PHYSICS COLLEGE

Which volcanoes are formed by pyroclastic deposits? Select all that apply. A. cinder cone B. stratovolcano C. shield volcano

Answers

Answer 1

Answer:

Answer:

Its a cinder cone cause after it all falls down to make deposits.

Answer 2

Answer: cinder cone and stratovolcano

Related Questions

Tonya picks up a leaf from the ground and holds it at arm’s length. She lets go, and the leaf falls to the ground. What force pulled the leaf to the ground?
Some snakes have special sense organs that allow them to see the heat emitted from warm blooded animals what kind of an electromagnetic waves does the sense organs detect?A. Visible light wavesB. Ultraviolet light wavesC. Infrared WavesD. Microwaves
If half the kinetic energy of the initially moving object (m1m1) is transferred to the other object (m2m2), what is the ratio of their masses? Express your answers using three significant figures separated by a comma.
Which statement accurately describes the relationship between force and momentum?A. As the mass of an object increases its momentum increases, and it takes more force to change its motion. B. As the velocity of an object increases, its momentum decreases and it takes less force to change its motion.C. As the mass of an object increases its momentum decreases and it takes less force to change it motion D. As the velocity of an object decreases its momentum increases and it takes more force to change its motion
A bullet with a mass of 20 g and a speed of 960 m/s strikes a block of wood of mass 4.5 kg resting on a horizontal surface. The bullet gets embedded in the block. The speed of the block immediately after the collision is:________. A) cannot be found because we don't know whether the surface is frictionless. B) is 0.21 km/s. C) is 65 m/s. D) is 9.3 m/s. E) None of these is correct

Imagine two billiard balls on a pool table. Ball A has a mass of 2 kilograms and ballB has a mass of 3 kilograms. The initial velocity of ball A is 9 meters per second to
the right, and the initial velocity of the ball B is 6 meters per second to the left. The
final velocity of ball A is 9 meters per second to the left, while the final velocity of
ball B is 6 meters per second to the right.

1. Explain what happens to each ball after the collision. Why do you think this
occurs? Which of Newton’s laws does this represent?

Answers

This is an example of an elastic collision. The two objects collide and return to their original shapes and move separately. In such a collision, kinetic energy is conserved. I think we can agree that this represents Newton's third law by demonstrating conservation of momentum.

Answer:

Yes, the law of conservation of momentum is satisfied. The total momentum before the collision is 1.5 kg • m/s and the total momentum after the collision is 1.5 kg • m/s. The momentum before and after the collision is the same.

Explanation:

A weather balloon is designed to expand to a maximum radius of 16.2 m when in flight at its working altitude, where the air pressure is 0.0282 atm and the temperature is â65âC. If the balloon is filled at 0.873 atm and 21âC, what is its radius at lift-off?

Answers

Answer:

5.78971 m

Explanation:

$P_1$ = Initial pressure = 0.873 atm

$P_2$ = Final pressure = 0.0282 atm

$V_1$ = Initial volume

$V_2$ = Final volume

$r_1$ = Initial radius = 16.2 m

$r_2$ = Final radius

Volume is given by

$(4)/(3)\pi r^3$

From the ideal gas law we have the relation

$(P_1V_1)/(T_1)=(P_2V_2)/(T_2)\n\Rightarrow (0.873* (4)/(3)\pi r_1^3)/(294.15)=(0.0282(4)/(3)\pi r_2^3)/(208.15)\n\Rightarrow (0.873r_1^3)/(294.15)=(0.0282* 16.2^3)/(208.15)\n\Rightarrow r_1=(0.0282* 16.2^3* 294.15)/(208.15* 0.873)\n\Rightarrow r_1=5.78971\ m$

The radius of balloon at lift off is 5.78971 m

Final answer:

To find the radius of the weather balloon at lift-off, the ideal gas law can be used. Using the equation P1V1 = P2V2, where P1 and V1 are the initial pressure and volume, and P2 and V2 are the pressure and volume at lift-off, the radius at lift-off can be calculated to be approximately 4.99 m.

Explanation:

To find the radius of the weather balloon at lift-off, we can use the ideal gas law, which states that PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.

In this case, we know that the number of moles is constant, as the balloon is filled with the same amount of helium at lift-off and in flight. Therefore, we can write the equation as P1V1 = P2V2, where P1 and V1 are the initial pressure and volume, and P2 and V2 are the pressure and volume at lift-off.

Plugging in the given values, we have (0.873 atm)(V1) = (0.0282 atm)(16.2 m)^3. Solving for V1, we find that the volume at lift-off is approximately 110.9 m^3. The radius can then be calculated using the formula for the volume of a sphere: V = (4/3) * π * r^3, where r is the radius.

Therefore, the radius at lift-off is approximately 4.99 m.

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Ball 1 is launched with an initial vertical velocity v1 = 146 ft/sec. Ball 2 is launched 2.3 seconds later with an initial vertical velocity v2. Determine v2 if the balls are to collide at an altitude of 234 ft. At the instant of collision, is ball 1 ascending or descending?

Answers

Answer:

Initial velocity of ball 2 = 38.21 m/s = 125.36 ft/s

At the time of collision velocity of ball one is descending.

