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alexandr402 [8]
2 years ago
7

A 580-mm long tungsten wire, with a 0.046-mm-diameter circular cross section, is wrapped around in the shape of a coil and used

as a filament in an incandescent light bulb. When the light bulb is connected to a battery, a current of 0.526 A is measured through the filament. (Note: tungsten has a resistivity of 4.9 × 10-8 ? • m.) How many electrons pass through this filament in 5 seconds?
How many electrons pass through this filament in 5 seconds? 4.22 × 10-19 2.63 3.29 × 1018 1.64 × 1019

What is the resistance of this filament? What is the resistance of this filament? 4.28 ? 58.5 m? 17.1 ? 9.0 ?

What is the voltage of the battery that would produce this current in the filament? 31 mV 9.0 V 2.6 × 10-8 V 32 V

Physics
1 answer:
AleksAgata [21]2 years ago
3 0

Explanation:

Below is an attachment containing the solution.

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A ladybug sits at the outer edge of a turntable, and a gentleman bug sits halfway between her and the axis of rotation. The turn
Cerrena [4.2K]

Answer:

e. Not enough information to determine

Explanation:

This question is incomplete. Here is the complete question with my solution afterwards;

A ladybug sits at the outer edge of a turntable, and a gentleman bug sits halfway between her and the axis of rotation. The turntable (initially at rest) begins to rotate with its rate of rotation constantly increasing.

What is the first event that will occur?(Assume non-zero frictional force and the same coefficients of friction for both bugs.)

a. The ladybug begins to slide

b. The gentleman bug begins to slide

c. Both bugs begin to slide at the same time

d. Nothing ever happens

e. Not enough information to determine

The centripetal force acting on a rotating body or bugs can be written as,

F=mrw^2

m= mass of the corresponding bugs

r= corresponding radial distance of each bug

w= angular speed of the turntable

The centripetal force tries to slide the bugs in an outward direction and it is directly proportional to the products of its mass and radial distance from the axis of rotation of the turntable

F ∝ mr

Since the radial distance from the axis of rotation of the turntable for each bug is given, but the mass is not given, the given information is therefore not enough to determine which bugs will slide first.

Option "e" is correct.

7 0
2 years ago
Read 2 more answers
Wile E. Coyote wants to launch Roadrunner into the air using a long lever asshown below. The lever starts at rest before the Coy
algol13

Answer:

Explanation:

Find attached the solution

4 0
2 years ago
A mover hoists a 50 kg box from the ground to a height of 2 m. What was the change in the box's energy
SSSSS [86.1K]

Answer:

980 J

Explanation:

The change in box's energy is equal to its change in gravitational potential energy:

\Delta U = m g \Delta h

where

m = 50 kg is the mass of the box

g = 9.8 m/s^2 is the acceleration due to gravity

\Delta h= 2m is the change in height of the box

Substituting numbers, we find

\Delta U = (50 kg)(9.8 m/s^2)(2 m)=980 J

3 0
2 years ago
You and your friend throw balloons filled with water from the roof of a several story apartment house. You simply drop a balloon
Aleks [24]

Answer:

Height = 53.361 m

Explanation:

There are two balloons being thrown down, one with initial speed (u1) = 0 and the other with initial speed (u2) = 43.12

From the given information we make the following summary

u_{1} = 0m/s

t_{1} = t

u_{2} = 43.12m/s

t_{2} = (t-2.2)s

The distance by the first balloon is

D = u_{1} t_{1}  + \frac{1}{2} at_{1}^2

where

a = 9.8m/s2

Inputting the values

D = (0)t + \frac{1}{2} (9.8)t^2\\ D = 4.9t^2

The distance traveled by the second balloon

D = u_{2} t_{2}  + \frac{1}{2} at_{2}^2

Inputting the values

D = (43.12)(t-2.2)  + \frac{1}{2} (9.8)(t-2.2)^2

simplifying

D = 4.9t^2 + 21.56t -71.148

Substituting D of the first balloon into the D of the second balloon and solving

4.9t^2 = 4.9t^2 + 21.56t -71.148 \\ 21.56t = 71.148\\ t = 3.3s

Now we know the value of t. We input this into the equation of the first balloon the to get height of the apartment

D = 4.9(3.3)^2\\ D = 53.361 m

7 0
2 years ago
A guitar string has a linear density of 8.30 ✕ 10−4 kg/m and a length of 0.660 m. the tension in the string is 56.7 n. when the
Sedbober [7]
Ans: Beat Frequency = 1.97Hz

Explanation:
The fundamental frequency on a vibrating string is 

f =   \sqrt{ \frac{T}{4mL} }<span>  -- (A)</span>

<span>here, T=Tension in the string=56.7N,
L=Length of the string=0.66m,
m= mass = 8.3x10^-4kg/m * 0.66m = 5.48x10^-4kg </span>


Plug in the values in Equation (A)

<span>so </span>f = \sqrt{ \frac{56.7}{4*5.48*10^{-4}*0.66} }<span> = 197.97Hz </span>

<span>the beat frequency is the difference between these two frequencies, therefore:
Beat frequency = 197.97 - 196.0 = 1.97Hz
-i</span>
3 0
2 years ago
Read 2 more answers
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