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shepuryov [24]
3 years ago
5

If a rod attached to the approaching charge if the rod consists of "stiff" spring-like bonds for which atoms undergo small oscil

lations. What can we say, about these springlike bonds when the charge is first, furthest away and second, closest to the source charge
Physics
1 answer:
Hoochie [10]3 years ago
5 0

Answer: hello options related to your question is missing attached below is the missing part of your question

answer: No charge of the length of the bonds expected because the rod did not touch the charge source ( option A )

Explanation:

When the Charge is first, Furthest away and second  and closest to the source charge. <em>The spring like bonds can be said to have No charge of the length  of the bonds expected because the rod did not touch the charge source </em><em>when Furthest away the bond with charge will be less effective </em>

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How much energy or stopping power is needed to bring a car to a stop from 100 mph?
Fed [463]
I think 100 mph pushing the car the opposite direction
3 0
3 years ago
You are pulling a child in a wagon. The rope handle is inclined upward at a 60∘ angle. The tension in the handle is 20 N.
dem82 [27]
  • Angle (θ) = 60°
  • Force (F) = 20 N
  • Distance (s) = 200 m
  • Therefore, work done
  • = Fs Cos θ
  • = (20 × 200 × Cos 60°) J
  • = (20 × 200 × 1/2) J
  • = (20 × 100) J
  • = 2000 J

<u>Answer</u><u>:</u>

<u>2</u><u>0</u><u>0</u><u>0</u><u> </u><u>J</u>

Hope you could get an idea from here.

Doubt clarification - use comment section.

6 0
2 years ago
Read 2 more answers
A projectile is launched straight up from a height of 960 feet with an initial velocity of 64 ft/sec. Its height at time t is h(
Natasha2012 [34]

Answer:

a) t=2s

b) h_{max}=1024ft

c) v_{y}=-256ft/s

Explanation:

From the exercise we know the initial velocity of the projectile and its initial height

v_{y}=64ft/s\\h_{o}=960ft\\g=-32ft/s^2

To find what time does it take to reach maximum height we need to find how high will it go

b) We can calculate its initial height using the following formula

Knowing that its velocity is zero at its maximum height

v_{y}^{2}=v_{o}^{2}+2g(y-y_{o})

0=(64ft/s)^2-2(32ft/s^2)(y-960ft)

y=\frac{-(64ft/s)^2-2(32ft/s^2)(960ft)}{-2(32ft/s^2)}=1024ft

So, the projectile goes 1024 ft high

a) From the equation of height we calculate how long does it take to reach maximum point

h=-16t^2+64t+960

1024=-16t^2+64t+960

0=-16t^2+64t-64

Solving the quadratic equation

t=\frac{-b±\sqrt{b^{2}-4ac}}{2a}

a=-16\\b=64\\c=-64

t=2s

So, the projectile reach maximum point at t=2s

c) We can calculate the final velocity by using the following formula:

v_{y}^{2}=v_{o}^{2}+2g(y-y_{o})

v_{y}=±\sqrt{(64ft/s)^{2}-2(32ft/s^2)(-960ft)}=±256ft/s

Since the projectile is going down the velocity at the instant it reaches the ground is:

v=-256ft/s

5 0
3 years ago
Please answer D in the image with an explanation
puteri [66]

Answer:

The force is 274 N.

Explanation:

In figure 2:

(d) Let the tension in the string is T.

According to the Newton's second law,

Net force = mass x acceleration

Apply for 200N.

T - 200 sin 35 =\frac{200}{9.8}\times a \\T - 114.7 = 20.4 a..... (1)\\220 - T = \frac{220}{9.8}\times a\\220 - T = 22.45 a..... (2)\\Adding both the equations\\334.7 = 42.85 aa =7.81 m/s^{2}

Now put in (1)

T - 114.7 = 20.4 x 7.81

T = 274 N

4 0
2 years ago
Please help this is really confusing
hoa [83]
The correct answer is b , the the liquid with heat and cool faster
8 0
3 years ago
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