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stealth61 [152]

3 years ago

11

Two loudspeakers emit 600 Hz Hz notes. One speaker sits on the ground. The other speaker is in the back of a pickup truck. You h

ear 7.00 beats per second as the truck drives away from you. What is the truck'sspeed?

1 answer:

Firdavs [7]3 years ago

3 0

Answer:

The truck's speed is 4.04 m/s.

Explanation:

Given that,

Emit frequency = 600 Hz

Beat = 7.00 beat/sec

We need to calculate the truck's speed

Using formula of speed

$\text{frequency observed}=\text{frequency emitted}\times\dfrac{v}{v+v_{source}}$

Where, v = speed of sound

Put the value into the formula

$(600-7)=600\times(\dfrac{343}{343-v_{truck}})$

$v_{truck}=\dfrac{600\times343-593\times343}{593}$

$v_{truck}=4.04\ m/s$

Hence, The truck's speed is 4.04 m/s.

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A 12cm candle is placed 6cm from a converging lens with a focal length of 15cm. What is the height of the image of the candle? S

amm1812

Answer:

The height of the image of the candle is 20 cm.

Explanation:

Given that,

Size of the candle, h = 12 cm

Object distance from the candle, u = -6 cm

Focal length of converging lens, f = 15 cm

To find,

The height of the image of the candle.

Solution,

Firstly, we will find the image distance of the candle. Let it is equal to v. Using lens formula to find the image distance.

$\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}$

v is image distance

$\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}\\\\\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}\\\\\dfrac{1}{v}=\dfrac{1}{15}+\dfrac{1}{(-6)}\\\\v=-10\ cm$

If h' is the height of the image. Magnification is given by :

$m=\dfrac{h'}{h}=\dfrac{v}{u}$

$h'=\dfrac{vh}{u}\\\\h'=\dfrac{-10\times 12}{-6}\\\\h'=20\ cm$

So, the height of the image of the candle is 20 cm.

3 0

3 years ago

Which trends is indirectly proportional to effective nuclear charge

Minchanka [31]

Answer:

Atomic size

Explanation:

In the periodic table , atomic size is indirectly proportional to the effective nuclear charge .the atomic size reduces from left to right across the table. This is because electrons are added to the same shell.

4 0

4 years ago

Show that the expression pRT has the same units as the pressure, and thus that the ideal gas law is dimensionally consistent.

Gemiola [76]

Answer and Explanation:

We have given the expression $\rho RT$

Dimension of $\rho =ML^{-3}$

Dimension of R which is gas constant $=ML^2T^{-2}\Theta ^{-1}M^{-1}$

Dimension of temperature T $\Theta ^{-1}$

And dimension of pressure $ML^{-1}T^{-2}$

Now combine dimension of $\rho RT$ $=ML^{-3}ML^2T^{-2}\Theta ^{-1}M^{-1}\Theta ^{-1}=ML^{-1}T^{-2}$

So the dimension of $\rho RT$ and dimension P is same so there unit will also be same

From ideal gas equation we know that $PV=nRT$

$P=\frac{n}{V}RT=\rho RT$

As the both P and $\rho RT$ has same dimension so they are dimensionally constant

7 0

4 years ago

camcorder has a power rating of 14 watts. If the output voltage from its battery is 3 volts, what current does it use?

Aliun [14]

Power rating tells you the amount of electricity a particular device like a camcorder needs. The power, voltage and current are all related using the equation: P = IV.

Therefore,

P = IV
I = P/V = 14 watts/3 volts = 4.67 A

Thus, the current must be 4.67 Amperes for the camcorder.

3 0

4 years ago

A projectile of mass 6.8 kg kg is shot horizontally with an initial speed of 14.5 m/s from a height of 26.7 m above a flat deser

sleet_krkn [62]

Answer:

Explanation:

Given

mass of projectile $m=6.8\ kg$

initial horizontal speed $u_x=14.5\ m/s$

height $h=26.7\ m$

Considering vertical motion

velocity gained by projectile during 26.7 m motion

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

v=final velocity

u=initial velocity

a=acceleration

s=displacement

$v^2-(0)^2=2\times (9.8)\times (26.7)$

$v=\sqrt{523.32}$

$v=22.87\ m/s$

Horizontal velocity will remain same as there is no acceleration

final velocity $v_{net}=\sqrt{(v)^2+(u_x)^2}$

$v_{net}=\sqrt{733.57}=27.08\ m/s$

Initial kinetic Energy $K_i=\frac{1}{2}mu_x^2$

$K_i=\frac{1}{2}\times 6.8\times (14.5)^2=714.85\ J$

Final Kinetic Energy $K_f=\frac{1}{2}mv_{net}^2$

$K_f=\frac{1}{2}\times 6.8\times (27.08)^2$

$K_f=2493.30\ J$

Work done by all the force is equal to change in kinetic Energy of object

Work done by gravity is $W_g$

$W_g=\Delta K$

$W_g=2493.30-714.85=1778.45\ J$

8 0

3 years ago