 9.1 presents a model for solving this problem. The wheel of a car has a radius of 0.380 m. The engine of th

Vilka [71]

Answer:

The magnitude of the static frictional force is 1200 N

Explanation:

given information :

radius, r = 0.380 m

applied-torque, τ1 = 456 N

The car has a constant velocity, thus the acceleration is zero

α = 0

Στ = I α

τ1 - τ2 = I α

τ2 = counter-torque

τ1 - τ2 = 0

τ1 = τ2

r x $F_{s}$ = τ1

$F_{s}$ = the static frictional force (N)

$F_{s}$ = τ1 /r

= 456 N/0.380 m

= 1200 N

7 0

3 years ago

A golfer gives a ball a maximum initial speed of 51.5 m/s. how far does it go

nata0808 [166]

<u>Answer:</u>

Golf ball will go a maximum of 270.36 meter.

<u>Explanation:</u>

Projectile motion has two types of motion Horizontal and Vertical motion.

Vertical motion:

We have equation of motion, v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration and t is the time taken.

Considering upward vertical motion of projectile.

In this case, Initial velocity = vertical component of velocity = u sin θ, acceleration = acceleration due to gravity = -g $m/s^2$ and final velocity = 0 m/s.

0 = u sin θ - gt

t = u sin θ/g

Total time for vertical motion is two times time taken for upward vertical motion of projectile.

So total travel time of projectile = 2u sin θ/g

Horizontal motion:

We have equation of motion , $s= ut+\frac{1}{2} at^2$ , s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

In this case Initial velocity = horizontal component of velocity = u cos θ, acceleration = 0 $m/s^2$ and time taken = 2u sin θ /g

So range of projectile, $R=ucos\theta*\frac{2u sin\theta}{g} = \frac{u^2sin2\theta}{g}$

Now in the given problem

A golfer gives a ball a maximum initial speed of 51.5 m/s. how far does it go

u = 51.5 m/s, for maximum range θ = 45⁰

So maximum distance reached = $\frac{51.5^2sin(2*45)}{9.81}=270.36 meter$

So it will go a maximum of 270.36 meter.

5 0

3 years ago

Interactive Solution 8.29 offers a model for this problem. The drive propeller of a ship starts from rest and accelerates at 2.3

MAXImum [283]

Answer:

Δθ = 15747.37 rad.

Explanation:

The total angular displacement is the sum of three partial displacements: one while accelerating from rest to a certain angular speed, a second one rotating at this same angular speed, and a third one while decelerating to a final angular speed.
Applying the definition of angular acceleration, we can find the final angular speed for this first part as follows:

$\omega_{f1} = \alpha * \Delta t = 2.38*e-3rad/s2*2.04e3s = 4.9 rad/sec (1)$

Since the angular acceleration is constant, and the propeller starts from rest, we can use the following kinematic equation in order to find the first angular displacement θ₁:

$\omega_{f1}^{2} = 2* \alpha *\Delta\theta (2)$

Solving for Δθ in (2):

$\theta_{1} = \frac{\omega_{f1}^{2}}{2*\alpha } = \frac{(4.9rad/sec)^{2}}{2*2.38*e-3rad/sec2} = 5044.12 rad (3)$

The second displacement θ₂, (since along it the propeller rotates at a constant angular speed equal to (1), can be found just applying the definition of average angular velocity, as follows:

$\theta_{2} =\omega_{f1} * \Delta_{t2} = 4.9 rad/s * 1.48*e3 s = 7252 rad (4)$

Finally we can find the third displacement θ₃, applying the same kinematic equation as in (2), taking into account that the angular initial speed is not zero anymore:

$\omega_{f2}^{2} - \omega_{o2}^{2} = 2* \alpha *\Delta\theta (5)$

Replacing by the givens (α, ωf₂) and ω₀₂ from (1) we can solve for Δθ as follows:

$\theta_{3} = \frac{(\omega_{f2})^{2}- (\omega_{f1}) ^{2} }{2*\alpha } = \frac{(2.42rad/s^{2}) -(4.9rad/sec)^{2}}{2*(-2.63*e-3rad/sec2)} = 3451.25 rad (6)$

The total angular displacement is just the sum of (3), (4) and (6):
Δθ = θ₁ + θ₂ + θ₃ = 5044.12 rad + 7252 rad + 3451.25 rad
⇒ Δθ = 15747.37 rad.

4 0

3 years ago