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3 years ago

Please help me with this review question.

Physics

1 answer:

jolli1 [7]3 years ago

8 0

Answer:

28.7%

Explanation:

efficiency = work output /work input × 100

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Problem 28.3 Suppose that you are planning a trip in which a spacecraZ is to travel at a constant velocity for exactly six month

iragen [17]

Answer:

v = 0.999981c m/s

Explanation:

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3 years ago

A 50-kg box is being pushed along a horizontal surface. The coefficient of static friction between the box and the ground is 0.6

dsp73

To solve the exercise it is necessary to apply the concepts related to Newton's Second Law, as well as the definition of Weight and Friction Force.

According to the problem there is a movement in the body and it is necessary to make a sum of forces on it, so that

$\sum F = ma$

There are two forces acting on the body, the Force that is pushing and the opposing force that is that of friction, that is

$F - F_f = ma$

To find the required force then,

$F=F_f+ma$

By definition we know that the friction force is equal to the multiplication between the friction coefficient and the weight, that is to say

$F = \mu mg +ma$

$F = 0.35*50*0.8+50*1.2$

$F=(171.5N)+(50Kg)(1.2m/s^2)$

$F=231.5N$

$F\approx 230N$

Therefore the horizontal force applied on the block is B) 230N

6 0

4 years ago

If a material, such as concrete, can withstand a large applied weight but gives away (ruptures) quickly when a crack forms, then

MrMuchimi

The correct answer is brittle:)

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3 years ago

Read 2 more answers

A +35 µC point charge is placed 46 cm from an identical +35 µC charge. How much work would be required to move a +0.50 µC test c

Ira Lisetskai [31]

Answer:

512.5 mJ

Explanation:

Let the two identical charges be q = +35 µC and distance between them be r₁ = 46 cm. A charge q' = +0.50 µC located mid-point between them is at r₂ = 46 cm/2 = 23 cm = 0.23 m.

The electric potential at this point due to the two charges q is thus

V = kq/r₂ + kq/r₂

= 2kq/r₂

= 2 × 9 × 10⁹ Nm²/C² × 35 × 10⁻⁶ C/0.23 m

= 630/0.23 × 10³ V

= 2739.13 × 10³ V

= 2.739 MV

When the charge q' is moved 12 cm closer to either of the two charges, its distance from each charge is now r₃ = r₂ + 12 cm = 23 cm + 12 = 35 cm = 0.35 m and r₄ = r₂ - 12 cm = 23 cm - 12 cm = 11 cm = 0.11 cm.

So, the new electric potential at this point is

V' = kq/r₃ + kq/r₄

= kq(1/r₃ + 1/r₄)

= 9 × 10⁹ Nm²/C² × 35 × 10⁻⁶ C(1/0.35 m + 1/0.11 m)

= 315 × 10³(2.857 + 9.091) V

= 315 × 10³ (11.948) V

= 3763.62 × 10³ V

= 3.764 MV

Now, the work done in moving the charge q' to the point 12 cm from either charge is

W = q'(V' - V)

= 0.5 × 10⁻⁶ C(3.764 MV - 2.739 MV)

= 0.5 × 10⁻⁶ C(1.025 × 10⁶) V

= 0.5125 J

= 512.5 mJ

8 0

3 years ago