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3 years ago

What are electromagnetic waves?

Physics

1 answer:

adelina 88 [10]3 years ago

3 0

D: Sóng có thể truyền qua khoảng không của không gian,

với tốc độ ánh sáng.

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You are trying to climb a castle wall so, from the ground, you throw a hook with a rope attached to it at 24.1 m/s at an angle o

Serhud [2]

Answer:

The value is $h = 13.2 \ m$

Explanation:

From the question we are told that

The speed of the rope with hook is $u = 24 .1 \ m/s$

The angle is $\theta = 65.0^o$

The speed at which it hits top of the wall is $v = 16.3 m/s$

Generally from kinematic equation we have that

$v_y^2 = u_y ^2 + * 2 (-g)* h$

Here h is the height of the wall so

$[16.3 sin (65)]^2 = [24.1 sin (65)] ^2+ 2 (-9.8)* h$

=> $h = 13.2 \ m$

4 0

3 years ago

HELP ASAP PLZ!

Dafna1 [17]

1. is D. Scientific method cannot be replaced.
2. is B.
3. is B as well.

8 0

3 years ago

Read 2 more answers

2CO + 2NO → 2CO2 + N2 is it balannced

natita [175]

If the same atoms appear on both sides, then it's balanced.

In this reaction, there are 4 Oxygens, 2 Carbons, and 2 Nitrogens on each side. So numerically, <em>it's balanced</em>. But I don't know enough chemistry to say whether the reaction is possible.

4 0

3 years ago

Read 2 more answers

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 $=0.33\ m$ .

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3 years ago

In which of the following does personality psychology specialize?

ValentinkaMS [17]

B, Studying how ones' character changes over time... That's the only one that has to deal with personality.

5 0

3 years ago

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