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fgiga [73]
3 years ago
13

As a battery is used to charge a capacitor, does the overall charge inside the battery get smaller, greater, or stay the same?

Physics
1 answer:
irga5000 [103]3 years ago
5 0
The Charge Gets Smaller.
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1. What distance is required for a train to stop if its initial velocity is 23 m/s and its
Irina-Kira [14]

Answer:

x=?

dt=?

vi=23m/s

vf=0m/s (it stops)

d=0.25m/s^2

time =

vf=vi+d: 0=23m/s+(0.25m/s^2)t

t=92s

displacement=

vf^2=vi^2+2a(dx)

23^2=0^2+2(0.25m/s^2)x =-1058m

Explanation:

you can find time from vf = vi + a(Dt): 0 = 23 m/s + (0.25 m/s/s)t so t = 92 s and you can find the displacement from vf2 = vi2 + 2a(Dx) and find the answer in one step: 232 = 02 + 2(0.25 m/s/s)x so x = -1058 m

6 0
3 years ago
If two objects of unequal mass collide, both objects will continue to move in the direction of
gayaneshka [121]
Assuming this is an elastic collision, they go in the direction of the object with more mass
3 0
3 years ago
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Two forces F1 and F2 of equal magnitude are applied to a brick of mass 20kg lying on the floor as shown in the figure above. If
jeka57 [31]

Let F_{1}=F_{2}=F.

Normal force equals (using Newton's third law) N=mg+F\sin30^{o}-F\sin37^{o}.

F_{f}\leq \mu N = \mu(mg+F\sin30^{o}-F\sin37^{o}), but F_{f}\leq F(\cos30^o+\cos37^o) for all F_{f} (in order to start moving the break). Therefore F(\cos 30^o+\cos37^o)\geq \mu(mg+F\sin30^{o}-F\sin37^{o}), solving for F: F\geq \frac{\mu mg}{\cos30^o+\cos37^o-\mu\sin30^o+\mu\sin37^o}\approx 46,91\; \textbf{N}

7 0
3 years ago
An irrigation canal has a rectangular cross section. At one point whare the canal is 16.0 m wide, and the water is 3.8 m deep, t
Irina-Kira [14]

Answer:

The depth of the water at this point is 0.938 m.

Explanation:

Given that,

At one point

Wide= 16.0 m

Deep = 3.8 m

Water flow = 2.8 cm/s

At a second point downstream

Width of canal = 16.5 m

Water flow = 11.0 cm/s

We need to calculate the depth

Using Bernoulli theorem

A_{1}V_{1}=A_{2}V_{2}

Put the value into the formula

16.0\times3.8\times2.8=16.5\times x\times 11.0

x=\dfrac{16.0\times3.8\times2.8}{16.5\times11.0}

x=0.938\ m

Hence,  The depth of the water at this point is 0.938 m.

7 0
3 years ago
What force is needed to give a 4.5-kg bowling ball an acceleration of 9 m/s2?
Sloan [31]
F=ma  
f=4.5*9                                           

40.5 N
hope this help
8 0
3 years ago
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