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vredina [299]
3 years ago
8

You can also describe position by using coordinates of _____and _______

Physics
1 answer:
SOVA2 [1]3 years ago
7 0

Answer:

x and y

Explanation:

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Mya (m = 65 kg) floats 4 m from her spaceship (m = 1600 kg). What gravitational force would each mass exert on the other?
Aleksandr-060686 [28]

Answer:

ans: 4.34 × 10^(-9) N

Explanation:

mass of Mya say (m) = 65 kg

mass of spaceship say (M) = 1600 kg

universal gravitational constant(G) =6.67 × 10^(-11) Nm²/kg²

separation distance (d) = 4m

so,

gravitational force (F)= GMm/d²

=( 6.67 × 65 × 1600) / ( 10¹¹ × 4²)

= 4.34 × 10⁴ / 10¹³

= 4.34 × 10^(-9) N

8 0
3 years ago
(Bonus) Which of the following statements are part of The Law of Conservation of Energy?
Murrr4er [49]
The answers are B, C, G, I
Hope I helped
(Don’t know if they are right)
6 0
4 years ago
Read 2 more answers
A coil consists of 180 turns of wire. Each turn is a square of side d=30 cm, and a uniform magnetic field directed perpendicular
alexira [117]

Answer:

Emf induced i equal to 329.4 volt

Explanation:

Note : Here i think we have to find emf induced in the coil

Number of turns in the coil N= 180

Sides of square d = 30 cm = 0.3 m

So area of the square A=0.3\times 0.3=0.09m^2

Magnetic field is changes from 0 to 1.22 T

Therefore dB=1.22-0=1.22T

Time interval in changing the magnetic field dt = 0.06 sec

Induced emf is given by

e=-N\frac{d\Phi }{dt}=-NA\frac{dB}{dt}

e=-180\times 0.09\times \frac{1.22}{0.06}=329.4volt

8 0
4 years ago
Building codes usually limit the current carried by a No. 14 copper wire to 15 A. Many household circuits are wired with this si
Sedbober [7]

Answer:

The drift velocity of the electrons in this case is c) 5.52 x 10-4 m/s

Explanation:

Hi

First of all, we need to find the volume occupied by 63.3g of copper, so V=\frac{m}{\rho}= \frac{0.0633Kg}{8,900kg/m^{3}} =7.11 \times 10^{-6} m^{3}, then if we assume each atom of copper contributes with one free electron to the material body n=\frac{N_{A}}{V}=\frac{6.02 \times 10^{23} electrons}{7.11 \times 10^{-6} m^{3}} =8.46 \times 10^{28} electrons/m^{3}.

Finally, we apply v_{d}=\frac{I}{nqA}, where A is area so, A=\pi (\frac{d}{2})^{2}= \pi (\frac{0.0016m}{2})^{2}=2 \times 10^{-6} m^{2}, thus v_{d}=\frac{I}{nqA}=\frac{15C/s}{(8.46 \times 10^{28} m^{-3})(1.60 \times 10^{-19} C)(2 \times 10^{-6} m^{2})}=5.50 \times 10^{-4} m/s.

As we can see c. 5.52 x 10-4 m/s is the nearest one.

3 0
3 years ago
A passenger plane is flying above the ground.Describe the two components of its mechanical enserfy
Goryan [66]
The force and the air resistance depends on the mechanical enserfy.
3 0
4 years ago
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