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goldfiish [28.3K]
2 years ago
11

Why is cathode positive in leclanche cell??

Physics
1 answer:
laila [671]2 years ago
7 0

Answer:

Because the zinc is reluctant

Explanation:

A leclanche cell contains a conducting solution (electrolyte) of ammonium chloride, a cathode (positive terminal) of carbon, a depolarizer of manganese dioxide (oxidizer), and an anode (negative terminal) of zinc (reductant).

As the Zn2+ ions move away from the anode, leaving their electrons on its surface,

Zn → Zn2+ + 2e−

the anode becomes more negatively charged than the cathode. When the cell is connected to an external electrical circuit, the excess electrons on the zinc anode flow through the circuit to the carbon rod, the movement of electrons forming an electric current.

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5. Stopping a fast-moving object is harder than stopping a slow-moving<br> one.<br> True<br> False
Strike441 [17]
True because well it’s moving fast lol sometimes ur eyes have a hard time following its speed
6 0
2 years ago
Read 2 more answers
Firemen are shooting a stream of water at a burning building. A high-pressure hose shoots out the water with a speed of 26.0 m/s
alekssr [168]

Answer:

a) θ = 58.3º

b) vfh = 13.7 m/s

c) g = -9.8 m/s2

d) h = 22.2 m

e) vfb = 15.5 m/s

Explanation:

a)

  • Assuming that gravity is the only influence that causes an acceleration to the water, due to it is always downward, since both directions are independent each other, in the horizontal direction, the water moves at a constant speed.
  • Since the velocity vector has a magnitude of 26.0 m/s, we can find its horizontal component as follows:
  • vₓ₀ = v * cos θ (1)
  • where θ is the angle between the water and the horizontal axis (which we define as the x-axis, being positive to the right).
  • Applying the definition of average velocity, taking the end of the hose like the origin, and making t₀ = 0, we can write the following expression:

        x_{f} = v_{ox} * t = v_{o} * cos \theta * t  (2)

  • Replacing by the givens of xf = 41.0m, t = 3.00 s, and v=26.0 m/s, we can solve for the angle of elevation θ, as follows:

        cos \theta = \frac{x_{f} }{v*t} = \frac{41.0m}{26.0m/s*3.00s} = 0.526 (3)

  • ⇒θ = cos⁻¹ (0.526) = 58.3º (4)

b)

  • At the highest point in its trajectory, just before starting to fall, the vertical component of the velocity is just zero.
  • Since the horizontal component keeps constant during all the journey, we can conclude that the speed at this point is just v₀ₓ, that we can find easily from (1) replacing by the values of v and cos θ, as follows:
  • vₓ₀ = v * cos θ = 26.0 m/s * 0.526 = 13.7 m/s. (5)

c)

  • At any point in the trajectory, the only acceleration present is due to the action of gravity, which accepted value is -9.8 m/s2 (taking the upward direction on the vertical y-axis as positive)

d)

  • Since we know the time when the water strikes the building, it will be the same for the vertical movement, so, we can use the kinematic equation for vertical displacement, as follows:

       \Delta y = v_{oy} * t - \frac{1}{2} *g*t^{2} (6)

  • Our only unknown remains v₀y, which can be obtained in the same way than the horizontal component:
  • v₀y = v * sin θ = 26.0 m/s * 0.85 = 22.1 m/s (7)
  • Replacing (7) in (6), we get:

       \Delta y = 22.1 m/s* 3.0s - \frac{1}{2} *9.8m/s2*(3.00s)^{2} = 22.2 m (8)

e)

  • When the water hits the building the velocity vector, has two components, the horizontal vₓ and the vertical vy.
  • The horizontal component, since it keeps constant, is just v₀x:
  • v₀ₓ = 13.7 m/s
  • The vertical component can be found applying the definition of acceleration (g in this case), solving for the final velocity, as follows:

       v_{fy} = v_{oy} - g*t  (9)

  • Replacing by the time t (a given), g, and  v₀y from (7), we can solve (9) as follows:

       v_{fy} = 22.1 m/s - 9.8m/s2*3.00s = -7.3 m/s  (10)

  • Since we know the values of both components (perpendicular each other), we can find the magnitude of the velocity vector (the speed, i.e. how fast is it moving), applying the Pythagorean Theorem to v₀ₓ and v₀y, as follows:

       v_{f} = \sqrt{(13.7m/s)^{2} +(-7.3m/s)^{2}} = 15.5 m/s (11)

3 0
2 years ago
Table C: Average Speeds for Lower Racetrack
Blababa [14]

Answer:

Centripetal Acceleration = v^2/r

= (circumference/time)^2/r

= (2*pi*r/t)²)/r

= ((2³.14*50/14.3)²)/50

= 9.64 m/s²

brainlist?

Explanation:

5 0
3 years ago
A roller coaster starts from rest at its highest point and then descends on its (frictionless) track. Its speed is 36 m/s when i
Ganezh [65]

Answer:

its speed when its height was half that of its starting point is 25.46 m/s

Explanation:

Given;

final speed of the roller coaster, v = 36 m/s

Applying general equation of motion;

V² = U² + 2gh

where;

V is the final speed of the roller coaster

U is the initial speed of the roller coaster = 0

h is the height attained at a given velocity

36² = 0 + (2 x 9.8)h

1296 = 19.6 h

h = 1296/19.6

h = 66.1224 m

when its height was half that of its starting point, h₂ = ¹/₂ h

h₂ = ¹/₂(66.1224 m) = 33.061 m

At h = 33.061 m, V = ?

V² = U² + 2gh

V² = 0 + 2 x 9.8 x 33.061

V² = 648

V = √648

V = 25.46 m/s

Therefore, its speed when its height was half that of its starting point is 25.46 m/s

8 0
3 years ago
"The wave" is a particular type of pulse that can
scoray [572]

Answer:

t = 36π seconds

Explanation:

For resolving this problem, we are going to consider a representative stadium of the United States. The Mercedes-Benz Stadium located in Atlanta, Georgia has an average radius of 90 m.

Then, its circumference measures:

L = 2πr

L = 2π(90)

L = 180π m

First, we estimate the wave's velocity: the average width of an person is 0.5 m, then the velocity is:

v = x/t

                 Where x: person's width

                             t: time

v = 0.5/0.1 = 5 m/s

The time required for the pulse to make one circuit around the stadium is:

t = x/v = 180π/5 = 36π seconds

4 0
3 years ago
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