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3 years ago

How would you best define moving charges

Physics

1 answer:

svet-max [94.6K]3 years ago

8 0

Current.A moving charge is an object that changes position to one particular obsever.

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A ball is shot straight up from the surface of the

koban [17]

A ) The displacement:
d = v o t - (gt²) / 2 =
= 19.6 m/s × 1 s - ( 9.8 m/s² x 1 s² ) / 2 =
= 19.6 m - 4.9 m = 14.7 m
b ) v = v o - g t
0 = 19.6 m/s - 9.8 t ( when the ball is at the highest point )
9.8 t = 19.6
t = 19.6 : 9.8
t = 2 s
h = v o t - (gt²)/2 = 19.6 x 2 - ( 9.8 x 4 ) / 2 = 39.2 - 19.6
h = 19.6 m
c ) h = gt² / 2
19.6 = 9.8 t²/2
9.8 t² = 39.2
t² = 39.2 : 9.8
t² = 4
t 2 = 2 s ( and we know that t 1 = 2 s )
t = t 1 + t 2 = 2 s + 2 s = 4 s

3 0

3 years ago

What is the name of Al(ClO2)3?

Masja [62]

It would be aluminum chlorite.

4 0

3 years ago

Calculating If the average or normal temperature decrease with altitude in the troposphere is 6.5°C/km, calculate the approximat

grigory [225]

Answer:

- The temperature of 10°C will be experienced at an altitude of 2.52 km

- The temperature of 0°C will be experienced at an altitude of 4.15 km

Explanation:

Lapse Rate = -6.5°C/km of ascent.

Lapse Rate = Temperature difference/altitude difference

For the 10°C,

Temperature difference = 10 - 27 = -17°C

-6.5 = -17/(difference in altitude between the two points)

Difference in altitude = 17/6.5 = 2.52 km

For 0°C,

Temperature difference = 0 - 27 = - 27°C

-6.5 = -27/(difference in altitude between the two points)

Altitude difference = 27/6.5 = 4.15 km

5 0

3 years ago

A baseball player hits a homerun, and the ball lands in the left field seats, which is 103m away from the point at which the bal

Sati [7]

(a) The ball has a final velocity vector

$\mathbf v_f=v_{x,f}\,\mathbf i+v_{y,f}\,\mathbf j$

with horizontal and vertical components, respectively,

$v_{x,f}=\left(20.5\dfrac{\rm m}{\rm s}\right)\cos(-38^\circ)\approx16.2\dfrac{\rm m}{\rm s}$

$v_{y,f}=\left(20.5\dfrac{\rm m}{\rm s}\right)\sin(-38^\circ)\approx-12.6\dfrac{\rm m}{\rm s}$

The horizontal component of the ball's velocity is constant throughout its trajectory, so $v_{x,i}=v_{x,f}$ , and the horizontal distance x that it covers after time t is

$x=v_{x,i}t=v_{x,f}t$

It lands 103 m away from where it's hit, so we can determine the time it it spends in the air:

$103\,\mathrm m=\left(16.2\dfrac{\rm m}{\rm s}\right)t\implies t\approx6.38\,\mathrm s$

The vertical component of the ball's velocity at time t is

$v_{y,f}=v_{y,i}-gt$

where g = 9.80 m/s² is the magnitude of the acceleration due to gravity. Solve for the vertical component of the initial velocity:

$-12.6\dfrac{\rm m}{\rm s}=v_{y,i}-\left(9.80\dfrac{\rm m}{\mathrm s^2}\right)(6.38\,\mathrm s)\implies v_{y,i}\approx49.9\dfrac{\rm m}{\rm s}$

So, the initial velocity vector is

$\mathbf v_i=v_{x,i}\,\mathbf i+v_{y,i}\,\mathbf j=\left(16.2\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\left(49.9\dfrac{\rm m}{\rm s}\right)\,\mathbf j$

which carries an initial speed of

$\|\mathbf v_i\|=\sqrt{{v_{x,i}}^2+{v_{y,i}}^2}\approx\boxed{52.4\dfrac{\rm m}{\rm s}}$

and direction θ such that

$\tan\theta=\dfrac{v_{y,i}}{v_{x,i}}\implies\theta\approx\boxed{72.0^\circ}$

(b) I assume you're supposed to find the height of the ball when it lands in the seats. The ball's height y at time t is

$y=v_{y,i}t-\dfrac12gt^2$

so that when it lands in the seats at t ≈ 6.38 s, it has a height of

$y=\left(49.9\dfrac{\rm m}{\rm s}\right)(6.38\,\mathrm s)-\dfrac12\left(9.80\dfrac{\rm m}{\mathrm s^2}\right)(6.38\,\mathrm s)^2\approx\boxed{119\,\mathrm m}$

6 0

4 years ago

Is bleach heterogeneous

irga5000 [103]

No, bleach is not heterogeneous. Bleach is actually homogeneous.
Heterogeneous is if it the same throughout.
Homogeneous is if it is different (has chunks and clumps of different materials)

6 0

4 years ago