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3 years ago

What are all the kinds of kinematic equations

1 answer:

4 0

There are four (4) kinematic equations, which relate to displacement, D, velocity, v, time, t, and acceleration, a. Kinematic Equations Formula Questions

v=v0+at. v = v 0 + at.
d=12(v0+v)t d = 1 2 ( v 0 + v ) t or alternatively vaverage=dt. v average = d t.
d=v0t+(at22)
v2=v20+2ad.

Hope this helps ;)

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At a rock concert, the sound intensity 1.0 m in front of the bank of loudspeakers is 0.10 W/m². A fan is 30 m from the loudspeak

Klio2033 [76]

To solve this problem we will apply the concepts related to the Area, the power and the proportionality relationships between intensity and distance.

The expression for sound power is,

$P = AI$

Here,

A = Area

I = Intensity

P = Power

At the same time the area can be written as,

$A = \frac{\pi d^2}{4}$

Now the intensity is inversely proportional to the square of the distance from the source, then

$I \propto \frac{1}{r^2}$

The expression for the intensity at different distance is

$\frac{I_1}{I_2}= \frac{r^2_2}{r_1^2}$

Here,

$I_1$ = Intensity at distance 1

$I_2$ = Intensity at distance 2

$r_1$ = Distance 1 from light source

$r_2$ = Distance 2 from the light source

If we rearrange the expression to find the intensity at second position we have,

$I_2 = I_1 (\frac{r_1^2}{r_2^2})$

If we replace with our values at this equation we have,

$I_2 = (0.10W/m^2)(\frac{1.0m^2}{30.0m^2})$

$I_2 = 1.11*10^{-4} W/m^2$

Now using the equation to find the area we have that

$A = \frac{\pi (8.4*10^{-3}m)^2}{4}$

$A = 5.5*10^{-5}m^2$

Finally with the intensity and the area we can find the sound power, which is

$P = AI$

$P = (5.5*10^{-5}m^2)(1.11*10^{-4}W/m^2)$

$P = 6.1*10^{-9}J/s$

Power is defined as the quantity of Energy per second, then

$E = 6.1*10^{-9}J$

8 0

3 years ago

Two stars have the same radius but have very different temperatures. the red star has a surface temperature of 3,000 k and the b

dangina [55]

I would say that insofar as the two stars temperatures are presumably closely related to their luminosity, that the blue star at 156,100 k compared to 3000k for the red star then the blue star has a luminosity of 52 times that of the red star.

4 0

3 years ago

An electron moving parallel to a uniform electric field increases its speed from 2.0 × 10^7 m/s to 4.0 × 10^7 m/s over a distanc

julia-pushkina [17]

Answer:

$E = 2.84 * 10^5 N/C$

Explanation:

The speed increased from 2.0 * 10^7 m/s to 4.0 * 10^7 m/s over a 1.2 cm distance.

Let us find the acceleration:

$v^2 = u^2 + 2as$

$(4.0 * 10^7)^2 = (2.0 * 10^7)^2 + 2 * a * 0.012\\\\(4.0 * 10^7)^2 - (2.0 * 10^7)^2 = 0.024a\\\\1.2 * 10^{15}= 0.024a\\\\a = 1.2 * 10^{15} / 0.024\\\\a = 5 * 10^{16} m/s^2$