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My name is Ann [436]
2 years ago
8

A rotating, funnel-shaped column of air that extends down to the ground is called a(n) . If it does not reach the ground it is c

alled a(n)
Physics
1 answer:
Rama09 [41]2 years ago
6 0

black women are hideous

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Two 2.0 kg bodies, A and B, collide. The velocities before the collision are ~vA = (15ˆi + 30ˆj) m/s and ~vB = (−10ˆi + 5.0ˆj) m
AleksandrR [38]

Answer:

Part a)

10\hat i + 15\hat j = \vec v

Part b)

\Delta K = 500 J

Explanation:

As we know that there is no external force on the system of two masses so here total momentum of the system will remains conserved

so we can say

m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}

(2kg)(15\hat i + 30 \hat j) + (2 kg)(-10\hat i  + 5\hat j) = 2kg(-5\hat i + 20\hat j) + 2\vec v

5\hat i + 35\hat j = (-5\hat i + 20\hat j) +\vec v

10\hat i + 15\hat j = \vec v

Part b)

magnitude of the initial speed of A = \sqrt{15^2 + 30^2} = 33.54 m/s

magnitude of the initial speed of B = \sqrt{10^2 + 5^2} = 11.18 m/s

magnitude of final speed of A = \sqrt{5^2 + 20^2} = 20.61 m/s

magnitude of final speed of B = \sqrt{10^2 + 15^2} = 18.03 m/s

Now change in total kinetic energy is given as

\Delta K = (\frac{1}{2}m_1v_{1i}^2 + \frac{1}{2}m_2v_{2i}^2) - (\frac{1}{2}m_1v_{1f}^2 + \frac{1}{2}m_2v_{2f}^2)

\Delta K = (\frac{1}{2}2(33.54)^2 + \frac{1}{2}2(11.18)^2) - (\frac{1}{2}2(20.61)^2 + \frac{1}{2}2(18.03)^2)

\Delta K = 500 J

6 0
3 years ago
Does the orbital period of a planet depend on the mass of the planet or on the mass of the star that it orbits?
jasenka [17]

Answer:

The orbital period of a planet depends on the mass of the planet.

Explanation:

A less massive planet will take longer to complete one period than a more massive planet.

8 0
2 years ago
A rectangular conducting loop of wire is approximately half-way into a magnetic field B (out of the page) and is free to move. S
Black_prince [1.1K]

Answer:

. The loop is pushed to the right, away from the magnetic field

Explanation

This decrease in magnetic strength causes an opposing force that pushes the loop away from the field

8 0
3 years ago
An archer tests various arrowheads by shooting arrows at a pumpkin that is suspended from a tree branch by a rope, as shown to t
erik [133]

Answer:

Bounce 1 ,  pass 3,   emb2

Explanation:

(By the way I am also doing that question on College board physics page) For the Bounce arrow, since it bumps into the object and goes back, it means now it has a negative momentum, which means a larger momentum is given to the object. P=mv, so the velocity is larger for the object, and larger velocity means a larger kinetic energy which would result in a larger change in the potential energy. Since K=0.5mv^2=U=mgh, a larger potential energy would have a larger change in height which means it has a larger angle θ with the vertical line. Comparing with the "pass arrow" and the "Embedded arrow", the embedded arrow gives the object a larger momentum, Pi=Pf (mv=(M+m)V), it gives all its original momentum to the two objects right now. (Arrow and the pumpkin), it would have a larger velocity. However for the pass arrow, it only gives partial of its original momentum and keeps some of them for the arrow to move, which means the pumpkin has less momentum, means less velocity, and less kinetic energy transferred into the potential energy, and means less change in height, less θangle.  So it is  Bounce1, pass3, emb2.  

6 0
3 years ago
Light in the air is incident at an angle to the surface of (12.0 A) degrees on a piece of glass with an index of refraction of (
Orlov [11]

The question is incomplete. You dis not provide values for A and B. Here is the complete question

Light in the air is incident at an angle to a surface of (12.0 + A) degrees on a piece of glass with an index of refraction of (1.10 + (B/100)). What is the angle between the surface and the light ray once in the glass? Give your answer in degrees and rounded to three significant figures.

A = 12

B = 18

Answer:

18.5⁰

Explanation:

Angle of incidence i = 12.0 + A

A = 12

= 12.0 + 12

= 14

Refractive index u = 1.10 + B/100

= 1.10 + 18/100

= 1.10 + 0.18

= 1.28

We then find the angle of refraction index u

u = sine i / sin r

u = sine24/sinr

1.28 = sine 24 / sine r

1.28Sine r = sin24

1.28 sine r = 0.4067

Sine r = 0.4067/1.28

r = sine^-1(0.317)

r = 18.481

= 18.5⁰

4 0
3 years ago
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