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lora16 [44]
3 years ago
13

Astronomers have discovered several volcanoes on io, a moon of jupiter. one of them, named loki, ejects lava to a maximum height

of 2.00 â 105 m. suppose another volcano on a different moon ejected lava at a height of 1.89 â 105 m where the acceleration of gravity is 1.72 m/s2.
Physics
1 answer:
r-ruslan [8.4K]3 years ago
7 0
The question seems to be incomplete. However, I can think of a possible logical question this problem could have. The equation for the maximum height attained by any object thrown upwards is:

H = v²/2g

I think the question would be determining the gravity in Io assuming that the initial velocity of the lava is the same. Then, the solution is as follows:

Let's use the other volcano to find v.
1.89×10⁵ m = v²/2(1.72 m/s²)
Solving for v,
v = 806.325 m/s

So, we use this to find g in Io.
2×10⁵ m = (806.325)²/2(g)
Solving for g,
<em>g = 1.6254 m/s²</em>
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Answer:

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The next application much simpler but here on Earth. There are many hydro-electric power stations in use all over the world. Water is stored at a high level and released falling 100s of metres to a turbine where it generates electricity.

Hope that helps.

Explanation:

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A car is traveling at 100 km/h when the driver sees an accident 80 m ahead and slams on the brakes. what minimum constant decele
Effectus [21]
Assuming the driver starts slamming the brakes immediately, the car moves by uniformly decelerated motion, so we can use the following relationship
2aS = v_f^2 - v_i^2 (1)
where 
a is the deleceration
S is the distance covered after a time t
v_f is the velocity at time t
v_i=100 km/h = 27.8 m/s is the initial speed of the car

The accident is 80 m ahead of the car, so the minimum deceleration required to avoid the accident is the value of a such that S=80 m and v_f=0 (the car should stop exactly at S=80 m to avoid the accident). Using these data, we can solve  the equation (1) to find a:
a=- \frac{v_i^2}{2 S}= -\frac{(27.8 m/s)^2}{2 \cdot 80 m} =-4.83 m/s^2
And the negative sign means it is a deceleration.

4 0
3 years ago
An electron moving parallel to a uniform electric field increases its speed from 2.0 × 10^7 m/s to 4.0 × 10^7 m/s over a distanc
julia-pushkina [17]

Answer:

E = 2.84 * 10^5 N/C

Explanation:

The speed increased from 2.0 * 10^7 m/s to 4.0 * 10^7 m/s over a 1.2 cm distance.

Let us find the acceleration:

v^2 = u^2 + 2as

(4.0 * 10^7)^2 = (2.0 * 10^7)^2 + 2 * a * 0.012\\\\(4.0 * 10^7)^2 - (2.0 * 10^7)^2 = 0.024a\\\\1.2 * 10^{15}= 0.024a\\\\a = 1.2 * 10^{15} / 0.024\\\\a = 5 * 10^{16} m/s^2

Electric force is given as the product of charge and electric field strength:

F = qE

where q = electric charge

E = Electric field strength

Force is generally given as:

F = ma

where m = mass

a = acceleration

Equating both:

ma = qE

E = ma / q

For an electron:

m = 9.11 × 10^{-31} kg

q = 1.602 × 10^{-19} C

Therefore, the electric field strength of the electron is:

E = \frac{9.11 * 10^{-31} * 5 * 10^{16}}{1.602 * 10^{-19}} \\\\E = 2.84 * 10^5 N/C

8 0
3 years ago
How fast is an object moving if it has 10000J of kinetic energy and a mass of 5kg​
nignag [31]

Answer:

this is for the ppl who delete my answer

suck 8--

Explanation:

4 0
3 years ago
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