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GalinKa [24]
1 year ago
6

A shopping mall is putting up lights for the holidays. When they plug the holiday lights in they do not light up.

Physics
1 answer:
mash [69]1 year ago
3 0

Answer:

Explanation:

Remark

Anybody who has dealt  with Christmas Tree lights knows what can be done. You must take a bulb you know to be good and try each bulb in turn until the whole string lights up. If two bulbs are dead, give up and go buy another string. Two bulbs create a lot of combinations.

The circuit is  a series circuit.

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Along a horizontal snow-covered track, a sled, of mass m = 105 kg, slides by the action of a horizontal force of 230 N. The coef
Andrew [12]

Answer:

Explanation:

The only thing I can figure you need here is the accleration of the sled. The equation we need to find this is Newton's Second Law that says that sum of the forces acting on an object is equal to the object's mass times its acceleration. For us, that looks like this because of the friction working against the sled:

F - f = ma but of course it's much more involved than that simple equation! We have the F value as 230 N, and we have the mass as 105, but we do not have the frictional force, f, and we need it to solve for a in the above equation. We know that

f = μF_n where μ is the coefficient of friction, and F_n is the normal force, aka weight of the object. We will use the coefficient of friction and find the weight in order to fill in for f:

F_n=mg so

F_n=(105)(9.8) so the weight of the sled is

F_n= 1.0 × 10³ with the correct number of sig dig there. Now to find f:

f = (.025)(1.0 × 10³) so

f = 25 to the correct number of sig fig. Now on to our "real" equation:

F - f = ma and

230 - 25 = 105a. We have to do the subtraction first, round, and then divide since the rules for addition and subtraction are different from the rules for dividing and multiplying.

230 - 25 will round to the tens place giving us 210. Then

210 = 105a. 210 has 2 sig figs in it while 105 has 3, so we will divide and round to 2 sig fig:

a = 2.0 m/sec²

3 0
2 years ago
Where on this diagram does the ball have the highest point of gravitational potential energy?
mixer [17]
It should be at the very top since it has more space to fall which gives it more potential energy
3 0
2 years ago
How do you think overpumping groudwater is related to the formation of sinkholes?
vladimir2022 [97]
Ground water keept the ground at a stable level when it is gone the cavern it was in has no support and is at risk of callaps
3 0
3 years ago
A flat loop of wire consisting of a single turn of cross-sectional area 8.00 cm2 is perpendicular to a magnetic field that incre
Evgen [1.6K]

Complete Question

A flat loop of wire consisting of a single turn of cross-sectional area 8.00 cm2 is perpendicular to a magnetic field that increases uniformly in magnitude from 0.500 T to 1.60 T in 0.99 s. What is the resulting induced current if the loop has a resistance of 1.20  \ \Omega

Answer:

The current is   I =  0.0007 41 \ A

Explanation:

From the question we are told that

   The  area is  A = 8.00 \ cm^2  = 8.0 *10^{-4} \  m^2

   The initial magnetic field at t_o = 0 \ seconds  is B_i = 0.500 \ T

   The magnetic field at t_1 = 0.99 \ seconds is  B_f  = 1.60 \ T

     The resistance is  R = 1.20  \ \Omega

Generally the induced emf is mathematically represented as

      \epsilon  =  A * \frac{B_f - B_i }{ t_f - t_o }

=>   \epsilon  =  8.0 *10^{-4} * \frac{1.60  - 0.500 }{ 0.99- 0  }

=>   \epsilon  = 0.000889 \ V

Generally the current induced is mathematically represented as

     I = \frac{\epsilon}{R }

=>  I = \frac{0.000889}{ 1.20 }  

=>  I =  0.0007 41 \ A  

6 0
2 years ago
A rocket, initially at rest on the ground, accelerates straight upward from rest with constant acceleration 44.1m/s^2 . The acce
ollegr [7]
The acceleration and distance is related to the following expression:
y=v0*t + a*t^2/2 ; v0=0 
y=44.1*100/2 = 2205m 
hence, the speed will be 
v=0 + a*t = 441m/s 
from that height it will just be subjected to the gravitational acceleration 
0=v_acc^2 -2g*y_free 
y_free = v_acc^2/2g = 9922.5m 
<span>y_max = y_acc+y_free = 441+9922.5 =10363.5m</span>
7 0
2 years ago
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