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Murljashka [212]
3 years ago
7

Describe How the frequency of a wave changed s as it’s wavelength changes

Physics
1 answer:
MatroZZZ [7]3 years ago
4 0
When the frequency of a wave increases, the wavelength decreases. This is because the faster the frequency, the more waves in a period of time there is, making each wave length shorter.
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A 6.60-kg block slides with an initial speed of 1.56 m/s up a ramp inclined at an angle of 28.4° with the horizontal. The coeffi
Vlad [161]

Answer:

The distance travel by block before coming to rest is 0.122 m

Explanation:

Given:

Mass of block m = 6.60 kg

Initial speed of block v _{i} = 1.56 \frac{m}{s}

Final speed of block v_{f} = 0 \frac{m}{s}

Coefficient of kinetic friction \mu _{k} = 0.62

Ramp inclined at angle \theta = 28.4°

Using conservation of energy,

Work done by frictional force is equal to change in energy,

  \mu _{k} mgd \cos 28.4 =  \Delta K - \Delta U

Where \Delta U = mg d\sin 28.4

\mu _{k} mgd \cos 28.4 =  \frac{1}{2}mv_{i} ^{2} - mgd\sin 28.4

\mu _{k} mgd \cos 28.4 +mgd\sin 28.4  =  \frac{1}{2}mv_{i} ^{2}

d(6.60 \times 9.8 \times 0.62 \times 0.879 + 6.60 \times 9.8 \times 0.475) = \frac{1}{2} \times 6.60 \times (1.56)^{2}

 d = 0.122 m

Therefore, the distance travel by block before coming to rest is 0.122 m

7 0
3 years ago
A test tube is in uniform circular motion in a centrifuge, which is a piece of scientific equipment used to separate solids from
Molodets [167]
Since direction is part of velocity, and the object is moving in a circle, its velocity is constantly changing.
3 0
2 years ago
When an electric current is flowing through a wire, the force deflecting the charged particles is the greatest when the wire is
Makovka662 [10]
The force is greatest when the wire is perpendicular to the magnetic field.

In fact, the expression for the magnetic force is
F=ILB \sin \theta
where I is the current, L the length of the wire, B the magnetic field intensity and \theta is the angle between the directions of I and B.

We can see that F is maximum when \sin \theta=1, so when \theta = 90^{\circ}, so when the wire and the magnetic field are perpendicular to each other.
5 0
2 years ago
Read 2 more answers
A block of mass m=2.20m=2.20 kg slides down a 30.0^{\circ}30.0
Xelga [282]

Answer:

v_m \approx -4.38\; \rm m \cdot s^{-1} (moving toward the incline.)

v_M \approx 4.02\; \rm m \cdot s^{-1} (moving away from the incline.)

(Assumption: g = 9.81\; \rm m \cdot s^{-2}.)

Explanation:

If g = 9.81\; \rm m \cdot s^{-2}, the potential energy of the block of m = 2.20\; \rm kg would be m \cdot g\cdot h = 2.20\; \rm kg \times 9.81\; \rm m \cdot s^{-2} \times 3.60\; \rm m \approx 77.695\; \rm J when it was at the top of the incline.

If friction is negligible, all these energies would be converted to kinetic energy when this block reaches the bottom of the incline. There shouldn't be any energy loss along the horizontal surface, either. Therefore, the kinetic energy of this m = 2.20\; \rm kg\! block right before the collision would also be approximately 77.695\; \rm J.

Calculate the velocity of that m = 2.20\; \rm kg based on its kinetic energy:

\displaystyle v_m(\text{initial}) = \sqrt{\frac{2\times (\text{Kinetic Energy})}{m}} \approx \sqrt{\frac{2 \times 77.695\; \rm J}{2.20\; \rm kg}} \approx 8.4043\; \rm m \cdot s^{-1}}.

A collision is considered as an elastic collision if both momentum and kinetic energy are conserved.

Initial momentum of the two blocks:

p_m = m \cdot v_m(\text{initial}) \approx 2.20\; \rm kg \times 8.4043\; \rm m \cdot s^{-1} \approx 18.489\; \rm kg \cdot m \cdot s^{-1}.

p_M = M \cdot v_M(\text{initial}) \approx 2.20\; \rm kg \times 0\; \rm m \cdot s^{-1} \approx 0\; \rm kg \cdot m \cdot s^{-1}.

Sum of the momentum of each block right before the collision: approximately 18.489\; \rm kg \cdot m \cdot s^{-1}.

Sum of the momentum of each block right after the collision: (m\cdot v_m + m \cdot v_M).

For momentum to conserve in this collision, v_m and v_M should ensure that m\cdot v_m + m \cdot v_M \approx 18.489\; \rm kg \cdot m \cdot s^{-1}.

Kinetic energy of the two blocks right before the collision: approximately 77.695\; \rm J and 0\; \rm J. Sum of these two values: approximately 77.695\; \rm J\!.

Sum of the energy of each block right after the collision:

\displaystyle \left(\frac{1}{2}\, m \cdot {v_m}^2 + \frac{1}{2}\, M \cdot {v_M}^2\right).

Similarly, for kinetic energy to conserve in this collision, v_m and v_M should ensure that \displaystyle \frac{1}{2}\, m \cdot {v_m}^2 + \frac{1}{2}\, M \cdot {v_M}^2 \approx 77.695\; \rm J.

Combine to obtain two equations about v_m and v_M (given that m = 2.20\; \rm kg whereas M = 7.00\; \rm kg.)

\left\lbrace\begin{aligned}& m\cdot v_m + m \cdot v_M \approx 18.489\; \rm kg \cdot m \cdot s^{-1} \\ & \frac{1}{2}\, m \cdot {v_m}^2 + \frac{1}{2}\, M \cdot {v_M}^2 \approx 77.695\; \rm J\end{aligned}\right..

Solve for v_m and v_M (ignore the root where v_M = 0.)

\left\lbrace\begin{aligned}& v_m \approx -4.38\; \rm m\cdot s^{-1} \\ & v_M \approx 4.02\; \rm m \cdot s^{-1}\end{aligned}\right..

The collision flipped the sign of the velocity of the m = 2.20\; \rm kg block. In other words, this block is moving backwards towards the incline after the collision.

6 0
2 years ago
The refractive index of water is 1.33
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Explanation:

light travel slower in daimond

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2 years ago
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