All categories

3 years ago

The waves with the lowest energy and lowest frequencies of the electromagnetic spectrum are the

1 answer:

5 0

<span>The waves with the lowest energy and lowest frequencies of the electromagnetic spectrum are the "Radio waves"

So, option B is your answer

Hope this helps!
</span>

You might be interested in

un motor electric efectueaza un lucru mecanic de 864 j in 0.5 min.tensiunea la bornele sale este de 12v si este parcurs de un cu

GrogVix [38]

Answer:

96%

Explanation:

To find the values of the motor efficiency you use the following formula:

$E=\frac{P_o}{P_i}100$

P_o: output power = 864J/0.5min=864J/30s=28.8W

P_i: input power = I*V = (3A)(12V) = 36W

By replacing this values you obtain:

$E=\frac{28.8W}{30W}*100=96\%$

hence, the motor efficiency is about 96%

traslation:

Pentru a găsi valorile eficienței motorului, utilizați următoarea formulă:

P_o: putere de ieșire = 864J / 0.5min = 864J / 30s = 28.8W

P_i: putere de intrare = I * V = (3A) (12V) = 36W

Înlocuind aceste valori obțineți:

prin urmare, eficiența motorului este de aproximativ 96%

3 0

3 years ago

In general, the ________ of a simple machine is the ratio of the distance over which the force is applied to the distance over w

aksik [14]

Mechanical advantage, i hope i helped you!

3 0

3 years ago

Read 2 more answers

Phantasy [73]

Answer:

Explanation:

lll

5 0

3 years ago

Read 2 more answers

What is the average speed of an object that travels 6 meters in 2 seconds and then travels 3 meters in 1 second?

olya-2409 [2.1K]

Answer:

3 m/s

Explanation:

Average Speed = $\frac{Total Distance}{Total time}$

Plug in the numbers, it will be (6m + 3m) divided by (2s + 1s), which is 9m/3s, which equals to 3m/s.

4 0

2 years ago

A package of mass m is released from rest at a warehouse loading dock and slides down a 3.0-m-high frictionless chute to a waiti

LuckyWell [14K]

Answer:

The speed of the package of mass m right before the collision $= 7.668\ ms^-1$

Their common speed after the collision $= 2.56\ ms^-1$

Height achieved by the package of mass m when it rebounds $= 0.33\ m$

Explanation:

Have a look to the diagrams attached below.

a.To find the speed of the package of mass m right before collision we have to use law of conservation of energy.

$K_{initial} + U_{initial} = K_{final}+U_{final}$

where $K$ is Kinetic energy and $U$ is Potential energy.

$K= \frac{mv^2}{2}$ and $U= mgh$

Considering the fact $K_{initial} = 0\ and U_{final} =0$ we will plug out he values of the given terms.

So $V_{1}{(initial)} =\sqrt{2gh} = \sqrt{2\times9.8\times3} = 7.668\ ms^-1$

Keypoints:

Sum of energies and momentum are conserved in all collisions.
Sum of KE and PE is also known as Mechanical energy.
Only KE is conserved for elastic collision.
for elastic collison we have $e=1$ that is co-efficient of restitution.

<u>KE = Kinetic Energy and PE = Potential Energy</u>

b.Now when the package stick together there momentum is conserved.

Using law of conservation of momentum.

$m_1V_1(i) = (m_1+m_2)V_f$ where $V_1{i} =7.668\ ms^-1$ .

Plugging the values we have

$m\times 7.668 = (3m)\times V_{f}$

Cancelling m from both sides and dividing 3 on both sides.

$V_f = 2.56\ ms^-1$

Law of conservation of energy will be followed over here.

c.Now the collision is perfectly elastic $e=1$

We have to find the value of $V_{f}$ for m mass.

As here $V_{f}=-2.56\ ms^-1$ we can use that if both are moving in right ward with $2.56$ then there is a $-2.56$ velocity when they have to move leftward.

The best option is to use the formulas given in third slide to calculate final velocity of object $1$ .

$V_{1f} = \frac{m_1-m_2}{m_1+m_2} \times V_{1i}= \frac{m-2m}{3m} \times7.668=\frac{-7.668}{3} = -2.56\ ms^-1$

Now using law of conservation of energy.

$K_{initial} + U_{initial} = K_{final}+U_{final}$

$\frac{m\times V(f1)^2}{2} + 0 = 0 +mgh$

$\frac{v(f1)^2}{2g} = h$

$h= \frac{(-2.56)^2}{9.8\times 3} =0.33\ m$

The linear momentum is conserved before and after this perfectly elastic collision.

So for part a we have the speed $=7.668\ ms^-1$ for part b we have their common speed $=2.56\ ms^-1$ and for part c we have the rebound height $=0.33\ m$ .

3 0

3 years ago