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s344n2d4d5 [400]

3 years ago

14

The ball has 7.35 joules of potential energy at position B. At position A, all of the energy changes to kinetic energy. The velo

city of the ball at position A is meters/second. Assume there’s no air resistance. Use g = 9.8 m/s2 , PE = m × g × h, and . mass is 1.5 kilograms and g= 9.8 and the hight of position b is 0.5 meters.

1 answer:

Lina20 [59]3 years ago

3 0

I assume that the ball is stationary (v=0) at point B, so its total energy is just potential energy, and it is equal to 7.35 J.
At point A, all this energy has converted into kinetic energy, which is:
$K= \frac{1}{2}mv^2$
And since K=7.35 J, we can find the velocity, v:
$v= \sqrt{ \frac{2K}{m} }= \sqrt{ \frac{2 \cdot 7.35 J}{1.5 kg} }=3.1 m/s$

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3 years ago

Two loudspeakers, 5.5 m apart and facing each other, play identical sounds of the same frequency. You stand halfway between them

zhuklara [117]

Answer:

a) f = 809.5 Hz , b) f = 1619 Hz

Explanation:

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λ = 0.21 2 π/π

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Now let's use the ratio of the speed of sound and the wavelength and the frequency

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3 0

3 years ago

goldfiish [28.3K]

Answer:

Electric force = $3.15\times 10^{-11}\textrm{ N}$ .

Explanation:

Given:

Charge on one grain, $q_{1}=6\times 10^{-10}\textrm{ C}$

Charge on second grain, $q_{2}=2.3\times 10^{-15}\textrm{ C}$

Distance between the two grains is, $r=2\textrm{cm}=0.02\textrm{ m}$

Electric force acting between two charges $q_{1}\textrm{ and }q_{2}$ separated by a distance $r$ is:

$F_{e}=\frac{kq_{1}q_{2}}{r^2}$

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Now, plug in the given values and solve for $F{e}$ .

$F{e}=\frac{9\times 10^9\times 6.0\times 10^{-10}\times 2.3\times 10^{-15}}{(0.02)^2}\\\\F_{elec}=3.15\times 10^{-11}\textrm{ N}$

Therefore, the electric force between them is $3.15\times 10^{-11}\textrm{ N}$ .

5 0

3 years ago

In 1977, Kitty O'Neil drove a hydrogen peroxide-powered rocket dragster for a record time interval (3.22 s) and final speed (663

Tom [10]

Answer:

a) 0.05719 km/s^2

b) 0.00921 km/s^2

Explanation:

average acceleration can be calcuated as follows

$a=\frac{\Delta v}{\Delta t} = \frac{v_f-v_i}{t_f-t_i}$

a)

- In the initial time ( $t_i=0$ ), just before of race start the rocket is at rest. So $v_i=0$ .

- After 3.22 s the rocket reach 663 km/h, So $t_f=3.22 s$ and the $v_f=663 km/h$ . We convert units from km/h to km/s to have the same time units (seconds) the velocity would be:

$663 \frac{km}{h}*\frac{1 h}{3600 s} = 0.18467 km/s$

(take into accoun that 1 h=3600 s)

and acceleration is:

$a= \frac{0.184167 km/s-0 km/s}{3.22s-0s}= 0.05719 km/s^2$

b)

acceleration while stopping is calculate with the same formula but now

-we define the initial time ( $t_i=0$ ) the instant just before begining the stopping. At this instant the velocity is 663 km/h which is the same as 0.18467 km/s.

-After 20 seconds the rocket stops. So tex]t_f=20 s[/tex] and $v_f=0$

So, acceleration is:

$a= \frac{0km/s-0.184167 km/s}{20s-0s}= 0.00921 km/s^2$

3 0

3 years ago