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3 years ago

Is it using energy when someone stands against a locker that doesn’t move

1 answer:

5 0

When someone stands against a locker and is does not moving at all, then there will be no displacement and since displacement = 0

Work done also becomes equal to zero.

Work done is usually defined as change in energy. Since the work done is zero there has been no energy used.

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Two loudspeakers emit sound waves of the same frequency along the x-axis. The amplitude of each wave is a. The sound intensity i

leonid [27]

Answer:

Explanation:

To find the amplitude of the sound, we must first determine the wavelength and the phase difference between the two speakers.

For the wavelength;

Recall that, the separation between two successive max. and min. intensity points are $\dfrac{\lambda}{2}$

Thus; for both speakers; the wavelength of the sound is:

$\dfrac{\lambda}{2} = (10+30) cm$

$\dfrac{\lambda}{2} = (40) cm$

λ = 80 cm

The relation between the path difference(Δx) and the phase difference(Δ∅) is:

$\Delta \phi = \dfrac{2 \pi}{\lambda}\Delta x + \Delta \phi_o$

where;

Δx = 10 cm

λ = 80 cm

Δ∅ = π rad

∴

$\Delta \phi = \dfrac{2 \pi}{\lambda}\Delta x + \Delta \phi_o$

$\pi \ rad = \dfrac{2 \pi}{80 \ cm}(10 \ cm) + \Delta \phi_o$

$\pi \ rad = \dfrac{2 \pi}{8}+ \Delta \phi_o$

$\pi \ rad = \dfrac{ \pi}{4}+ \Delta \phi_o$

$\Delta \phi_o = \pi -\dfrac{ \pi}{4}$

$\Delta \phi_o = \dfrac{ 4\pi - \pi}{4}$

$\Delta \phi_o = \dfrac{ 3\pi}{4} \ rad$

Suppose both speakers are placed side-by-side, then the path difference between the two speakers is: Δx = 0 cm

Thus, we have:

$\Delta \phi = \dfrac{2 \pi}{\lambda}\Delta x + \Delta \phi_o$

$\Delta \phi = \dfrac{2 \pi}{\lambda}(0 \ cm ) + \dfrac{3 \pi}{4} \ rad$

$\Delta \phi = \dfrac{3 \pi}{4} \ rad$

∴

The amplitude of the sound wave if the two speakers are placed side-by-side is:

$A = 2a \ cos \bigg (\dfrac{\Delta \phi }{2} \bigg)$

$A = 2a \ cos \bigg (\dfrac{\dfrac{3 \pi}{4} }{2} \bigg)$

$A = 2a \ cos \bigg ({\dfrac{3 \pi}{8} } \bigg)$

A = 0.765a

7 0

3 years ago

How long does it take to drive from one end of california to the other?

evablogger [386]

California is the third largest state and the only two bigger states than California are Alaska and Texas so it really depends on how you want to cross it. There are two routes to cross California depending on how you plan your visit and places you need to see. Depending on the route you take crossing California can take from twelve to almost sixteen hours of drive.

8 0

3 years ago

. A 40.0-kg child standing on a frozen pond throws a 0.500-kg stone to the east with a speed of 5.00 m/s. Neglecting friction be

tamaranim1 [39]

Answer:

Explanation:

A 40kg child throw stone of 0.5kg

At a direction of 5m/s

Recoil can be calculated using recoil of a gun formula

m_1•v_1 + m_2•v_2

m_1•v_1 = -m_2•v_2

The negative sign show that the momentum of the boy is directed oppositely to that of the stone

m_1 Is mass of boy

v_1 is the recoil velocity of the boy

m_2 is mass of stone

v_2 is the velocity of stone

Then,

m_1•v_1 = -m_2•v_2

40•v_1 = -0.5 × 5

40•v_1 = -2.5

v_1 = -2.5 / 40

v_1 = -0.0625 m/s

The recoil velocity of the boy is 0.0625 m/s

6 0

3 years ago

Identify the formula for the binary ionic compound, aluminum oxide.

vesna_86 [32]

Answer:

my bad ion even know what it is i just need sum points

Explanation:

3 0

3 years ago

Read 2 more answers

A string under a tension of 50.4 N is used to whirl a rock in a horizontal circle of radius 2.51 m at a speed of 21.1 m/s. The s

Leokris [45]

Answer:

619.8 N

Explanation:

The tension in the string provides the centripetal force that keeps the rock in circular motion, so we can write:

$T=m\frac{v^2}{r}$

where

T is the tension

m is the mass of the rock

v is the speed

r is the radius of the circular path

At the beginning,

T = 50.4 N

v = 21.1 m/s

r = 2.51 m

So we can use the equation to find the mass of the rock:

$m=\frac{Tr}{v^2}=\frac{(50.4)(2.51)}{21.1^2}=0.284 kg$

Later, the radius of the string is decreased to

r' = 1.22 m

While the speed is increased to

v' = 51.6 m/s

Substituting these new data into the equation, we find the tension at which the string breaks:

$T'=m\frac{v'^2}{r'}=(0.284)\frac{(51.6)^2}{1.22}=619.8 N$

5 0

3 years ago