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damaskus [11]
3 years ago
7

Which will most likely occur before a volcanic eruption?

Physics
1 answer:
liraira [26]3 years ago
6 0
Volcanic eruption most likely to happen when the plates are moving intense in the ground. Maybe an earthquake are most likely for a volcanic eruption to occur. 
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After leaving the end of a ski ramp, a ski jumper lands downhill at a point that is displaced 68.3 m horizontally from the end o
inessss [21]

Answer:

a)52.58 m/s

b)56.13°

Explanation:

assume the upward direction as positive

x-component of the velocity = 29.3×cos33.6°=24.40 m/s (remain constant)

y-component of the velocity which is -29.3sin33.6°= -16.21 m/s

time of flight = 68.3/24.40= 2.7991 seconds

now, we can obtain final velocity in y-direction

v_f_y= v_i_y-gt

v_f_y=-16.21-(-9.8)×2.7991

=43.66 m/s

v_0=\sqrt{29.3^2+43.66^2}

=52.58 m/s

for direction

tan^{-1}\frac{43.66}{29.3}

56.13° from the horizontal

3 0
3 years ago
Question 11 of 11 | Page 11 of 11
KiRa [710]

Answer:

Decreases the time period of revolution

Explanation:

The time period of Cygnus X-1 orbiting a massive star is 5.6 days.

The orbital velocity of a planet is given by the formula,

                                        v = √[GM/(R + h)]

In the case of rotational motion, v = (R +h)ω

                                         ω = √[GM/(R + h)] /(R +h)

Where 'ω' is the angular velocity of the planet

The time period of rotational motion is,

                                     T = 2π/ω

By substitution,

                                      <em>T = 2π(R +h)√[(R + h)/GM] </em>

Hence, from the above equation, if the mass of the star is greater, the gravitational force between them is greater. This would reduce the time period of revolution of the planet.

3 0
3 years ago
Calculate the true mass (in vacuum) of a piece of aluminum whose apparent mass is 4.5000 kgkg when weighed in air. The density o
spin [16.1K]

Answer:

The true weight of the aluminium is m_{alu} = 4.5021 kg

Explanation:

Given data

m_{app} = 4.5 kg

\rho_{air} = 1.29 \frac{kg}{m^{3} }

\rho_{al} = 2.7× 10^{3} \frac{kg}{m^{3} }

The true mass of the aluminium is given by

m_{alu} = \frac{\rho_{alu}m_{app}}{\rho_{alu} -\rho_{air} }

Put all the values in above equation we get

m_{alu} = \frac{(2700)(4.5)}{2700-1.29}

m_{alu} = 4.5021 kg

Therefore the true weight of the aluminium is m_{alu} = 4.5021 kg

6 0
3 years ago
How much work does it take to lift 345 boxes to a height of 6.00 m of each box has a mass of 7.89 kg
Art [367]

Given the value of the mass of each boxes, the work done in lifting the boxes to the given height is 1.6 × 10⁵J.

<h3>Work done</h3>

Work done is simply defined as the energy transfer that takes place when an object is either pushed or pulled over a certain distance by an external force. It is expressed as;

W = F × d

Where F is force applied or Weight and d is distance

Also Force = Weight = mass × acceleration due to gravity.

Since gravity is acting on the boxes as it been lift

W = Weight × height from ground level

W = mg × d

Where m is mass of the boxes, g is accelration due to gravity( g = 9.8m/s² ) and d is distance from ground level.

Given the data in the question;

  • Since each box has a mass of 7.89 kg
  • Mass of the 345 boxes = 345 × 7.89 kg = 2722.05kg
  • Distance or height d = 6.0m
  • Work done W = ?

To determine the work done, we substitute our values into the expression above.

W = mg × d

W = 2722.05kg × 9.8m/s² × 6.0m

W = 160056.5kgm²/s²

W = 160056.5J

W = 1.6 × 10⁵J

Therefore,  Given the value of the mass of each boxes, the work done in lifting the boxes to the given height is 1.6 × 10⁵J.

Learn more about work done here: brainly.com/question/26115962

3 0
3 years ago
A stone is dropped from the upper observation deck of a tower, "500" m above the ground. (Assume g = 9.8 m/s2.) (a) Find the dis
Yuri [45]

Answer:

h₍₁₎ = 495,1 meters

h₍₂₎ = 480,4 m

h₍₃₎ = 455,9 m

...

..

Explanation:

The exercise is "free fall". t = \sqrt{\frac{2h}{g} }

Solving with this formula you find the time it takes for the stone to reach the ground (T) = 102,04 s

The heights (h) according to his time (t) are found according to the formula:

h(t) = 500 - 1/2 * g * t²

Remplacing "t" with the desired time.

4 0
3 years ago
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