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NeX [460]
4 years ago
7

The diagram below shows a wave with its wavelength indicated in red.

Physics
2 answers:
lisabon 2012 [21]4 years ago
5 0

C. It will decrease


il63 [147K]4 years ago
3 0

The answer is it will decrease.

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A flare is shot from a snowmobile with an initial velocity of 4.3 m/s at 90*, while the snowmobile is moving at 8.5 m/s at 0.0*.
Sedaia [141]

The driver is tooling along in his snowmobile, pointed north,
at 8.5 m/s.

He's carrying the flares with him, so the flares are also moving north
at 8.5 m/s.

When he fires the flare straight up, it has a vertical velocity of 4.3 m/s
straight up, and a horizontal velocity of 8.5 m/s towards the north.

The magnitude of the net velocity is √(4.3² + 8.5²) .
That's about  9.53 m/s, at some angle between straight up
and straight north.

The angle above horizontal is the angle that has a tangent of  4.3/8.5 .
I'll let you work out the angle.
 
8 0
3 years ago
The transfer of thermal energy is a major aspect of which branch of physics?
scZoUnD [109]
Mechanics is dealing with forces that are effecting some body, electrostatics is about electrical fields of not moving bodies, and quantum mechanics is dealing with quantum states of atoms. 

Thermodynamics as the word say, is dealing with thermal energy that is moving (transferring from one body to another or even better from one medium to another).

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3 0
3 years ago
Read 2 more answers
Why is visual constancy important?
weqwewe [10]

Answer:

C.

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Visual constancy is a key mechanism that allows the perception to remain.the same even as images change

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4 0
2 years ago
If an object doubles in mass, what happens to it momentum? Explain why.
Snezhnost [94]

Let's look at the equation for momentum (p)

p=mv

Where p is the momentum, m is the mass and v is the velocity.

If an object doubles in mass then we have 2m and so:

p=(2m)v\\p=2mv\\\\p=2(mv)

If it doubles in mass, then the momentum is also doubled.

Brainliest will be much appreciated!

8 0
3 years ago
Two particles are fixed on an x axis. Particle 1 of charge 44.9 μC is cated at х- 24.5 cm; particle 2 of charge Q is located at
Kryger [21]

The force that is being exerted on particle 3 by particle 1 is equal to:

F_{13}=\frac{K*Q_1*Q_3}{r13^2}

r_{13} = \sqrt{(0.245m)^2 + (0.245m)^2}=0.3465m^2

F_{13}=\frac{9*10^9Nm^2/C^2*44.9*10^-6C*38.9*10^-6C}{0.3465m^2}=45.37N

As both particles has positive charges, the particles will repel each other and the resultant force will have the direction of the vector r_{13}. Therefore, F_{13} will have x and y components equal to:

F{13x}=F{13}*\frac{r13x}{|r13|}=45.37N*\frac{0.245m}{0.3465m}=32.08 N

F{13y}=F{13}*\frac{r13y}{|r13|}=45.37N*\frac{0.245m}{0.3465m}=32.08 N

In order to calculate Force between particles 2 and 3, we first assume Q2 to be possitive (if it's negative the result will have a negative value, so this doesn't matter). We follow the same line of reasoning we used to calculate F13, just that Q2 will be unknown.

F_{23}=\frac{K*Q_2*Q_3}{r13^2}

r_{23} = \sqrt{(-0.0753m)^2 + (0.245m)^2}=0.2563m^2

F_{23}=\frac{9*10^9Nm^2/C^2*Q_2*38.9*10^-6C}{0.2563m^2}=1.36*10^6Q_2

F{23x}=F{23}*\frac{r23x}{|r23|}=1.36*10^6Q_2*\frac{-0.0753m}{0.2563m}= -401286.15Q_2

F{23y}=F{23}*\frac{r23y}{|r23|}=1.36*10^6Q_2*\frac{0.245m}{0.2563m}=1.305*10^6 Q_2

a) For incise a, F13y + F23y has to be equal to 0:

F{13y}+F{23y}=0

32.08 N+1.305*10^6 Q_2=0

Q_2=\frac{-32.08}{1.305*10^6}=-24.6*10^{-6}C =-24.6uC

b) For incise b, F13x + F23x has to be equal to 0:

F{13x}+F{23x}=0

32.08 N - 401286.15 Q_2=0

Q_2=\frac{32.08}{401286.15}=80*10^{-6}C =80uC

7 0
3 years ago
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