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Leno4ka [110]
2 years ago
9

if a volcano spews a 500-kg rock vertically upward a distance of 500m. what was its velocity when it left the volcano? if the vo

lcano spews the equivalent of 1000 rocks of this size every minute, what is its power output?
Physics
1 answer:
Alenkinab [10]2 years ago
7 0
This can be solve using the formula v = sqrt( 2gy)where v is the velocityg is the acceleration due to gravityy is the vertical distance.
this is the formula of velocity of impact upon free fall. impact velocity is equal to its initial velocity
v = sqrt( 2 ( 9.81 m/s2) ( 500 m))v = 99 m/s
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A circular surface with a radius of 0.057 m is exposed to a uniform external electric field of magnitude 1.44 × 104 N/C. The mag
klio [65]

Answer:

57.94°

Explanation:

we know that the expression of flux

\Phi =E\times S\times COS\Theta

where Ф= flux

           E= electric field

           S= surface area

        θ = angle between the direction of electric field and normal to the surface.

we have Given Ф= 78 \frac{Nm^{2}}{sec}

                          E=1.44\times 10^{4}\frac{Nm}{C}

                          S=\pi \times 0.057^{2}

                         COS\Theta =\frac{\Phi }{S\times E}

 =   \frac{78}{1.44\times 10^{4}\times \pi \times 0.057^{2}}

 =0.5306

 θ=57.94°

4 0
1 year ago
No person who thinks scientifically places any faith in the predictions of astrologers. Nevertheless there are many people who r
Lilit [14]
Therefore, it can be reasonably concluded according to your
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7 0
2 years ago
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An object that weighs 2.450 N is attached to an ideal massless spring and undergoes simple harmonic oscillations with a period o
Viktor [21]

Answer:

Spring constant, k = 24.1 N/m

Explanation:

Given that,

Weight of the object, W = 2.45 N

Time period of oscillation of simple harmonic motion, T = 0.64 s

To find,

Spring constant of the spring.

Solution,

In case of simple harmonic motion, the time period of oscillation is given by :

T=2\pi\sqrt{\dfrac{m}{k}}

m is the mass of object

m=\dfrac{W}{g}

m=\dfrac{2.45}{9.8}

m = 0.25 kg

k=\dfrac{4\pi^2m}{T^2}

k=\dfrac{4\pi^2\times 0.25}{(0.64)^2}

k = 24.09 N/m

or

k = 24.11 N/m

So, the spring constant of the spring is 24.1 N/m.

6 0
1 year ago
Suppose that an asteroid traveling straight toward the center of the earth were to collide with our planet at the equator and bu
MArishka [77]

Complete Question:

Suppose that an asteroid traveling straight toward the center of the earth were to collide with our planet at the equator and bury itself just below the surface. What would have to be the mass of this asteroid, in terms of the earth’s mass M, for the day to become 25.0% longer than it presently is as a result of the collision? Assume that the asteroid is very small compared to the earth and that the earth is uniform throughout.

Answer:

m = 0.001 M

For the whole process check the following page: https://www.slader.com/discussion/question/suppose-that-an-asteroid-traveling-straight-toward-the-center-of-the-earth-were-to-collide-with-our/

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2 years ago
During chemistry class, Carl performed several lab tests on two white solids. The results of three tests are seen in the data ta
Nastasia [14]

Answer: it’s c) ionic

Explanation:

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