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olga_2 [115]
3 years ago
6

The first active volcano observed outside the earth was discovered in 1979 on io, one of the moons of jupiter. suppose a volcano

on another moon is observed to be ejecting material to a height of about 1.52 ✕ 105 m. given that the acceleration of gravity on this moon is 1.91 m/s2 find the initial velocity of the ejected material.
Physics
1 answer:
inna [77]3 years ago
4 0

Gravitation due to the planet's mass is always against the ejected particle velocity. This will result in deceleration of the ejected particle and eventually bring the velocity to zero at the topmost point of the trajectory before pulling it back to the ground.

By the kinematic equation

v^2 - u^2 = 2.a.s

where s is the distance travelled = 1.52 x 10^5 m

a is the acceleration due to gravity = -1.91 m/s^2

v and u are initial and final velocities, so v = 0.

Therefore

u = \sqrt{s.a.s}

u = 761.99 m/s

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A wave has a frequency of 1000 Hz. If the wavelength is 0.5 m, how fast is the wave<br> traveling?
Soloha48 [4]

Answer: 1000 Hz · 0,5 m = 500 m/s

Explanation: speed = frequency · wavelength

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3 years ago
A boat of mass 225 kg drifts along a river at a speed of 21 m/s to the west. what impulse is required to decrease the speed of t
zzz [600]
The impulse required to decrease the speed of the boat is equal to the variation of momentum of the boat:
J=\Delta p=m \Delta v
where
m=225 kg is the mass of the boat
\Delta v=v_f-v_i=15 m/s-21 m/s=-6 m/s is the variation of velocity of the boat
By substituting the numbers into the first equation, we find the impulse:
J=m\Delta v=(225 kg)(-6 m/s)=-1350 N s
and the negative sign means the direction of the impulse is against the direction of motion of the boat.
8 0
3 years ago
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mario62 [17]

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3 years ago
A train at a constant 60.0 km/h moves east for 40.0 min, then in a direction 50.0 degrees east of due north for 20.0 min, and th
son4ous [18]

Answer:

Part a)

v = 7.57 km/h

Part b)

\theta = 67.5 degreeNorth of East

Explanation:

Speed of train towards East = 60 km/h

displacement towards East is given as

d_1 = 40 km

now it turns towards 50 degree East of North

so its distance is given as

d_2 = 20 km(sin50 \hat i + cos50\hat j)

d_2 = 15.3 \hat i + 12.8 \hat j

then finally it moves towards west for 50 min

d_3 = -50 \hat i

Now the total displacement of the train is given as

d = d_1 + d_2 + d_3

d = (40 + 15.3 - 50)\hat i + 12.8 \hat j

d = 5.3\hat i + 12.8 \hat j

now total time duration of the motion is given as

T = 40 min + 20 min + 50 min

T = 1.83 h

now average velocity is given as

v_{avg} = \frac{5.3\hat i + 12.8\hat j}{1.83}

v_{avg} = 2.89\hat i + 6.99\hat j

Part a)

magnitude of the average velocity is given as

v = \sqrt{v_x^2 + v_y^2}

v = \sqrt{2.89^2 + 6.99^2}

v = 7.57 km/h

Part b)

Direction of the velocity is given as

tan\theta = \frac{v_y}{v_x}

tan\theta = \frac{6.99}{2.89}

\theta = 67.5 degreeNorth of East

6 0
4 years ago
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