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GenaCL600 [577]
2 years ago
5

If I = 2.0 A in the circuit segment shown below, what is the potential difference VB - VA?

Physics
1 answer:
Tom [10]2 years ago
7 0

Answer:

10 V

Explanation:

The potential difference between two points is the amount of work required to carry a unit charge from one point to the other point. This would result in a potential difference between this two points.

The difference between the potential across two points B and A is V_{BA}=V_B-V_A

From the image attached:

V_B-V_A=10-10I+20\\\\But\ I = 2A,hence:\\\\V_B-V_A=10-10(2)+20\\\\V_B-V_A=10-20+20\\\\V_B-V_A=10\ V

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A particular interaction force does work wint inside a system. the potential energy of the interaction is u. which equation rela
larisa86 [58]

ΔU = -Wint

Consdier the work of of interaction is W =m*g*h - equation -1

and the Potential energy U.

Final Potential energy Uf =0 , And the Initial Potential Energy Ui =m*g*h

<span>Now we will write the equation for a Change in Potential energy ΔU,</span>

ΔU = Uf - Ui

= 0-m*g*h

<span>  ΔU = -m*g*h --Equation 2</span>

Now compare the both equation

<span>Wint = -ΔU</span>

we can rewrite the above equation

ΔU = -W.

<span>So our Answer is ΔU = -W. .</span>

<span> </span>

5 0
3 years ago
A cube that has a volume of 1.00 m³ contains N distinguishable particles.
grandymaker [24]

Answer:

1 P = 0.5

2 P = 0.3

3 P = 0.01

Explanation:

The probability formula is

          P =V^N

Where P is the probability  V is the volume while N is the number of distinguishing particles

So for N = 1 and V = 0.500m^3

           P = (0.500m^3)^1\\

               = 0.5

For  N = 1 and V =  0.300m^3

           P = (0.300m^3)^1

               = 0.3

For   N = 1 and V = 0.0100m^3

     P =(0.0100m^3)^1

         = 0.01

         

6 0
2 years ago
Suppose a 1300 kg car is traveling around a circular curve in a road at a constant
Scrat [10]

Answer:

F = 4212 N

Explanation:

Given that,

Mass of a car, m = 1300 kg

Speed of car on the road is 9 m/s

Radius of curve, r = 25 m

We need to find the magnitude of the unbalanced force that steers the car out of its natural straight-  line path. The force is called centripetal force. It can be given by :

F=\dfrac{mv^2}{r}\\\\F=\dfrac{1300\times 9^2}{25}\\\\F=4212\ N

So, the force has a magnitude of 4212 N

4 0
2 years ago
if you are in Birmingham AL, and you want to use your cell phone to talk to your cousin in Houston, TX, what must occur in order
lesantik [10]

Answer:

For communication to be communicated between you and your cousin who is in Houston,Texas, when a call is made by you, a request is made to the specific phone, and the telephone tower will get the request from the mobile phone. Then a signal is sent via a transmitter underground, by this the satellite communicates with the local receiver in Houston, that is linked to the local tower over there, the tower would request for your cousin's number and connects the two of you, once a link is set.

Explanation:

From the example stated, what is required for such for a far distance, is a communication satellite link.

When a call is made by you, the a connection request is sent to the specified phone.The telephone tower receives the request from The mobile phone. The local tower(Birmingham,Al) is linked to a ground transmitter by the means of a Fiber optical cable.

A signal is sent to satellite via the ground transmitter.The satellite then set's off the local receiver in (Houston,Texas) which on it's end is connected to the local tower there. This tower then ask for your cousin's mobile for a call that will be incoming, a link is set, once he/she receives the call, from there a conversation can be done.

3 0
3 years ago
Two streams merge to form a river. One stream has a width of 8.4 m, depth of 3.5 m, and current speed of 2.2 m/s. The other stre
hoa [83]

Answer:

Explanation:

We shall solve this problem on the basis of pinciple that water is incompressible so volume of flow will be equal at every point .

rate of volume flow of one stream

= cross sectional area x velocity

= 8.4 x 3.5 x 2.2 = 64.68 m³ /s

rate of volume flow of other stream

= 6.6 x 3.6 x 2.7

= 64.15 m³ /s

rate of volume flow of rive , if d be its depth

= 11.2 x d x 2.8

= 31.36 d

volume flow of river = Total of volume flow rate of two streams

31.36 d  = 64.15 + 64.68

31.36 d  = 128.83

d = 4.10 m /s .

6 0
2 years ago
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