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Whitepunk [10]
3 years ago
12

What caused day and night on a planet

Physics
1 answer:
Volgvan3 years ago
7 0

Answer:

The rotation of a planet around it's sun

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Suppose that the dipole moment associated with an iron atom of an iron bar is 2.8 × 10-23 J/T. Assume that all the atoms in the
masya89 [10]

To solve this exercise it is necessary to apply the equations related to the magnetic moment, that is, the amount of force that an image can exert on the electric currents and the torque that a magnetic field exerts on them.

The diple moment associated with an iron bar is given by,

\mu = \alpha *N

Where,

\alpha = Dipole momento associated with an Atom

N = Number of atoms

\alpha y previously given in the problem and its value is 2.8*10^{-23}J/T

L = 5.8cm = 5.8*10^{-2}m

A = 1.5cm^2 = 1.5*10^{-4}m^2

The number of the atoms N, can be calculated as,

N = \frac{\rho AL}{M_{mass}}*A_n

Where

\rho = Density

M_{mass} = Molar Mass

A = Area

L = Length

A_n =Avogadro number

N = \frac{(7.9g/cm^3)(1.5cm)(5.8cm^2)}{55.9g/mol}(6.022*10^{23}atoms/mol)

N = 7.4041*10^{23}atoms

Then applying the equation about the dipole moment associated with an iron bar we have,

\mu = \alpha *N

\mu = (2.8*10^{-23})*(7.4041*10^{23})

\mu = 20.72Am^2

PART B) With the dipole moment we can now calculate the Torque in the system, which is

\tau = \mu B sin(90)

\tau = (20.72)(2.2)

\tau = 45.584N.m

<em>Note: The angle generated is perpendicular, so it takes 90 ° for the calculation made.</em>

3 0
3 years ago
Chris is about to do an experiment to measure the density of water at several temperatures. His teacher has him prepare and sign
Marina86 [1]
Inspect the glassware for cracks or chips prior to beginning the lab.
7 0
3 years ago
Read 2 more answers
A ball has a mass of 1.5kg and is thrown straight up with a speed of 60m/s, what is the ball’s momentum:
madam [21]

Answer:

Assumption: the air resistance on this ball is negligible. Take g = 10\; \rm m \cdot s^{-2}.

a. The momentum of the ball would be approximately 60\;\rm kg \cdot m \cdot s^{-1} two seconds after it is tossed into the air.

b. The momentum of the ball would be approximately \rm \left(-45\; \rm kg \cdot m \cdot s^{-1}\right) three seconds after it reaches the highest point (assuming that it didn't hit the ground.) This momentum is smaller than zero because it points downwards.

Explanation:

The momentum p of an object is equal its mass m times its velocity v. That is: \vec{p} = m \cdot \vec{v}.

Assume that the air resistance on this ball is negligible. If that's the case, then the ball would accelerate downwards towards the ground at a constant g \approx -10\; \rm m \cdot s^{-2}. In other words, its velocity would become approximately 10\; \rm m \cdot s^{-1} more negative every second.

The initial velocity of the ball is 60\; \rm m \cdot s^{-1}. After two seconds, its velocity would have become 60\;\rm m \cdot s^{-1} + 2\; \rm s \times \left(-10\;\rm m \cdot s^{-1}\right) = 40\; \rm m \cdot s^{-1}. The momentum of the ball at that time would be around p = m \cdot v \approx 60\; \rm kg \cdot m \cdot s^{-1}.

When the ball is at the highest point of its trajectory, the velocity of the ball would be zero. However, the ball would continue to accelerate downwards towards the ground at a constant g \approx -10\; \rm m \cdot s^{-2}. That's how the ball's velocity becomes negative.

After three more seconds, the velocity of the ball would be 0\; \rm m \cdot s^{-1} + 3\; \rm s \times \left(-10\; \rm m \cdot s^{-2}\right) = -30 \; \rm m \cdot s^{-1}. Accordingly, the ball's momentum at that moment would be p = m \cdot v \approx \left(-45\; \rm kg \cdot m \cdot s^{-1}\right).

3 0
3 years ago
A copper wire 1.0 meter long and with a mass of .0014 kilograms per meter vibrates in two segments when under a tension of 27 Ne
Furkat [3]

Answer:

the frequency of this mode of vibration is 138.87 Hz

Explanation:

Given;

length of the copper wire, L = 1 m

mass per unit length of the copper wire, μ = 0.0014 kg/m

tension on the wire, T = 27 N

number of segments, n = 2

The frequency of this mode of vibration is calculated as;

F_n = \frac{n}{2L} \sqrt{\frac{T}{\mu} } \\\\F_2 = \frac{2}{2\times 1} \sqrt{\frac{27}{0.0014} }\\\\F_2 = 138.87 \ Hz

Therefore, the frequency of this mode of vibration is 138.87 Hz

7 0
3 years ago
Reese can run 10 meter in 5 seconds.What is her speed?
Bad White [126]

Answer:

2 meters per second

Explanation:

4 0
3 years ago
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