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3 years ago

A sound wave is an example of a

2 answers:

Studentka2010 [4]3 years ago

4 0

Longitudinal waves....
sounds propagates through moving waves of compression and ratification of a medium which allows longitudinal propagation

kap26 [50]3 years ago

4 0

I just did this test the answer is

b. longitudinal wave.

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An experimental apparatus has two parallel horizontal metal rails separated by 1.0 m. A 3.0 Ω resistor is connected from the lef

Blizzard [7]

Answer:

The induced current and the power dissipated through the resistor are 0.5 mA and $7.5\times10^{-7}\ Watt$ .

Explanation:

Given that,

Distance = 1.0 m

Resistance = 3.0 Ω

Speed = 35 m/s

Angle = 53°

Magnetic field $B=5.0\times10^{-5}\ T$

(a). We need to calculate the induced emf

Using formula of emf

$E = Blv\sin\theta$

Where, B = magnetic field

l = length

v = velocity

Put the value into the formula

$E=5.0\times10^{-5}\times1.0\times35\sin53^{\circ}$

$E=1.398\times10^{-3}\ V$

We need to calculate the induced current

$E =IR$

$I=\dfrac{E}{R}$

Put the value into the formula

$I=\dfrac{1.398\times10^{-3}}{3.0}$

$I=0.5\ mA$

(b). We need to calculate the power dissipated through the resistor

Using formula of power

$P=I^2 R$

Put the value into the formula

$P=(0.5\times10^{-3})^2\times3.0$

$P=7.5\times10^{-7}\ Watt$

Hence, The induced current and the power dissipated through the resistor are 0.5 mA and $7.5\times10^{-7}\ Watt$ .

6 0

3 years ago

Read 2 more answers

According to the law of conservation of matter,the number of ______ is not changed by a chemical reaction

abruzzese [7]

The answer is mass. I have to comment more than 20 characters.

6 0

3 years ago

Read 2 more answers

A particle has a charge of -4.25 nC.

SpyIntel [72]

Answer:

-611.32 N/C

0.43723 m

Explanation:

k = Coulomb constant = $8.99\times 10^{9}\ Nm^2/C^2$

q = Charge = -4.25 nC

r = Distance from particle = 0.25 m

Electric field is given by

$E=\dfrac{kq}{r^2}\\\Rightarrow E=\dfrac{8.99\times 10^9\times -4.25\times 10^{-9}}{0.25^2}\\\Rightarrow E=-611.32\ N/C$

The magnitude is 611.32 N/C

The electric field will point straight down as the sign is negative towards the particle.

$E=\dfrac{kq}{r^2}\\\Rightarrow r=\sqrt{\dfrac{kq}{E}}\\\Rightarrow r=\sqrt{\dfrac{8.99\times 10^9\times 4.25\times 10^{-9}}{13}}\\\Rightarrow r=1.71436\ m$

The distance from the electric field is 1.71436 m

4 0

3 years ago

Read 2 more answers

Suppose that two objects attract each other with a gravitational force of 50N. If the mass

vagabundo [1.1K]

Answer:

112.5 N

Explanation:

50 = GMm/r^2

Let F be the new force of attraction

F/50 = ( G(3M)(3m)/(2r)^2 ) / (GMm/r^2)

[Elimiating G,M,m,r]

F = 112.5 N

7 0

3 years ago

What is the formula for calculating the efficiency of a heat engine? Efficiency = StartFraction T Subscript h Baseline minus T S

Anna007 [38]

Efficiency of heat engine is determined by the ratio of difference in temperature of cold from hot reservoir to the temperature of hot reservoir over temperature of hot reservoir.

Answer: Option A

<u>Explanation:</u>

Efficiency is defined as the output from the input. So it is the ratio of energy output to the energy input. In case of temperature, it is the change in temperature from hot reservoir to cold reservoir with the overall hot reservoir temperature.

$Efficiency =\frac{T_{h} -T_{c} }{T_{h} }$

Thus, option A is the most suitable as the heat will be transferred from high temperature to low temperature. So the hot reservoir will be releasing the energy. So the conversion of hot reservoir temperature to cool reservoir temperature is defined as the efficiency. Thus, option A is the most suitable.

9 0

3 years ago