Explanation:

Velocity of ball 1 = 146 ft/sec = 44.50m/s

The balls are to collide at an altitude of 234 ft

H = 234 ft = 71.32 m

We have equation of motion

v² = u² + 2as

v² = 44.50² + 2 x (-9.81) x 71.32

v = ±24.10 m/s.

Time for each velocity can be calculated using equation of motion

v = u + at

24.10 = 44.50 - 9.81 t , t = 2.07 s

-24.10 = 44.50 - 9.81 t , t = 6.99 s

Since the second ball throws after 2.3 seconds we can avoid case with t = 2.07 s.

So at the time of collision velocity of ball one is descending.

The collision occurs at t = 6.99 s.

Time of flight of ball 2 = 6.99 - 2.3 = 4.69 seconds.

Height traveled by ball 2 = 71.32 m

We need to find velocity

We have

s = ut + 0.5 at²

71.32 = u x 4.69 - 0.5 x 9.81 x 4.69²

u = 38.21 m/s = 125.36 ft/s

Initial velocity of ball 2 = 38.21 m/s = 125.36 ft/s

Answer:

v2=139 ft

Explanation:

First we just look at the motion of the first particle. It is moving vertically in a gravitational field so is decelerating with rate g = 9.81 m/s^2 = 32.18 ft/s^2. We can write it's vertical position as a function of time.

h_1=v_1*t-(a*t/2)

We set this equal to 234 ft to find when the body is passing that point, a solve the quadratic equation for t.

t_1,2=v_1±(√v_1^2-4*a/2*h_1)/a=2.57 s, 7.44 s

Since we know the second ball was launched after 2.3 seconds, we know that the time we are looking for is the second one, when the first ball is descending. The second ball will have 2.3 seconds less so the time we further use is t_c = 7.44 - 2.3 = 5.14 s. With this the speed of the second ball needed for collision at given height, can be found.

Solving a similar equation, but this time for v2 to obtain the result.

h_2=234 ft=v2*t_c-(a*t_c^2/2)--->v2=139 ft

Two wooden crates rest on top of one another. The smaller top crate has a mass of m1 = 25 kg and the larger bottom crate has a mass of m2 = 91 kg. There is NO friction between the crate and the floor, but the coefficient of static friction between the two crates is µs = 0.79 and the coefficient of kinetic friction between the two crates is µk = 0.62. A massless rope is attached to the lower crate to pull it horizontally to the right (which should be considered the positive direction for this problem).1)The rope is pulled with a tension T = 234 N (which is small enough that the top crate will not slide). What is the acceleration of the small crate?2)In the previous situation, what is the magnitude of the frictional force the lower crate exerts on the upper crate?3)What is the maximum tension that the lower crate can be pulled at before the upper crate begins to slide?4)The tension is increased in the rope to 1187 N causing the boxes to accelerate faster and the top box to begin sliding. What is the acceleration of the upper crate?m/s25)As the upper crate slides, what is the acceleration of the lower crate?

Answers

Answer:

Explanation:

Let the bigger crate be in touch with the ground which is friction less. In the first case both m₁ and m₂ will move with common acceleration because m₁ is not sliding over m₂.

1 ) Common acceleration a = force / total mass

= 234 / ( 25 +91 )

= 2.017 m s⁻².

2 ) Force on m₁ accelerating it , which is nothing but friction force on it by m₂

= mass x acceleration

= 25 x 2.017

= 50.425 N

The same force will be applied by m₁ on m₂ as friction force which will act in opposite direction.

3 ) Maximum friction force that is possible between m₁ and m₂

= μ_s m₁g

= .79 x 25 x 9.8

= 193.55 N

Acceleration of m₁

= 193 .55 / 25

= 7.742 m s⁻²

This is the common acceleration in case of maximum tension required

So tension in rope

= ( 25 +91 ) x 7.742

= 898 N

4 ) In case of upper crate sliding on m₂ , maximum friction force on m₁

= μ_k m₁g

= .62 x 25 x 9.8

= 151.9 N

Acceleration of m₁

= 151.9 / 25

= 6.076 m s⁻².

The International Space Station (ISS) is a space station orbiting the earth above the ground. If the radius of the earth is 3,958.8 miles, mass of earth is 5.972 x 10 24 kg, the period of the ISS at the orbit around the earth is 7.84 hours, can you calculate what is the distance from the ISS to the surface of the earth, in unit of miles

Answers

Answer:

8488 miles

Explanation:

The orbital period around an earth is given as:

$T=2\pi \sqrt{(r^3)/(Gm) }$

Where G = constant = 6.67 x 10ˉ¹¹ N m² kgˉ², m = mass of object, T = period taken to round the earth, r = distance from the center of the earth to the orbiting object = radius of earth + orbital altitude.

Given that T = 7.84 hours = 28224 seconds, m = 5.972 x 10²⁴ kg, radius of earth = 3,958.8 miles = 6371071 m

$T=2\pi \sqrt{(r^3)/(Gm) }\n\nsquaring:\n\nT^2=4\pi^2 ((r^3)/(Gm) )\n\nr^3=(GmT^2)/(4\pi^2) \n\nr=\sqrt[3]{(GmT^2)/(4\pi^2) } \n\nr=\sqrt[3]{(6.67*10^(-11)*5.972*10^(24)*(28224)^2)/(4\pi^2) } \n\nr=20031232.62\ meters$