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4 years ago

Why can’t we see microwaves

Physics

1 answer:

motikmotik4 years ago

8 0

Answer:

The human retina can only detect incident light that falls in waves 400 to 720 nanometers long, so we can't see microwave or ultraviolet wavelengths. This also applies to infrared lights which has wavelengths longer than visible and shorter than microwaves, thus being invisible to the human eye.

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Toon Train is traveling at the speed of 10 m/s at the top of a hill. Five seconds later it reaches the bottom of the hill and is

Naddika [18.5K]

Answer:

the rate of acceleration of the train is 4 m/s²

Explanation:

Given;

initial velocity of the train, u = 10 m/s

change in time of motion, dt = 5 s

final velocity of the train, v = 30 m/s

The rate of acceleration of the train is calculated as;

$a = \frac{dv}{dt} = \frac{v-u}{dt} = \frac{30-10}{5} = \frac{20}{5} = 4 \ m/s^2$

Therefore, the rate of acceleration of the train is 4 m/s²

5 0

3 years ago

What is newton's first low of motion?

sergey [27]

Newton's first law of motion is that an object in motion will tend to stay in motion unless an external force acts upon it.

8 0

3 years ago

Read 2 more answers

You pull on a spring whose spring constant is 22 N/m, and stretch it from its equilibrium length of 0.3 m to a length of 0.7 m.

Liono4ka [1.6K]

Answer:

W= 4.4 J

Explanation

Elastic potential energy theory

If we have a spring of constant K to which a force F that produces a Δx deformation is applied, we apply Hooke's law:

F=K*x Formula (1): The force F applied to the spring is proportional to the deformation x of the spring.

As the force is variable to calculate the work we define an average force

$F_{a} =\frac{F_{f}+F_{i} }{2}$ Formula (2)

Ff: final force

Fi: initial force

The work done on the spring is :

W = Fa*Δx

Fa : average force

Δx : displacement

$W = F_{a} (x_{f} -x_{i} )$ :Formula (3)

$x_{f}$ : final deformation

$x_{i}$ :initial deformation

Problem development

We calculate Ff and Fi , applying formula (1) :

$F_{f} = K*x_{f} =22\frac{N}{m} *0.7m =15.4N$

$F_{i} = K*x_{i} =22\frac{N}{m} *0.3m =6.6N$

We calculate average force applying formula (2):

$F_{a} =\frac{15.4N+6.2N}{2} = 11 N$

We calculate the work done on the spring applying formula (3) : :

W= 11N*(0.7m-0.3m) = 11N*0.4m=4.4 N*m = 4.4 Joule = 4.4 J

Work done in stages

Work is the change of elastic potential energy (ΔEp)

W=ΔEp

ΔEp= Epf-Epi

Epf= final potential energy

Epi=initial potential energy

$E_{pf} =\frac{1}{2} *k*x_{f}^{2}$

$E_{pi} =\frac{1}{2} *k*x_{i}^{2}$

$E_{pf} =\frac{1}{2} *22*0.7^{2} = 5.39 J$

$E_{pf} =\frac{1}{2} *22*0.3^{2} = 0.99 J$

W=ΔEp= 5.39 J-0.99 J = 4.4J

4 0

3 years ago

HELP

sergiy2304 [10]

Answer:

Explanation:

a charged and uncharged object attratct eachother that is the answer your welcome

8 0

3 years ago

A car of mass 1800 kg collides with a truck of mass 5200 kg, and just after the collision the car and truck slide along, stuck t

kupik [55]

Explanation:

Elastic collision is said to occur if the total kinetic energy is not conserved and if there is a rebound after collision

Step one

Analysis of the problem

Immediately after impact the car's velocity was zero making it a perfect elastic collision

Step two

Given

Mass of car M1=1800kg

Mass of truck M2=5200kg

Initial velocity of Car U1=44m/s

Initial velocity of truck U2=21m/s

Final velocity of car V1= 0m/s

Final velocity of truck V2=20m/s

Step three

According to the principle of conservation of momentum

Total momentum before collision

=M1U1+M2U2

Total momentum after impact

=M1V1+M2V2

M1U1+M2U2 =M1V1+M2V2

Substituting our data into the expression we have 1500*44+5800*21=1500*0+5800*20

=66000+1218000=11600

1284000=11600

From the solution the momentum before impact is 1284000Ns

Momentum after impact is 11600Ns

This is indicating that after impact there was loss in momentum as a result of the car having a velocity of zero

8 0

4 years ago