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Shtirlitz [24]
3 years ago
9

Explain what is meant by “In Phase” and “Out of Phase?

Physics
2 answers:
frozen [14]3 years ago
8 0
They are at a positive peek coming out together
LekaFEV [45]3 years ago
5 0
In phase would mean both waves are at a positive peak, out of phase would mean one is at a positive whilst the other is at a negative. Out of phase would mean the waves cancel each other out
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otez555 [7]

Answer: D and friction

Explanation:

PLz brainliest

7 0
3 years ago
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In this type of bond, electrons are lost or gained by atoms, and the atoms are held together by electrical attraction.
expeople1 [14]
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8 0
3 years ago
A 3.20 kg block starts at rest and slides a distance d down a frictionless β = 30.0 ◦ incline, where it runs into a spring with
Virty [35]

Answer:

Explanation:

a ) work done by gravitational force

= mg sinθ ( d + .21)

Potential energy stored in compressed spring

= 1/2 k x²

= .5 x 431 x ( .21 )²

= 9.5

According to conservation of energy

mg sinθ ( d + .21)  = 9.5

3.2 x 9.8 x sin 30( d + .21 ) = 9.5

d = 40 cm

b )

As long as mg sin30 is greater than kx ( restoring force ) , there will be acceleration in the block.

mg sin30 = kx

3.2 x 9.8 x .5 = 431 x

x =  3.63 cm

When there is compression of 3.63 cm in the spring , block will have maximum velocity. there after its speed will start decreasing.

7 0
3 years ago
Sam, whose mass is 75 kg, straps on his skis and starts down a 50-m-high, 20 frictionless slope. A strong headwind exerts a hori
LenaWriter [7]

Answer:

Explanation:

Sam mass=75kg

Height is 50m

20° frictionless slope

Horizontal force on Sam is 200N

According to the work energy theorem, the net work done on Sam will be equal to his change in kinetic energy.

Therefore

Wg - Ww =∆K.E

Note initial the body was at rest at top of the slope.

Then, ∆K.E is K.E(final) - K.E(initial)

K.E Is given as ½mv²

Since initial velocity is zero then, K.E(initial ) is zero

Therefore, ∆K.E=½mVf²

Wg is work done by gravity and it is given by using P.E formulas

Wg=mgh

Wg=75×9.8×50

Wg=36750J

Ww is work done by wind and it's is given by using formulae for work

Work=force × distance

Ww=horizontal force × horizontal distance

Using Trig.

TanX=opposite/adjacent

Tan20=h/x

x=h/tan20

x=50/tan20

x=137.37m

Then,

Ww=F×x

Ww=200×137.37

We=27474J

Now applying the formula

Wg - Ww =∆K.E

36750 - 27474 =½×75×Vf²

9276=37.5Vf²

Vf²=9275/37.5

Vf²= 247.36

Vf=√247.36

Vf=15.73m/s

3 0
3 years ago
A frog hops 5m east and 2m north. What is the magnitude of the frogs total displacement in m?
TiliK225 [7]

As per the question a frog jumps 5 m towards east.

Frog again jumps 2 m north.

Let the displacement along east is denoted by vector A and the displacement towards north is denoted as vector B.

Hence magnitude of A = 5 m

           Magnitude of B = 2 m

We are asked to calculate the total displacement.

Here the angle between them is 90 degree as A is towards east and B is towards north.

As per parallelogram law of vector addition,the magnitude of total displacement [R] will be-

                            R=\sqrt{ A^{2} +B^{2}+2AB\cos\theta}

                            =\sqrt{ 5^{2}+ 2^{2}+2*5*2 cos 90}

                            =\sqrt{25+4+0} m  [cos90= 0]

                            =\sqrt{29} m

                            = 5.38516 m   [ans]

                     


7 0
3 years ago
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