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uysha [10]
3 years ago
7

a cart has 30J or potential energy at the top of a 20m high hill. Assuming no energy is transformed to another form of energy, h

ow much kinetic energy does the cart have when it is halfway down the hill?
Physics
1 answer:
topjm [15]3 years ago
3 0

Answer:

i would say 20J

Explanation:

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It is dangerous to stand near the railway leak when train passing by,why?​
4vir4ik [10]

Answer:

Because a person may be pulled in the direction of the moving train. Thereby causing accident

Explanation:

According to Daniel Bernoulli's theorem, he was widely known as a Mathematician. He stated that due to the higher velocity of a moving train, there is higher kinetic energy in terms of volume around it, while the air pressure between the person and the train becomes lower.

As a result, a person near a moving train may be pulled in the direction of the moving train. Thereby causing accidents that may lead to death.

3 0
3 years ago
We estimate the mass of the galaxy by observations of its ________ curve.
seropon [69]
The appropriate response is the rotation. There are most likely no less than 100 billion planets in the Milky Way. The Solar System is situated inside the circle, around 26,000 light-years from the Galactic Center, on the inward edge of one of the winding molded centralizations of gas and tidies called the Orion Arm.
8 0
3 years ago
The pedals on a bicycle give a mechanical advantage by allowing you to turn the pedals a __________ distance to turn the _______
denpristay [2]
Short distance to turn the circumference of the wheels. Mechanical advantage is load/effort I think
6 0
3 years ago
Read 2 more answers
Parallel Circuits:
Ber [7]

Answer:

D.)

Explanation:

the current separates on each branch according to the resistance it experience.

8 0
3 years ago
Two particles are fixed to an x axis: particle 1 of charge −8.00 ✕ 10⁻⁷ C at x = 6.00 cm, and particle 2 of charge +8.00 ✕ 10⁻⁷
victus00 [196]

Answer:

0 N/C

Explanation:

Parameters given:

q_1 = -8.00 * 10^{-7} C

x_1 = 6.00 cm

q_2 = +8.00 * 10^{-7} C

x_2 = 21 cm

The distance between q_1 and q_2 is

21 - 6 = 15cm

Electric field is given as

E = \frac{kq}{r^2}

r = 15/2 = 7.5cm = 0.075m

The electric field at their midpoint due to q_1 is:

E = \frac{9 * 10^9 * -8.0 * 10^{-7}}{0.075^2}

E_1 = -1.28 * 10^6 N/C

The electric field at the midpoint due to q_2 is:

E = \frac{9 * 10^9 * 8.0 * 10^{-7}}{0.075^2}

E_2 = 1.28 * 10^6 N/C

The net electric field will be:

E = E_1 + E_2

E = -1.28 * 10^6 + 1.28 * 10^6

E = 0 N/C

7 0
4 years ago
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