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Fofino [41]
3 years ago
8

A flashlight bulb is connected to a square loop of wire that measures cm on a side as shown in the figure below. assume the bulb

will only light if the potential across its terminals is at least v. the loop is oriented perpendicular to a uniform magnetic field of t. the magnetic field then decreases to zero over a period of time such that the bulb is just barely lit
Physics
1 answer:
LUCKY_DIMON [66]3 years ago
4 0
Mmmmmmmm Ф↓⇄㏑㏒,,⇆↑aaaaaaa
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Road rage is an aggressive driving incident where the driver has:
WARRIOR [948]
I would say that it would be A.) Been insulted. This is because you are angered by someone else. It does not necessarily mean that something has de-escelated, you've lost control of your car, or you've nothing to lose. Hopefully that helps. :)

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Given that the atomic weight of hydrogen is approximately 1 gram per mole, use the method above to estimate the average molecula
denpristay [2]

Answer:

2

Explanation:

6 0
3 years ago
An 8.0 cm object is 40.0 cm from a concave mirror that has a focal length of 10.0 cm. Its image is 16.0 cm in front of the mirro
svet-max [94.6K]
 We can rearrange the mirror equation before plugging our values in. 
1/p = 1/f - 1/q. 
1/p = 1/10cm - 1/40cm
1/p = 4/40cm - 1/40cm = 3/40cm
40cm=3p  <-- cross multiplication
13.33cm = p

Now that we have the value of p, we can plug it into the magnification equation.

M=-16/13.33=1.2
1.2=h'/8cm
9.6=h'

So the height of the image produced by the mirror is 9.6cm.
6 0
2 years ago
What is the only thing that can pull a beam of light towards itself ? ​
Elan Coil [88]

Answer:

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2 years ago
Read 2 more answers
A mass weighting 16 lbs stretches a spring 3 inches. The mass is in a medium that exerts a viscous resistance of 20 lbs when the
const2013 [10]

Answer:

The equation for the object's displacement is u(t)=0.583cos11.35t

Explanation:

Given:

m = 16 lb

δ = 3 in

The stiffness is:

k=\frac{m}{\delta } =\frac{16}{3} =5.33lb/in

The angular speed is:

w=\sqrt{\frac{k}{m} } =\sqrt{\frac{5.33*386.4}{16} } =11.35rad/s

The damping force is:

F_{D} =cu

Where

FD = 20 lb

u = 4 ft/s = 48 in/s

Replacing:

c=\frac{F_{D} }{u} =\frac{20}{48} =0.42lbs/in

The critical damping is equal:

c_{c} =\frac{2k}{w} =\frac{2*5.33}{11.35} =0.94lbs/in

Like cc>c the system is undamped

The equilibrium expression is:

u(t)=u(o)coswt+u'(o)sinwt\\u(o)=7=0.583\\u'(o)=0\\u(t)=0.583coswt\\u(t)=0.583cos11.35t

3 0
3 years ago
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