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4vir4ik [10]
2 years ago
7

Explain the uses of Electrophorus ​

Physics
1 answer:
Maslowich2 years ago
8 0

Answer:

Explanation:

Comment

Interesting creature.

These creatures are the only survivors from an era long past, where nothing else has survived. They are eels capable of discharging a large electrical current. A pack of them (they can hunt in groups) can kill just about anything.

Tribes use and tame them to guard the tribe's tame Electrophorus. They are bread and maintained in schools. While the charge is huge, once discharged, these creatures take a very long time to recharge, so they are not entirely invincible.

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Which statement correctly identifies a difference between the function of most arteries and veins?
olasank [31]

Answer:a

Explanation:

8 0
3 years ago
Identify the names, charges, and location of three types of subatomic particles that make up an atom.
Eva8 [605]

Answer:Proton (charge of +e, in the nucleus), Neutron (0 charge, in the nucleus), and Electron (charge of –e, outside the nucleus).Nov 13, 2015

Explanation: YA

3 0
2 years ago
Read 2 more answers
Mirrors produce images by doing which of the following to light?
Gemiola [76]

Answer: C

Reflection

Explanation: Light travels in a straight line. Reflection is one of the properties of light. And this is the property in which mirror make use of. The ability of light to bounce back. It's this bouncing back characteristics of light ray that eventually produce the image of an object by the mirror.

If the light ray is absorbed, no image will be produced.

4 0
3 years ago
A cannon, positioned on a hill, shoots a cannonball horizontally at 23 m/s. The cannonball hits the stone wall 1.96 m below the
irina [24]

Answer: 14. 49 m

Explanation:

We can solve this problem with the following equations:

x=V_{o} cos \theta t (1)

y-y_{o}=V_{o} sin \theta t-\frac{1}{2}gt^{2} (2)

Where:

x is the horizontal distance between the cannon and the ball

V_{o}=23 m/s is the cannonball initial velocity

\theta=0\° since the cannonball was shoot horizontally

t is the time

y=0 is the final height of the cannonball

y_{o}=1.96 m is the initial height of the cannonball

g=9.8 m/s^{2} is the acceleration due gravity

Isolating t from (2):

t=\sqrt{-\frac{2(y-y_{o})}{g}} (3)

t=\sqrt{-\frac{2(0 m-1.96 m)}{9.8 m/s^{2}}} (4)

t=0.63 s (5)

Substituting (5) in (1):

x=(23 m/s) cos(0\°) 0.63 s (6)

Finally:

x=14.49 m

5 0
3 years ago
A 1200 kg car reaches the top of a 100 m high hill at A with a speed vA. What is the value of vA that will allow the car to coas
Pavel [41]

Complete question is:

A 1200 kg car reaches the top of a 100 m high hill at A with a speed vA. What is the value of vA that will allow the car to coast in neutral so as to just reach the top of the 150 m high hill at B with vB = 0 m/s. Neglect friction.

Answer:

(V_A) = 31.32 m/s

Explanation:

We are given;

car's mass, m = 1200 kg

h_A = 100 m

h_B = 150 m

v_B = 0 m/s

From law of conservation of energy,

the distance from point A to B is;

h = 150m - 100 m = 50 m

From Newton's equations of motion;

v² = u² + 2gh

Thus;

(V_B)² = (V_A)² + (-2gh)

(negative next to g because it's going against gravity)

Thus;

(V_B)² = (V_A)² - (2gh)

Plugging in the relevant values;

0² = (V_A)² - 2(9.81 × 50)

(V_A) = √981

(V_A) = 31.32 m/s

3 0
3 years ago
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