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Marianna [84]
3 years ago
14

A light wave has a wavelength of 0.015 m. What is the frequency of this light wave? (1 point)

Physics
2 answers:
Natalija [7]3 years ago
6 0

Answer:

Did you get an answer?

Explanation:

LuckyWell [14K]3 years ago
5 0

Answer:

2 x 10^10 Hz

Explanation:

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Motivation is best described as the _____. ability to be successful with your exercise way you see your exercise desire to conti
Olin [163]

Answer:

The desire to continue with your exercise would be the correct answer! :D

6 0
3 years ago
Read 2 more answers
2. A truck speeds up from a velocity of 6 m/s to 14 m/s in 4 seconds. What is the trucks acceleration? SHOW YOUR WORK
zavuch27 [327]

#2

As it is given here

initial speed is

v_i = 6 m/s

After 4 seconds the final speed is

v_f = 14 m/s

so here we can use the formula of acceleration using kinematics

a = \frac{v_f - v_i}{t}

a = \frac{14 - 6}{4}

a = 2 m/s^2

so here it will accelerate at 2 m/s^2 rate.

#3

As it is given here

initial it starts from rest

v_i = 0 m/s

After 2.5 seconds the final speed is

v_f = 15 m/s

so here we can use the formula of acceleration using kinematics

a = \frac{v_f - v_i}{t}

a = \frac{15 - 0}{2.5}

a = 6 m/s^2

so here it will accelerate at 6 m/s^2 rate.


#4

i think question is not correct as in first line it is saying about a bag of trash and then in next line it is asking for the position of Jumper and bridge.

7 0
2 years ago
Which of the following describe an electrical motor? Check all that apply
weeeeeb [17]

Answer:

changes electrical energy into mechanical energy

5 0
3 years ago
(a) Calculate the force (in N) needed to bring a 1100 kg car to rest from a speed of 85.0 km/h in a distance of 125 m (a fairly
nasty-shy [4]

(a) -2451 N

We can start by calculating the acceleration of the car. We have:

u=85.0 km/h = 23.6 m/s is the initial velocity

v = 0 is the final velocity of the car

d = 125 m is the stopping distance

So we can use the following equation

v^2 - u^2 = 2ad

To find the acceleration of the car, a:

a=\frac{v^2-u^2}{2d}=\frac{0-(23.6 m/s)^2}{2(125 m)}=-2.23 m/s^2

Now we can use Newton's second Law:

F = ma

where m = 1100 kg to find the force exerted on the car in order to stop it; we find:

F=(1100 kg)(-2.23 m/s^2)=-2451 N

and the negative sign means the force is in the opposite direction to the motion of the car.

(b) -1.53\cdot 10^5 N

We can use again the equation

v^2 - u^2 = 2ad

To find the acceleration of the car. This time we have

u=85.0 km/h = 23.6 m/s is the initial velocity

v = 0 is the final velocity of the car

d = 2.0 m is the stopping distance

Substituting and solving for a,

a=\frac{v^2-u^2}{2d}=\frac{0-(23.6 m/s)^2}{2(2 m)}=-139.2 m/s^2

So now we can find the force exerted on the car by using again Newton's second law:

F=ma=(1100 kg)(-139.2 m/s^2)=-1.53\cdot 10^5 N

As we can see, the force is much stronger than the force exerted in part a).

8 0
3 years ago
What is an experience from everyday life that appears to support the geocentric model
fomenos
<span>One everyday life experience that seems to support the geocentric model is the rising and setting of the Sun and Moon. The Moon rises and falls because it does revolve around the Earth and so it is easy to assume the same is true for the Sun.</span>
3 0
3 years ago
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