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brilliants [131]
4 years ago
15

Which of these is NOT an example of a reference direction?

Physics
1 answer:
Vikentia [17]4 years ago
8 0

Answer:

A

Explanation:

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If an object was accelerating at 10 m/s2, and a mass of 1 kg, what was size of the force acting on the object?
ryzh [129]

Answer:

10 N

Explanation:

f = ma

= 10m/s^2 * 1 kg

=10N

7 0
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Nearly all physics problems will use the unit m/s squared. Why are the seconds squared?
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Seconds squared is the time unit of acceleration.  It represents the change in distance units per second per second. For example, 3 m/sec² means a distance covering 3 meters in the first second, then 9 meters in the 2nd second, and 37 meters in the third second. (3^1, 3^2, 3^3).

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4 years ago
Why do you like the full moon ?
mr_godi [17]

Answer:

<h3>The Moon brings perspective. Observing the Moon, and I mean really looking – sitting comfortably, or lying down on a patch of grass and letting her light fill your eyes, it's easy to be reminded of how ancient and everlasting the celestial bodies are. When I do this, it always puts my life into perspective.</h3>
4 0
3 years ago
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Two point charges are separated by 10 cm, with an attractive force between them of 15 N. Find the force between them when they a
suter [353]

Answer:

(a) the force is 8.876 N

(b) the magnitude of each charge is 4.085 μC

Explanation:

Part (a)

Given;

coulomb's constant, K = 8.99 x 10⁹ N.m²/C²

distance between two charges, r = 10 cm = 0.1 m

force between the two charges, F = 15 N

when the distance between the charges changes to 13 cm (0.13 m)

force between the two charges, F = ?

Apply Coulomb's law;

F = \frac{Kq_1q_2}{r^2} \\\\let \ Kq_1q_2 = C\\\\F =\frac{C}{r^2} \\\\C = Fr^2\\\\F_1r_1^2 = F_2r_2^2\\\\F_2 =\frac{F_1r_1^2}{r_2^2} \\\\F_2 = \frac{15*0.1^2}{0.13^2} \\\\F_2 = 8.876 \ N

Part (b)

the magnitude of each charge, if they have equal magnitude

F = \frac{KQ^2}{r^2}

where;

F is the force between the charges

K is Coulomb's constant

Q is the charge

r is the distance between the charges

F = \frac{KQ^2}{r^2} \\\\Q = \sqrt{\frac{Fr^2}{K} } \\\\Q =  \sqrt{\frac{15*(0.1)^2}{8.99*10^9} } = 4.085 *10^{-6} \ C\\\\Q = 4.085 \ \mu C

4 0
3 years ago
When light is reflected by a mirror, the angle of incidence is always A. equal to the angle of reflection. B. less than the angl
ankoles [38]
When light is reflected by a mirror, the angle of incidence is always <span>A. equal to the angle of reflection. We know this by the Law of Reflection.</span>
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