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Alexxx [7]
3 years ago
13

9. True or False: Sawmills and gristmills are machines that do not need

Physics
1 answer:
gizmo_the_mogwai [7]3 years ago
4 0

Answer:

true can i get brainliest :)

Explanation:

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Why is cathode positive in leclanche cell??
laila [671]

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.

7 0
3 years ago
A 2.1W iPod is used for 30 minutes. How much energy does it use? (3
shutvik [7]
It us 200 energy if it the right awnser


3 0
3 years ago
Read 2 more answers
How are the magnetic domains of a magnet different from the domains of an ordinary piece of metal?
My name is Ann [436]
Answer:

In a magnet, the domains all point in the same direction; in an ordinary piece of metal, they're all jumbled up.

Explanation:

In a magnet, the domains all point toward the north pole; in an ordinary piece of metal, they all point to the south pole.



Side note:
Hope this helps!
Please give Brainliest!
6 0
2 years ago
How much force is needed to stop a 90-kg soccer player if he decelerates at 15 m/s^2?
svet-max [94.6K]
We have: F = m×a
Here, m = 90 Kg
a = 15 m/s²

Substitute their values into the expression:
F = 90 × 15
F = 1350 N

In short, Your Answer would be Option D

Hope this helps!
8 0
4 years ago
A hunter aims at a deer which is 40 yards away. Her cross- bow is at a height of 5ft, and she aims for a spot on the deer 4ft ab
shutvik [7]

Answer:

a)  θ₁ = 0.487º , b)   t = 0.400 s ,        x = 11.73 ft

Explanation:

For this exercise let's use the projectile launch relationships.

The initial height is I = 5 ft and the final height y = 4 ft

            y = y₀ + v_{oy} t - ½ g t²

The distance to the band is x = 40 yard (3 ft / 1 yard) = 120 ft

            x = v₀ₓ t

            t = x / v₀ₓ

We replace

             y –y₀ = v_{oy} x / v₀ₓ - ½ g x² / v₀ₓ²

             v_{oy} = v₀ sin θ

             v₀ₓ = vo cos θ

             

             y –y₀ = x tan θ - ½ g x² / v₀² cos² θ

                5-4 = 120 tan θ - ½ 32 120 / (300 2 cos2 θ)

                1 = 120 tan θ - 0.0213 sec² θ

Let's use the trigonometry relationship

               Sec² θ = 1 - tan² θ

                 1 = 120 tan θ - 0.0213 (1 –tan²θ)

                 0.0213 tan²θ + 120 tanθ -1.0213 = 0

                 

We change variables

          u = tan θ

          u² + 5633.8 u - 48.03 = 0

We solve the second degree equation

          u = [-5633.8 ±√(5633.8 2 + 4 48.03)] / 2

          u = [- 5633.8 ± 5633.82] / 2

           u₁ = 0.0085

           u₂= -5633.81

           u = tan θ

           θ = tan⁻¹ u

For u₁

           θ₁ = tan⁻¹ 0.0085

           θ₁ = 0.487º

For u₂

           θ₂ = -89.99º

The launch angle must be 0.487º

b) let's look for the time it takes for the arrow to arrive

         x = v₀ₓ t

         t = x / v₀ cos θ

         

         t = 120 / (300 cos 0.487)

         t = 0.400 s

The deer must be at a distance of

           v = 20 mph (5280 ft / 1 mi) (1 h / 3600s) = 29.33 ft / s

           x = v t

           x = 29.33 0.4

           x = 11.73 ft

3 0
3 years ago
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