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Brums [2.3K]
3 years ago
10

How does molten rock and gas leave the volcano

Physics
1 answer:
9966 [12]3 years ago
4 0
The vent the opening through which molten rock and gas leave a volcano
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3 years ago
Brainliest for first answer that’s not a link or a troll
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5 0
3 years ago
Absolute differences between each observed mass and the accepted value,divided by the accepted value, multiplied by 100%
timama [110]

Answer:

The percentage error    e% = Δm/m 100    

Explanation:

The absolute error is the assessment with which measurements are made, it is given by the minimum reading value of a given instrument.

The accepted value is the value measured in an instrument. Or the one calculated by an experiment.

The percentage error is the ratio of the absolute error between the value of the average per 100

      e% = Δm/m 100

3 0
3 years ago
A small lead ball, attached to a 1.10-m rope, is being whirled in a circle that lies in the vertical plane. The ball is whirled
hjlf

Answer:

1.84 m

Explanation:

For the small lead ball to be balanced at the tip of the vertical circle just before it is released, the reaction force , N equal the weight of the lead ball W + the centripetal force, F. This normal reaction ,N also equals the tension T in the string.

So, T = mg + mrω² = ma where m = mass of small lead ball, g = acceleration due to gravity = 9.8 m/s², r = length of rope = 1.10 m and ω = angular speed of lead ball = 3 rev/s = 3 × 2π rad/s = 6π rad/s = 18.85 rad/s and a = acceleration of normal force. So,

a = g + rω²

= 9.8 m/s² + 1.10 m × (18.85 rad/s)²

= 9.8 m/s² + 390.85 m/s²

= 400.65 m/s²

Now, using v² = u² + 2a(h₂ - h₁)  where u = initial velocity of ball = rω = 1.10 m × 18.85 rad/s = 20.74 m/s, v = final velocity of ball at maximum height = 0 m/s (since the ball is stationary at maximum height), a = acceleration of small lead ball = -400.65 m/s² (negative since it is in the downward direction of the tension), h₁ = initial position of lead ball above the ground = 1.3 m and h₂ = final position of lead ball above the ground = unknown.

v² = u² + 2a(h₂ - h₁)

So, v² - u² = 2a(h₂ - h₁)

h₂ - h₁ =  (v² - u²)/2a

h₂ =  h₁ + (v² - u²)/2a

substituting the values of the variables into the equation, we have

h₂ =  1.3 m + ((0 m/s)² - (20.74 m/s)²)/2(-400.65 m/s²)

h₂ =  1.3 m + [-430.15 (m/s)²]/-801.3 m/s²

h₂ =  1.3 m + 0.54 m

h₂ =  1.84 m

7 0
2 years ago
As the ball rolls down the ramp, there are changes in kinetic and potential energy. How much kinetic energy does the ball have w
Trava [24]

Answer: C. 50j

Explanation:

5 0
3 years ago
Read 2 more answers
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