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3 years ago

Calculate the velocity of a wave that has a frequency of 60 Hz and wavelength of 2.0 m/s

Physics

1 answer:

mote1985 [20]3 years ago

3 0

Answer:We have , a relation in frequency f and wavelength λ of a wave having the velocity v as ,

v=fλ ,

given f=60Hz , λ=20m ,

therefore velocity of wave , v=60×20=1200m/s

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Everyone's favorite flying sport disk can be approximated as the combination of a thin outer hoop and a uniform disk, both of di

Umnica [9.8K]

Answer:

The length of the boomerang is 0.364 m

Explanation:

The moment of inertia is:

$I=\frac{1}{2} m_{d} r^{2} +m_{h} r^{2}$

Where

md = 0.05 kg

mh = 0.12 kg

r = d/2 = 0.273/2 = 0.1365 m

$I=\frac{0.05*(0.1365)^{2} }{2} +(0.12)*(0.1365)^{2} =2.7x10^{-3} kgm^{2}$

The length of the boomerang is:

$L_{b} =\sqrt{\frac{12I}{m} } =\sqrt{\frac{12*2.7x10^{-3} }{0.245} } =0.364m$

5 0

3 years ago

A panda's coarse fur is made of cells, as is its blood, its skin, its nerves, and its brain. This is an example of which charact

Allisa [31]

D.cellular organization 
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4 0

3 years ago

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

Ax = 44.4 m and Ay = 25.1 m Find the magnitude of the vector.

max2010maxim [7]

Answer:

The magnitude of the vector A is 51 m.

Explanation:

Given:

The horizontal component of a vector A is given as:

$A_x=44.4\ m$

The vertical component of a vector A is given as:

$A_y=25.1\ m$

Now, we know that, a vector A can be resolved into two mutually perpendicular components; one along the x axis and the other along the y axis. The magnitude of the vector A can be written as the square root of the sum of the squares of each component.

Therefore, the magnitude of vector A is given as:

$|\overrightarrow A|=\sqrt{A_{x}^2+A_{y}^2}$

Now, plug in 44.4 for $A_x$ , 25.1 for $A_y$ and solve for the magnitude of A. This gives,

$|\overrightarrow A|=\sqrt{(44.4)^2+(25.1)^2}\\|\overrightarrow A|=\sqrt{1971.36+630.01}\\|\overrightarrow A|=\sqrt{2601.37}\\|\overrightarrow A|=51\ m$

Therefore, the magnitude of the vector A is 51 m.

6 0

3 years ago

I NEED HELP ASAP!! PLEASE

Kisachek [45]

Answer:

Acceleration = -3m/s²

Explanation:

Given the following data;

Initial velocity = 19m/s

Final velocity = 3m/s

Time = 2secs

To find the acceleration;

In physics, acceleration can be defined as the rate of change of the velocity of an object with respect to time.

This simply means that, acceleration is given by the subtraction of initial velocity from the final velocity all over time. Mathematically, acceleration is given by the equation;

$Acceleration (a) = \frac{final \; velocity - initial \; velocity}{time}$

Substituting into the equation, we have;

$a = \frac{3 - 9}{2}$

$a = \frac{-6}{2}$

Acceleration = -3m/s²

The value for her acceleration is negative because she is decelerating i.e her final velocity is lower than her initial velocity.

7 0

3 years ago