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Ket [755]
1 year ago
10

A 5OKG mass is taken to the moon and the mass determined there, comment on the likely result of the measurement

Physics
1 answer:
vodomira [7]1 year ago
7 0

Answer:

Mass will be the same

Explanation:

Mass does not change with gravity ....   WEIGHT will be different, but not mass.

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A spring with k = 500 N/m stores 704 J. How far is it extended from the equilibrium position
kaheart [24]

Given that:

k = 500 n/m,

work (W) = 704 J

spring extension (x) = ?

         we know that,

                      Work = (1/2) k x²

                          704 = (1/2) × 500 × x²

                            x = 1.67 m

A spring stretched for 1.67 m distance.

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what is the mass of a cannon ball if a force 2500 N gives the cannon ball an acceleration of 200 m/s squared?
nignag [31]

Using Newton's second law of motion:

F=ma ;  [ F = force (N: kgm/s^2);m= mass (kg); a = acceleration (m/s^2)


Given:                      Find:                   Formula:                  Solve for m:

F: 2500N                 mass:?                F=ma Eq.1              m=F/a  Eq. 2

a= 200m/s^2  


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565,900 seconds into days​
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3 years ago
Two 10-cm-diameter charged rings face each other, 21.0 cm apart. Both rings are charged to +40.0 nC. What is the electric field
Katyanochek1 [597]

Complete question:

Two 10-cm-diameter charged rings face each other, 21.0 cm apart. Both rings are charged to +40.0 nC. What is the electric field strength  at the midpoint between the two rings ?

Answer:

The electric field strength at the mid-point between the two rings is zero.

Explanation:

Given;

diameter of each ring, d = 10 cm = 0.1 m

distance between the rings, r = 21.0 cm = 0.21 m

charge of each ring, q = 40 nC = 40 x 10⁻⁹ C

let the midpoint between the two rings = x

The electric field strength  at the midpoint between the two rings is given as;

E_{mid} = E_{right} +E_{left}\\\\E_{right}  = \frac{KQ}{(x^2 + r^2)^\frac{2}{3} } \\\\E_{leftt}  = -\ \frac{KQ}{(x^2 + r^2)^\frac{2}{3} }\\\\E_{mid} = \frac{KQ}{(x^2 + r^2)^\frac{2}{3} }  - \frac{KQ}{(x^2 + r^2)^\frac{2}{3} } = 0

Therefore, the electric field strength at the mid-point between the two rings is zero.

7 0
2 years ago
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