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3 years ago

How do you calculate the speed of a wave?

Physics

2 answers:

Anton [14]3 years ago

4 0

Answer:

Wave speed is the distance a wave travels in a given amount of time, such as the number of meters it travels per second. Wave speed is related to wavelength and wave frequency by the equation: Speed = Wavelength x Frequency. This equation can be used to calculate wave speed when wavelength and frequency are known.

Explanation:

Maksim231197 [3]3 years ago

4 0

Answer:

Speed = Wavelength x Frequency

Explanation:

I hope this helps!

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The slope of the following graph represents

Harlamova29_29 [7]

Can you please provide an image of the graph?

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3 years ago

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PLEASE HELP ME WITH THIS ONE QUESTION

Vlad [161]

Answer: 44.57°C

Explanation:

The following can be deduced from the question:

Specific heat of water = 4.186 J/kg

From the question, we can infer that 625 × 4.186 joules of heat will be lost when there's a 1°C drop of water.

We then calculate the amount if degrees that it'll take to cool for 7.96 x 10⁴J. This will be:

= 7.96 × 10⁴ /(625 × 4.186)

= 79600/(625 x 4.186)

= 79600/2616.25

= 30.43°C

The final temperature will then be:

= 75.0°C - 30.43°C

= 44.57°C

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3 years ago

A water-skier is being pulled by a tow rope attached to a boat. As the driver pushes the throttle forward, the skier accelerates

Novosadov [1.4K]

Answer:

4334.4 J

Explanation:

Work done equals to kinetic energy change

KE=½mv²

Change in KE is given by

∆KE=½m(v²-u²)

Where m is mass of water-skier, KE is kinetic energy, ∆KE is the change in kinetic energy, v is final velocity and u is initial velocity.

Substituting 72 kg for m, 12.1 m/s for v and 5.10 m/s for u then

∆KE=½*72(12.1²-5.10²)=4334.4J

Therefore, the work done by the net external force acting on the skier is equal to 4334.4 J

6 0

3 years ago

Which image shows an example of the electromagnetic force in action?

yarga [219]

Answer:

Where are the images?

Explanation:

3 0

3 years ago

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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.

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3 years ago