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julsineya [31]
3 years ago
11

What is the earths slant called

Physics
2 answers:
ValentinkaMS [17]3 years ago
5 0
Earths slant is called its axis
dalvyx [7]3 years ago
3 0
Earth's slant is also known as it's axis
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Which type of galaxy has arms that contain sites of active star formation and start close to a bulge in the center?
Serhud [2]

Answer:

normal spiral I think. I hope this helps

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Newton's _____ law explains why my hands hurt when I clap loudly
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Newtons third law (inertia) is to blame
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A rubber rod rubbed with fur acquires a charge of -4,8×10(-9)C. What is the charge on the fur? How much mass is transferred to r
mote1985 [20]
1) The charge left on the fur is equal and opposite to the charge transferred to the rod:
Q=+4.8 \cdot 10^{9} C
In fact, when the rod is rubbed with the fur, a net charge of Q=-4.8 \cdot 10^{-9} C has been transferred to the rod, leaving it negatively charged. If we assume the fur was initially neutral, this means that we have now an excess of positive charges on the fur, and the amount of this charge must be equal (in magnitude, but with opposite sign) to the charge transferred to the rod.

2) The mass transferred to the rod is equal to the total mass of the electrons transferred to the rod.
The charge transferred to the rod is
Q=-4.8 \cdot 10^{-9} C
The charge of 1 electron is
e=-1.6 \cdot 10^{-19} C
So the number of electrons transferred is
N= \frac{Q}{e}= \frac{-4.8 \cdot 10^{-9} C}{-1.6 \cdot 10^{-19} C}=3.0 \cdot 10^{10}

The mass of 1 electron is m=9.1 \cdot 10^{-31} kg, therefore the total mass transferred to the rod is
M=Nm=(3 \cdot 10^{10})(9.1 \cdot 10^{-31} kg)=2.73 \cdot 10^{-20} kg

7 0
3 years ago
which type of landform develops at plate boundaries were one oceanic plate desends beneth another oceanic plate
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4 years ago
A boy sleds down a hill and onto a frictionless ice- covered lake at 10.0 m/s. In the middle of the lake is a 1000-kg boulder. W
mina [271]

Answer:

The speed of the sled is 9.2 m/s

The speed of the boulder is 0.82 m/s

Solution:

As per the question:

Mass of the boulder, m_{B} = 1000\ kg

Mass of the sled, m_{S} = 2.50\ kg

Mass of the boy, m_{b} = 40\ kg

Initial Velocity, v = 10.0 m/s

Now,

To calculate the speed of both the sled and the boulder after the occurrence of the collision:

m = m_{b} + m_{S} = 40 + 2.50 = 42.50\ kg

Initial velocity of the boulder, v_{B} = 0\ m/s

Since, the collision is elastic, both the energy and momentum rem,ain conserved.

Now,

Using the conservation of momentum:

mv + m_{B}v_{B} = mv' + m_{B}v'_{B}

where

v' = final velocity of the the system of boy and sled

v'_{B} = final velocity of the boulder

42.50\times 10 + m_{B}.0 = 42.50v' + 1000v'_{B}

42.50v' + 1000v'_{B} = 425            (1)

Now,

Using conservation of energy:

\frac{1}{2}mv^{2} + \frac{1}{2}m_{B}v_{B}^{2} = \frac{1}{2}mv'^{2} + \frac{1}{2}m_{B}v'_{B}^{2}

42.50\times 10^{2} + m_{B}.0 = 42.50v'^{2} + 1000v'_{B}^{2}

42.50v'^{2} + 1000v'_{B}^{2} = 4250         (2)

Now, from  eqn (1) and (2):

v' = \frac{m - m_{B}}{m + m_{B}}\times v

v' = \frac{42.50 - 1000}{42.5 + 1000}\times 10 = - 9.2\ m/s

Now,

v'_{B} = \frac{2m}{m + m_{B}}\times v

v'_{B} = \frac{2\times 42.50}{42.5 + 1000}\times 10 = 0.82\ m/s

5 0
3 years ago
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