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Grace [21]
3 years ago
7

Rebecca was at the playground and found a rusted nail. What happened to the nail? A) It went through a physical change because t

he oxygen in water mixed with air. B) It went through a phase change because the oxygen in water mixed with air. C) It went through a chemical change because the oxygen in water mixed with air. D) It went through both chemical and physical changes.
Physics
1 answer:
BaLLatris [955]3 years ago
8 0
Buff dude skid did di d I’d
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Can someone help me find the which direction is north and which is south on this solenoid?
arsen [322]

Explanation:

Now, looking down the solenoid tube determine what direction is the winding. If clockwise in relation to the positive wire then is the south pole, if anti-clockwise then is the north pole. So, to summarize the magnetic south pole is always clockwise in relation to the positive wire.

7 0
3 years ago
Is there a difference between radial velocity and angular velocity or are they the same thing?
Novosadov [1.4K]
Angular velocity is the rate of change of angle of a body, i.e. omega = v / r = (2*pi*r)/ r*t = (2*pi)/ T. where T is the time period of whatever is rotating and r is the radius of the circle. So if a circular disc is spinning at 1 m/s then the angular velocity of it is 2*pi radians/ second.
5 0
3 years ago
HEEELLPPPP ME LLELLLLEEAASEEEEEEEEEEEE
irina [24]

Answer:

Valley-river Landslide-Gravity Frost wedging- Glacier Canyon-Ice.

Explanation:

I think that's right

4 0
3 years ago
A hockey puck slides off the edge of a horizontal platform with an initial velocity of 28.0 m/shorizontally in a city where the
kozerog [31]

Answer:

θ = 12.60°

Explanation:

In order to calculate the angle below the horizontal for the velocity of the hockey puck, you need to calculate both x and y component of the velocity of the puck, and also you need to use the following formula:

\theta=tan^{-1}(\frac{v_y}{v_x})       (1)

θ: angle below he horizontal

vy: y component of the velocity just after the puck hits the ground

vx: x component of the velocity

The x component of the velocity is constant in the complete trajectory and is calculated by using the following formula:

v_x=v_o

vo: initial velocity = 28.0 m/s

The y component is calculated with the following equation:

v_y^2=v_{oy}^2+2gy         (2)

voy: vertical component of the initial velocity = 0m/s

g: gravitational acceleration = 9.8 m/s^2

y: height

You solve the equation (2) for vy and replace the values of the parameters:

v_y=\sqrt{2gy}=\sqrt{2(9.8m/s^2)(2.00m)}=6.26\frac{m}{s}

Finally, you use the equation (1) to find the angle:

\theta=tan^{-1}(\frac{6.26m/s}{28.0m/s})=12.60\°

The angle below the horizontal is 12.60°

7 0
3 years ago
A point charge q1 is at the center of a sphere of radius 20 cm. Another point charge q2 = 10 nC is located at a distance r = 10
jok3333 [9.3K]

Answer:

q₁  = -2.92 nC

Explanation:

Given;

first point charge, q₁ = ?

second point charge, q₂ = 10 nC

net flux through the surface of the sphere, Φ =  800 N.m²/C

According to Gauss’s law, the flux through any closed surface (Gaussian surface), is equal to the net charge enclosed divided by the permittivity of free space.

\phi = \frac{q_{enc.}}{\epsilon_o}

where;

Φ is net flux

q_{enc.} net charge enclosed

ε₀ is permittivity of free space.

q_{enc.} = Φε₀

       = 800 x 8.85 x 10⁻¹²

       = 7.08 x 10⁻⁹ C

q_{enc.} = 7.08 nC

q₁ + q₂ = q_{enc.}

q₁ = q_{enc.} - q₂

q₁  = 7.08nC -  10 nC

q₁  = -2.92 nC

4 0
3 years ago
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