A basketball player throws a ball horizontally toward another player 10 meters away. If the ball is

Two sound waves, from two different sources with the same frequency, 540 Hz, travel in the same direction at 330 m s . The sourc

oee [108]

Answer:

The value is $\Delta \phi = 4.12 \ rad$

Explanation:

From the question we are told that

The frequency of each sound is $f_1 = f_2 = f = 540 \ Hz$

The speed of the sounds is $v = 330 \ m/s$

The distance of the first source from the point considered is $a = 4.40 \ m$

The distance of the second source from the point considered is $b = 4.00 \ m$

Generally the phase angle made by the first sound wave at the considered point is mathematically represented as

$\phi_a = 2 \pi [\frac{a}{\lambda} + ft]$

Generally the phase angle made by the first sound wave at the considered point is mathematically represented as

$\phi_b = 2 \pi [\frac{b}{\lambda} + ft]$

Here b is the distance o f the first wave from the considered point

Gnerally the phase diffencence is mathematically represented as

$\Delta \phi= \phi_a - \phi_b = 2 \pi [\frac{ a}{\lambda} + ft ] - 2 \pi [\frac{b}{\lambda} + ft ]$

=> $\Delta \phi = \frac{2\pi [ a - b]}{ \lambda }$

Gnerally the wavelength is mathematically represented as

$\lambda = \frac{v}{f}$

=> $\lambda = \frac{330}{540}$

=> $\lambda = 0.611 \ m$

=> $\Delta \phi = \frac{2* 3.142 [ 4.40 - 4.0 ]}{ 0.611 }$

=> $\Delta \phi = 4.12 \ rad$

5 0

3 years ago

How do you calculate the refractive index of a material using the critical angle? (GCSE Physics)

Helga [31]

Answer:

µ = $\frac{1}{sinC}$

Explanation:

µ = 1/ sinC

µ -----> refractive index of medium

C ----> critical angle

Hope this helps!

4 0

3 years ago

Explain how power is determined.

ivanzaharov [21]

It is the amount of work you have done in a certain amount of time. <span />

3 0

3 years ago

Read 2 more answers

A hockey player strikes a puck, giving it an initial velocity of 14.0 m/s in the positive x-direction. The puck slows uniformly

Advocard [28]

a) $-1.71 m/s^2$

b) 7.7 m/s

c) 4.39 s

Explanation:

a)

The acceleration of an object is the rate of change of velocity of an object.

In this problem, the acceleration of the puck can be found using the following suvat equation:

$v^2-u^2=2as$

where:

v is the final velocity

u is the initial velocity

a is the acceleration

s is the distance travelled

For the puck in this problem:

u = 14.0 m/s

v = 6.50 m/s

s = 45.0 m

So, the acceleration is:

$a=\frac{v^2-u^2}{2s}=\frac{6.50^2-14.0^2}{2(45.0)}=-1.71 m/s^2$

b)

The velocity of the puck at time t can be found by using another suvat equation:

$v=u+at$

where

u is the initial velocity

a is the acceleration

t is the time elapsed

v is the final velocity

Here, we have:

u = 14.0 m/s

$a=-1.71 m/s^2$ (found in part a)

Therefore, the velocity of the puck after t = 3.70 s is:

$v=14.0+(-1.71)(3.70)=7.7 m/s$

c)

Here we want to find the time taken for the puck to travel a distance of

s = 45.0 m

To solve this part, we can use again the suvat equation:

$v=u+at$

Where in this case, we use:

u = 14.0 m/s is the initial velocity

v = 6.50 m/s is the final velocity when the puck has travelled 45.0 m (this information is given in the question)

$a=-1.71 m/s^2$ is the acceleration (found in part a)

Therefore, by re-arranging the equation, we find the time taken to cover 45.0 m:

$t=\frac{v-u}{a}=\frac{6.50-14.0}{-1.71}=4.39 s$

7 0

3 years ago

At the top of a pole vault, an athlete actually can do work pushing on the pole before releasing it. Suppose the pushing force t

Salsk061 [2.6K]

Answer:

The work done on the athlete is approximately 2.09 J

Explanation:

From the definition of the work done by a variable force:

$\displaystyle{\int_{x_i}^{x_j}F(x)dx}$

and substituting with the function of our problem:

$\displaystyle{\int_{0}^{0.19}(140x-190x^2)dx\approx2.09\mathrm{J}}$

5 0

4 years ago