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stepladder [879]
3 years ago
15

Calculate the gravitational potential energies of the melon and the pomegranate. Which one has potential energy

Physics
1 answer:
galben [10]3 years ago
6 0

Answer:

The mellon

Explanation:

I just know

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9. A plane starts at rest & accelerates along the ground before takeoff. It
Phoenix [80]

Answer:

  9.877 m/s^2

Explanation:

The acceleration can be computed from ...

  d = (1/2)at^2

  (1600 m) = (1/2)a(18 s)^2

  a = (1600/162) m/s^2 ≈ 9.877 m/s^2

6 0
3 years ago
In models of magnetic and electric fields, why are field vectors depicted by arrows?
taurus [48]

The arrows in models of magnetic and electric fields show  both their magnitude and direction.

In Physics, a vector refers to a quantity that has both magnitude and direction. Hence, a vector always points in a given direction. The direction in which the arrow points is the direction of the vector in space.

In models of magnetic and electric fields, field vectors depicted by arrows  because they represent both their magnitude and direction. The length of the arrow shows magnitude.

Learn more: brainly.com/question/102477

7 0
2 years ago
During 57 seconds of use, 330 C of charge flow through a microwave oven. Compute the size of the electric current.
swat32

Answer:

5.78amps

Explanation:

Given data

Time t= 57 seconds

Charge Q= 330C

Current I= ??

The expression for the electric current is given as

Q= It

Substituting we have

330= I*57

I= 330/57

I=5.78 amps

Hence the current is 5.78amps

3 0
3 years ago
What is the magnitude of the applied electric field inside an aluminum wire of radius 1.2 mm that carries a 3.0-a current? [ σal
galben [10]
Hello

1) First of all, since we know the radius of the wire (r=1.2~mm=0.0012~m), we can calculate its cross-sectional area
A=\pi r^2 = 3.14 \cdot (0.0012~m)^2=4.5\cdot10^{-6}~m^2

2)  Then, we can calculate the current density J inside the wire. Since we know the current, I=3~A, and the area calculated at the previous step, we have
J= \frac{I}{A}= \frac{3~A}{4.5\cdot10^{-6}~m^2} = 6.63\cdot10^5 ~A/m^2

3) Finally, we can calculate the electric field E applied to the wire. Given the conductivity \sigma=3.6\cdot10^7~ \frac{A}{Vm} of the aluminium, the electric field is given by
E= \frac{J}{\sigma}= \frac{ 6.63\cdot10^5 ~A/m^2}{3.6\cdot10^7~ \frac{A}{Vm} } = 0.018~V/m

4 0
3 years ago
Item 10
melisa1 [442]

Answer:

D. mass to see how it affected stretch length of a rubber band

5 0
3 years ago
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