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3 years ago

A particular form of electromagnetic radiation has a frequency of 5.42x10^15 Hz. What is the wavelength is nanometers? meters?

1 answer:

4 0

Electromagnetic radiation are represented in waves. Each type of wave has a certain shape and length. The distance between two peaks in a wave is called the wavelength. This value is equal to the speed of light divided by the frequency.
Wavelength = c/f
Wavelength = 3x10^8 / <span>5.42x10^15
</span><span>Wavelength = 5.54 x 10^-8 m = 55.35 nm</span>

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A remote controlled toy car starts from rest and begins to accelerate in a straight line. The figure below represents "snapshots

Nikolay [14]

(a). The car's average velocity between t = 1.0s to t = 1.5s will be - $1\;m/s$

(b). The car's acceleration at t = 1.5s will be - $0.4\;m/s^{2}$

(c). Car's speed is increasing with time.

We have a a remote controlled toy car that starts from rest and begins to accelerate in a straight line.

We have to determine -

The car's average velocity (in m/s) in the interval between -

t = 1.0 s to t = 1.5 s.

The car's acceleration at t = 1.5 s.
Determining whether car's speed increasing or decreasing with time.

<h3>What is Acceleration?</h3>

The rate of change of velocity with respect to time is called Acceleration. Mathematically -

$$a=\frac{dv}{dt}$

According to the question, we have the following data for the Car -

t = 0s → x = 0m

t = 0.5s → x = 0.1m

t = 1.0s → x = 0.4m

t = 1.5s → x = 0.9m

t = 2.0s → x = 1.6m

PART - A

The car's average velocity between t = 1.0s to t = 1.5s will be -

$$v_{avg} = \frac{0.9-0.4}{1.5-1}= 1 m/s$

PART - B

Velocity at t = 1.5 s will be -

$$v(1.5)=\frac{0.9}{1.5}= 0.6\;m/s$

The car's acceleration at t = 1.5s will be -

$$a(1.5) = \frac{v}{t} = \frac{0.6}{1.5} = 0.4\;m/s^{2}$

PART - C

Since, the acceleration of the car is positive, this means that the car is accelerating in the forward direction. Hence, its speed is increasing with time.

[ The following data was missing in your answer. The complete question would include this data also -

t = 0s → x = 0m

t = 0.5s → x = 0.1m

t = 1.0s → x = 0.4m

t = 1.5s → x = 0.9m

t = 2.0s → x = 1.6m ]

To solve more questions on Kinematics, visit the link below-

brainly.com/question/17272824

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7 0

1 year ago

Read 2 more answers

How are kinetic energy potential energy and thermal energy in a substance related?

asambeis [7]

The energy associated with an object's motion is called kinetic energy. ... This is also called thermal energy – the greater the thermal energy, the greater the kinetic energy of atomic motion, and vice versa.

3 0

3 years ago

A common carnival ride, called a gravitron, is a large r = 11 m cylinder in which people stand against the outer wall. the cylin

Elena-2011 [213]

Answer:

The minimum speed is 14.53 m/s.

Explanation:

Given that,

r = 11 m

Friction coefficient = 0.51

Suppose we need to find the minimum speed, that the cylinder must make a person move at to ensure they will stick to the wall

When frictional force becomes equal to or greater than the weight of person

Then, he sticks to the wall

We need to calculate the minimum speed

Using formula for speed

$f_{s}=\mu N\geq mg$

Where, $N =\dfrac{mv^2}{r}$

$\mu\times\dfrac{mv^2}{r}\gep mg$

$v^2\geq\dfrac{gr}{\mu}$

Put the value into the formula

$v=\sqrt{\dfrac{9.8\times11}{0.51}}$

$v=14.53\ m/s$

Hence, The minimum speed is 14.53 m/s.

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3 years ago

Which of the following situations describes a change that will result in a new kind of matter with different characteristics?

taurus [48]

Answer:

Chemical property - characteristic of something that allows it to change to something new.

Explanation:

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2 years ago

The particle, initially at rest, is acted upon only by the electric force and moves from point a to point b along the x axis, in

bezimeni [28]

1) Potential difference: 1 V

2) $V_b-V_a = -1 V$

Explanation:

When a charge moves in an electric field, its electric potential energy is entirely converted into kinetic energy; this change in electric potential energy is given by

$\Delta U=q\Delta V$

where

q is the charge's magnitude

$\Delta V$ is the potential difference between the initial and final position

In this problem, we have:

$q=4.80\cdot 10^{-19}C$ is the magnitude of the charge

$\Delta U = 4.80\cdot 10^{-19}J$ is the change in kinetic energy of the particle

Therefore, the potential difference (in magnitude) is

$\Delta V=\frac{\Delta U}{q}=\frac{4.80\cdot 10^{-19}}{4.80\cdot 10^{-19}}=1 V$

Here we have to evaluate the direction of motion of the particle.

We have the following informations:

- The electric potential increases in the +x direction

- The particle is positively charged and moves from point a to b

Since the particle is positively charged, it means that it is moving from higher potential to lower potential (because a positive charge follows the direction of the electric field, so it moves away from the source of the field)

This means that the final position b of the charge is at lower potential than the initial position a; therefore, the potential difference must be negative:

$V_b-V_a = - 1V$

8 0

2 years ago