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3 years ago

A wave has a frequency of 240 Hz and a wavelength of 3.0m what is the speed of the wave

Physics

2 answers:

r-ruslan [8.4K]3 years ago

8 0

Answer:

240 ms

Explanation:

Nastasia [14]3 years ago

7 0

Answer: 720 m/s

Hope this helps ♥️

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How do you measure the volume of irregular object? explain with diagram

masya89 [10]

Answer:

by using graphical meythod

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3 years ago

A 13.5 μF capacitor is connected to a power supply that keeps a constant potential difference of 22.0 V across the plates. A pie

-BARSIC- [3]

a) $3.27\cdot 10^{-3} J$

b) $11.60\cdot 10^{-3} J$

c) $8.33\cdot 10^{-3} J$

Explanation:

The energy stored in a capacitor is given by

$U=\frac{1}{2}CV^2$

where

C is the capacitance of the capacitor

V is the potential difference across the plates of the capacitor

For the capacitor in this problem, before insering the dielectric, we have:

$C=13.5 \mu F = 13.5\cdot 10^{-6}F$ is its capacitance

V = 22.0 V is the potential difference across it

Therefore, the initial energy stored in the capacitor is:

$U=\frac{1}{2}(13.5\cdot 10^{-6})(22.0)^2=3.27\cdot 10^{-3} J$

After the dielectric is inserted into the plates, the capacitance of the capacitor changes according to:

$C'=kC$

where

k = 3.55 is the dielectric constant of the material

C is the initial capacitance of the capacitor

Therefore, the energy stored now in the capacitor is:

$U'=\frac{1}{2}C'V^2=\frac{1}{2}kCV^2$

where:

$C=13.5\cdot 10^{-6}F$ is the initial capacitance

V = 22.0 V is the potential difference across the plate

Substituting, we find:

$U'=\frac{1}{2}(3.55)(13.5\cdot 10^{-6})(22.0)^2=11.60\cdot 10^{-3} J$

The initial energy stored in the capacitor, before the dielectric is inserted, is

$U=3.27\cdot 10^{-3} J$

The final energy stored in the capacitor, after the dielectric is inserted, is

$U'=11.60\cdot 10^{-3} J$

Therefore, the change in energy of the capacitor during the insertion is:

$\Delta U=11.60\cdot 10^{-3}-3.27\cdot 10^{-3}=8.33\cdot 10^{-3} J$

So, the energy of the capacitor has increased by $8.33\cdot 10^{-3} J$

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3 years ago

______ is the best way to determine your own correct intensity level.

scZoUnD [109]

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</span>

5 0

3 years ago

A block of mass m1 = 290 g is at rest on a plane that makes an angle θ = 30° above the horizontal. The coefficient of kinetic fr

GrogVix [38]

Answer:

The speed of the block when is fallen 30cm

v=0.726 $\frac{m}{s}$

Explanation:

∑F= (m2)g - ƒ - (m1)g*sin(θ) = (m1)a

g = 9.81 m/s²

ƒ = μN = μ(m1)g

$(m2)*g - u*(m1)*g - (m1)*g*sin(\alpha ) = (m1)*a$

(0.200)(9.81) - (0.1)(0.290)(9.81) - (0.290)(9.81)sin(30°) = (0.290)a

$(0.200kg)(9.81\frac{m}{s^{2}}) - (0.1)(0.290kg)(9.81\frac{m}{s^{2}}) - (0.290kg)(9.81\frac{m}{s^{2}})sin(30°) = (0.290kg)*a$

$0.511101=0.29*a\\a=0.879\frac{m}{s^{2} }$

$v_{f}^{2}=v_{o}^{2}+2*a(x_{f}^{2}-v_{o}^{2})\\v_{o}=0\\v_{o}=0\\v_{f}^{2}=2*a(x_{f})\\v_{f}=\sqrt{2*a(x_{f})}\\v_{f}=\sqrt{2*0.879\frac{m}{s^{2}}*0.30m} \\v_{f}=0.726 \frac{m}{s}$

5 0

4 years ago

A racquet ball with mass m = 0.256 kg is moving toward the wall at v = 11.8 m/s and at an angle of θ = 29° with respect to the h

icang [17]

Answer:

Part a)

$P = 5.72 kg m/s$

Part b)

$\Delta P = 2.93 kg m/s$

Part c)

$F = 44.4 N$

Part d)

$\Delta P = 5.02 kg m/s$

Part e)

$\Delta t = 0.113 s$

Part f)

$\Delta K = 0$

Explanation:

As we know that initial velocity of the ball is given as

$v = 11.8 cos29 \hat i + 11.8 sin29 \hat j$

$v_i = 10.3 \hat i + 5.72 \hat j$

Now final velocity of the system is given as

$v_f = 10.3\hat i - 5.72\hat j$

Part a)

now magnitude of initial momentum is given as

$P = mv$

$P = 0.256(11.8)$

$P = 5.72 kg m/s$

Part b)

Change in momentum is given as

$\Delta P = m(v_f - v_i)$

$\Delta P = 0.256(5.72 + 5.72)$

$\Delta P = 2.93 kg m/s$

Part c)

As we know that average force is defined as the rate of change in momentum

so here we have

$F = \frac{\Delta P}{\Delta t}$

$F = \frac{2.93}{0.066}$

$F = 44.4 N$

Part d)

Magnitude of change in momentum is given as

$\Delta P = m(v_f - v_i)$

$\Delta P = 0.256(7.8 + 11.8)$

$\Delta P = 5.02 kg m/s$

Part e)

As we know that in 2nd case the force is same as the initial force

so we will have

$\frac{\Delta P}{\Delta t} = F$

$\frac{5.02}{\Delta t} = 44.4$

$\Delta t = 0.113 s$

Part f)

Since this is elastic collision so change in kinetic energy must be ZERO

$\Delta K = 0$

8 0

3 years ago