After exercise your heart rate is

Two identical loudspeakers are placed on a wall 1.00 m apart. A listener stands 4.00 m from the wall directly in front of one of

natita [175]

Answer:

The phase difference is 0.659 rad.

Explanation:

Given that,

Distance between two identical loudspeakers d= 1.00 m

Distance between speakers and listener r= 4.00 m

Frequency = 300 Hz

Suppose we need to find the phase difference in radian between the waves from the speakers when they reach the observer

We need to calculate the r'

Using Pythagorean theorem

$r'=\sqrt{d^2+r^2}$

Where, d = distance between two identical loudspeakers

r = distance between speakers and listener

Put the value into the formula

$r'=\sqrt{(1.00)^2+(4.00)^2}$

$r'=\sqrt{1.00+16.00}$

$r'=4.12\ m$

We need to calculate the path difference

Using formula of path difference

$|r'-r|=4.12-4.00$

$|r'-r|=0.12\ m$

We need to calculate the wavelength

Using formula of wavelength

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

Where, v = speed of sound

f = frequency

Put the value into the formula

$\lambda=\dfrac{343}{300}$

$\lambda=1.143\ m$

We need to calculate the phase difference

Using formula of phase difference

$\phi=\dfrac{2\pi\times|r'-r| }{\lambda}$

$\phi=\dfrac{2\pi\times0.12}{1.143}$

$\phi=0.659\ rad$

Hence, The phase difference is 0.659 rad.

7 0

3 years ago

Two balls of clay, with masses M1 = 0.49 kg and M2 = 0.47 kg, are thrown at each other and stick when they collide. Mass 1 has a

malfutka [58]

Answer:

a) $p_i=1.568\hat{i}+0.752 \hat{j}$

b) $v_{fx}=1.668\ m.s^{-1}$

c) $v_{fy}=0.7999\ m.s^{-1}$

Explanation:

Given masses:

$m_1=0.49\ kg$

$m_2=0.47\ kg$

Velocity of mass 1, $v_1=3.2 \hat{i}\ m.s^{-1}$

Velocity of mass 2, $v_2=1.6 \hat{j}\ m.s^{-1}$

a)

Initial momentum:

$p_i=m_1.v_1+m_2.v_2$

$p_i=0.49\times 3.2 \hat{i}+0.47\times 1.6 \hat{j}$

$p_i=1.568\hat{i}+0.752 \hat{j}$

b)

magnitude of initial momentum:

$p_i=\sqrt{1.568^2+0.752 ^2}$

$p_i=1.739\ kg.m.s^{-1}$

From the conservation of momentum:

$p_f=p_i$

$m_f.v_f=1.739$

$v_f=\frac{1.739}{0.49+0.47}$

$v_f=1.85\ m.s^{-1}$ is the magnitude of final velocity.

Direction of final velocity will be in the direction of momentum:

$tan\theta=\frac{0.752 }{1.568}$

$\theta=25.62^{\circ}$

$\therefore v_{fx}=1.85\ cos25.62^{\circ}$

$v_{fx}=1.668\ m.s^{-1}$

c)

Vertical component of final velocity:

$v_{fy}=1.85\ sin 25.62^{\circ}$

$v_{fy}=0.7999\ m.s^{-1}$

6 0

3 years ago

what is the force of friction between an 80kg box and the ground on Earth if the coefficient is 0.2?

Reil [10]

Answer:

160N

Explanation: When 80kg mass is one group . It's reaction force acting on a ground.

Weight of the object = 80*10

= 800 N

Here we are given cofficient of static friction its 0.2. It should be smaller than 1

Friction force = Reaction * Friction Cofficient

Reaction = 800N ( Considering Vertical Equilibrium )

F = 800* 0.2

F = 160N

3 0

3 years ago

Read 2 more answers

The density of sodium is 968 g/cm3. Suppose you need 250 gm of sodium for an experiment. Which volume of sodium do you need?

Masja [62]

The formula for the density of a substance expressed in mass and volume is rho = mass/volume or p = m/v. Rearranging the formula to isolate volume gives the formula v = m/p. To solve for the problem given, this formula must be used. This gives a solution of:

v = m/p = 250 g/ 968 g/cm^3 = 0.258 cm^3 of sodium

4 0

3 years ago

An object is pulled to the left by a force of 50 N. The same amount of force pulls it to the right. The object will ____.

11111nata11111 [884]

<h2>Required Answer:</h2>

The body will stay at rest (Option D). It is because a force of magnitude 50 N is pulled towards left and another force is pulling it towards right with same magnitude 50 N. So, the direction of force is opposite and magnitude is same i.e. 50 N. So, they will cancel each other and net force is 0. Hence, there would be no acceleration.

Option A - Showing acceleration
Option B - Showing acceleration
Option C - Change of direction due to Net force

Hence, these options are incorrect because they are only possible when net external force is non-zero. Staying at rest i.e. Option D means there is no motion and hence no acceleration, this shows that net force is 0 $.$

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3 0

3 years ago