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kaheart [24]
4 years ago
14

Consider a particle on which several forces act, one of which is known to be constant in time: Fi = 3.00 i +4.00 ) N. As a resul

t, the particle moves along a straight path from a Cartesian coordinate of 0.00 m, 0.00 m) to (5.00 m, 6.00 m). What is the work done by Fi?
Physics
1 answer:
Leviafan [203]4 years ago
4 0

Given that.

F=3•i+4•j

And it from point (0,0)m to (5,6)m

dx=final position - initial position

dx=(5,6)-(0,0)

dx=(5,6)m

dx=5•i +6•j

Work done by the force is give by

W = F•dx

W=F•dx

Note that i•i=j•j=1 and i•j=j•i=0

Then,

W=(3i+4j)•(5i+6j)

Therefore,

W=3i•(5i+6j)+4j•(5i+6j)

W=15i•i+18i•j+20j•i+24j•j

W=15+0+0+24

W=39J

Then the work done by the force is 39 Joules

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3 years ago
17. (a) Will the electric field strength between two parallel conducting plates exceed the breakdown strength for air ( 3.0×106
sleet_krkn [62]

Answer:

Explanation:

Distance between plates d = 2 x 10⁻³m

Potential diff applied = 5 x 10³ V

Electric field = Potential diff applied /  d

= 5 x 10³  / 2 x 10⁻³

= 2.5 x 10⁶ V/m

This is less than  breakdown strength for air  3.0×10⁶ V/m

b ) Let the plates be at a separation of d .so

5 x 10³ / d = 3.0×10⁶ ( break down voltage )

d = 5 x 10³  / 3.0×10⁶

= 1.67 x 10⁻³ m

= 1.67 mm.

5 0
3 years ago
A light spring having a force constant of 115 N/m is used to pull a 9.00 kg sled on a horizontal frictionless ice rink. The sled
Ksivusya [100]

Answer:

Stretch in the spring = 0.1643 (Approx)

Explanation:

Given:

Mass of the sled (m) = 9 kg

Acceleration of the sled (a) = 2.10 m/s ²

Spring constant (k) = 115 N/m

Computation:

Tension force in the spring  (T) = ma

Tension force in the spring  (T) = 9 × 2.10

Tension force in the spring  (T) = 18.9 N

Tension force in the spring = Spring constant (k) × Stretch in the spring

18.9 N = 115 N  × Stretch in the spring

Stretch in the spring = 18.9 / 115

Stretch in the spring = 0.1643 (Approx)

8 0
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Which statement correctly describes variable stars?
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Answer:

D)

Explanation:

The Period-Luminosity relationship tells us that luminosity increases with the period, and of course the more luminosity a star has the more far away they can be seen, so from this we know that:

A) False since lower luminosities can be observed when they are close.

B) False since longer periods means higher luminosities

C) False since lower luminosities can be observed when they are close.

D) True: Variable stars with shorter periods have lower luminosities, so they can only be observed when they are close.

5 0
3 years ago
A negative velocity, approaching zero, represents a negative acceleration. True or False
ruslelena [56]

Answer:

False.

Explanation:

Lets assume our positive direction to the right (this reasoning works for any direction). A negative velocity would then be then directed to the left. If it varies as such that it aproaches to zero, it means that the variation is directed to the right, and that is where the direction of the acceleration must be pointing. In other words, its losing its velocity, so the acceleration must point opposite to the velocity. Then it means the acceleration is positive.

8 0
3 years ago
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