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Grace [21]
2 years ago
9

A periodic wave travels from one medium to another. Which pair of variables are likely to change in the process? A. velocity and

wavelength B. velocity and frequency C. frequency and wavelength D. frequency and phase E. wavelength and phase
Physics
1 answer:
dezoksy [38]2 years ago
5 0

Answer:

A. velocity and wavelength

Explanation:

  • When a wave travels from one medium to another it undergoes a change in direction and this is referred to as refraction.
  • Refraction is the bending of a wave or a change in direction of a wave as it travels from one medium to another. Refraction is accompanied by change in velocity and wavelength of a wave.
  • Velocity of a wave is proportional to the wavelength of the wave therefore, if the velocity of a wave changes the the wavelength will also change proportionally to the velocity.
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. A spring has a length of 0.200 m when a 0.300-kg mass hangs from it, and a length of 0.750 m when a 1.95-kg mass hangs from it
kap26 [50]

Answer:

29.4 N/m

0.1  

Explanation:

a) From the restoring Force we know that :  

F_r = —k*x  

the gravitational force :  

F_g=mg  

Where:

F_r is the restoring force .

F_g is the gravitational force

g is the acceleration of gravity

k is the constant force  

xi , x2 are the displacement made by the two masses.

Givens:

<em>m1 = 1.29 kg</em>

<em>m2 = 0.3 kg  </em>

<em>x1   = -0.75 m  </em>

<em>x2 = -0.2 m </em>

<em>g   = 9.8 m/s^2  </em>

Plugging known information to get :

F_r =F_g

-k*x1 + k*x2=m1*g-m2*g

k=29.4 N/m

b) To get the unloaded length 1:  

l=x1-(F_1/k)

Givens:

m1 = 1.95kg , x1 = —0.75m  

Plugging known infromation to get :

l= x1 — (F_1/k)  

= 0.1  

 

3 0
2 years ago
A 2.0-cm-diameter parallel-plate capacitor with a spacing of 0.50 mm is charged to 200 V. What are (a) the total energy stored i
Debora [2.8K]

Answer:

(A) Total energy will be equal to 0.044\times 10^{-5}J

(b) Energy density will be equal to 0.0175J/m^3

Explanation:

We have given diameter of the plate d = 2 cm = 0.02 m

So area of the plate A=\pi r^2=3.14\times 0.02^2=0.001256m^2

Distance between the plates d = 0.50 mm = 0.50\times 10^{-3}m

Permitivity of free space \epsilon _0=8.85\times 10^{-12}F/m

Potential difference V =200 volt

Capacitance between the plate is equal to C=\frac{\epsilon _0A}{d}=\frac{8.85\times 10^{-12}\times 0.001256}{0.50\times 10^{-3}}=0.022\times 10^{-9}F

(a) Total energy stored in the capacitor is equal to

E=\frac{1}{2}CV^2

E=\frac{1}{2}\times 0.022\times 10^{-9}\times 200^2=0.044\times 10^{-5}J

(b) Volume will be equal to V=Ad, here A is area and d is distance between plates

V=0.001256\times 0.02=2.512\times 10^{-5}m^3

So energy density =\frac{Energy}{volume}=\frac{0.044\times 10^{-5}}{2.512\times 10^{-5}}=0.0175J/m^3

7 0
2 years ago
A 21200 kg sailboat experiences an eastward force 42700 N due to the tide pushing its hull while the wind pushes the sails with
My name is Ann [436]

Answer:

2.95 m/s^{2}

Explanation:

The resultant force F

F=\sqrt {F_{1}^{2}+F_{2}^{2}+2F_{1}F_{2}cos135^{o}} where F_{1} is eastward force, F_{2} is force directed towards the North

F=\sqrt {42700^{2}+85000^{2}+(2*42700*85000)cos135^{o}}=62573.17217 N

F=62573.2 N

The magnitude of acceleration of sailboat is given by

a=\frac {F}{m}=\frac {62573.2}{21200}=2.95 m/s^{2}

7 0
2 years ago
To exercise, a man attaches a 4.0 kg weight to the heel of his foot. When his leg is stretched out before him, what is the torqu
Masja [62]

Answer:

B. τ = 16 Nm

Explanation:

In order to find the torque exerted by the weight attached to the heel of man's foot, when his leg is stretched out. We use following formula:

τ = Fd

here,

τ = Torque = ?

F = Force exerted by the weight = Weight = mg

F = mg = (4 kg)(10 m/s²) = 40 N

d = distance from knee to weight = 40 cm = 0.4 m

Therefore,

τ = (40 N)(0.4 m)

<u>B. τ = 16 Nm</u>

8 0
2 years ago
A 25 gram bullet is fired from a gun with a speed of 230 m/s. If the gun has a mass of 0.9 kg what is the recoil speed of gun ?
Step2247 [10]

Given data:

mass of the bullet (m) = 25 g = 0.025 kg,

mass of the gun    (M) = 0.9 kg,

speed of the bullet (v) =230 m/s,

speed of the bullet (V) = ?

From the given data it is clear that, the momentum is conserved. According to "<em>law of conservation of momentum" </em>the total momentum before and after the collision is equal.

In this problem the momentum before collision (bullet+gun) is zero.

Therefore, after the gun fires a bullet, the momentum must be zero.

Mathematically,

          M × V + m × v = 0

           where,

                     M × V = momentum of the gun

                     m × v = momentum of the bullet

            (0. 9 × V) + (0.025 × 230) = 0

             0.9 V = -5.75

                    V = -5.75/0.9

                        = -6.39 m/s  

<em>The gun recoils with a speed of 6.39 m/s</em>

8 0
2 years ago
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