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4 years ago

What provides the path for a circuit

2 answers:

5 0

Answer:The exit point is called the "return" because electrons always end up at the source when they complete the path of an electrical circuit.

Explanation:

ra1l [238]4 years ago

3 0

The exit point is called the "return" because electrons always end up at the source when they complete the path of an electrical circuit. The part of an electrical circuit that is between the electrons' starting point and the point where they return to the source is called an electrical circuit's "load".

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The human nervous system can propagate nerve impulses at about 102 m>s. Estimate the time it takes for a nerve impulse to tra

Olin [163]

Answer:

t = 0.196 s

Explanation:

The speed of a pulse is determined by the characteristics of the medium, its density and its resistance to stress, as long as these remain the speed will be constant for which we can use the kinetic expressions of the uniform movement

v = x / t

t = x / v

calculate

t = 2/102

t = 0.196 s

4 0

3 years ago

A boy throws a ball vertically up it returns the ground after 10 seconds find the maximum height reached by the ball

Akimi4 [234]

Answer:

Approximately $122.625\; {\rm m}$ (assuming that $g = 9.81\; {\rm m\cdot s^{-2}}$ , the ball was launched from ground level, and that the drag on the ball is negligible.)

Explanation:

Let $v_{0}$ denote the velocity at which the ball was thrown upward.

If the drag (air friction) on the ball is negligible, the ball would land with a velocity of exactly $(-v_{0})$ . The velocity of the ball would be changed from $v$ to $(-v_{0})\!$ (such that $\Delta v = (-v_{0}) - v_{0} = (-2\, v_{0})$ ) within $t = 10\; {\rm s}$ .

Also because the drag on the ball is negligible, the acceleration of the ball would be $a = -g = -9.81\; {\rm m\cdot s^{-2}}$ . Thus:

$\Delta v = a\, t = 10\; {\rm s} \times (-9.81\; {\rm m\cdot s^{-2}}) = -98.1\; {\rm m\cdot s^{-1}}$ .

Since $\Delta v = (-2\, v_{0})$ :

$-2\, v_{0} = \Delta v = -98.1\; {\rm m\cdot s^{-1}$ .

$\begin{aligned}v_{0} &= \frac{-98.1\; {\rm m\cdot s^{-1}}}{-2}= 49.05\; {\rm m \cdot s^{-1}}\end{aligned}$ .

The ball reaches maximum height when its velocity is $v_{1} = 0\; {\rm m\cdot s^{-1}}$ . Apply the SUVAT equation $x = ({v_{1}}^{2} - {v_{0}}^{2}) / (2\, a)$ to find the displacement $x$ between the original position (ground level, where $v_{0} = 49.05\; {\rm m\cdot s^{-1}}$ ) and the max-height position of the ball (where $v_{1} = 0\; {\rm m\cdot s^{-1}}$ .)

$\begin{aligned}x &= \frac{(0\; {\rm m\cdot s^{-1}})^{2} - (49.05\; {\rm m\cdot s^{-1}})^{2}}{2 \times (-9.81\; {\rm m\cdot s^{-2}})} \\ &\approx 122.625\; {\rm m\cdot s^{-1}}\end{aligned}$ .

7 0

2 years ago

A 70 kilogram hockey player skating east on an ice rink is hit by a 0.1 kilogram hockey puck moving toward the west. The puck ex

guajiro [1.7K]

Given:

70 Kilogram hockey

Hit by 0.1 kilogram

50 Newton

Description:

So in this case this bacially means that one object is experiencing a force then another object. So the answer to the question will be 50N.

Answer: 50N (3rd Law)

Hope this helps.

4 0

3 years ago

A motorcycle is following a car that is traveling at constant speed on a straight highway. Initially, the car and the motorcycle

ra1l [238]

Answer:

a) t = 4.16 s

b) x = 141.51 m

Explanation:

Given

v = 21.5 m/s

x0 = 52.0 m

a = 6.0 m/s²

a) Motorcycle

x = v0*t + (a*t²/2)

x = 21.5t + (6*t²/2)

x = 21.5t + 3t² (I)

Car

x = x0 + v0*t

x = 52 + 21.5t (II)

then we can apply I = II

21.5t + 3t² = 52 + 21.5t

⇒ 3t² = 52

⇒ t = 4.16 s

b) We can use I or II, then

x = 52 + 21.5*(4.16)

⇒ x = 141.51 m

8 0

3 years ago

Two buses are driving along parallel freeways that are 5mi apart, one heading east and the other heading west. Assuming that eac

Oksanka [162]

Answer:

101.54m/h

Explanation:

Given that the buses are 5mi apart, and that they are both driving at the same speed of 55m/h, rate of change of distance can be determined using differentiation as;

Let l be the be the distance further away at which they will meet from the current points;

$l=\sqrt{13^2-5^2}=12m\\\\\frac{dl}{dt}=-(55m/h+55m/h})\\\\=-110m/h$ #The speed toward each other.

$\frac{dh}{dt}=0, \ \ \ \ h=constant\\\\h^2+l^2=b^2\\\\2h\frac{dh}{dt}+2l\frac{dl}{dt}=2b\frac{db}{dt}\\\\2\times5\times0+2\times12\times(-110)=2\times13\frac{db}{dt}\\\\\frac{db}{dt}=-101.54m/h$

Hence, the rate at which the distance between the buses is changing when they are 13mi apart is 101.54m/h

4 0

3 years ago