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sveticcg [70]
2 years ago
14

Circus a path along which electric current flows how would changing the battery in a circuit from 9 V to 1.5 V most likely affec

t the circuit
Physics
2 answers:
irina [24]2 years ago
7 0
Yeah what he said


















cirus














a










path










along











which















electric
Nimfa-mama [501]2 years ago
4 0
When we reduce the voltage in circuit from 9 volts from 1.5 volts the flow of the current will also get reduced in a circuit.
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ASAP pls answer right if can’t see picture don’t answer
Anna35 [415]

Answer:

Not sure but

F = m* a

32= 5 * a

a= 6.4 m/s^2

6 0
3 years ago
A series of pulses, each of amplitude 0.1 m, is sent down a string that is attached to a post at one end. The pulses are reflect
AleksAgata [21]

Answer:

 A_resulting = 0.2 m

Explanation:

Let's analyze the impact of the pulse with the pole, this is a fixed obstacle that does not move therefore by the law of action and reluctant, the force that the pole applies on the rope is of equal magnitude to the force of the rope on the pole (pulse), but opposite directional, so the reflected pulse reverses its direction and sense.

With this information we analyze a point on the string where the incident pulse is and each reflected with an amplitude A = 0.1 m, the resulting is

           A_res = 2A

           A_resultant = 2 .01

           A_resulting = 0.2 m

8 0
3 years ago
Two 110 kg bumper cars are moving toward each other in opposite directions. Car A is moving at 8 m/s and Car Z at –10 m/s when t
Marysya12 [62]
From the law of conservation of momentum
m1u1+ m2u2= m1v1+ m2v2
110*8+ 110*-10= 110*-10 + 110* v2
v2= 8 m/sec
8 0
3 years ago
What property do the following elements have in common? Li, C, and F A) They are poor conductors of electricity. B) Each element
Aloiza [94]

Answer: Option (D) is the correct answer.

Explanation:

The given elements Li, C and F are all second period elements. So, when we move from left to right across a period then there occurs increase in number of valence electrons as there occurs increase in total number of electrons.

So, it means more electrons are added to the same energy level.

Thus, we can conclude that a property of valence electrons for each element is located in the same energy level is common in the given elements.

7 0
3 years ago
Read 2 more answers
The electric field between two parallel plates is uniform, with magnitude 628 N/C. A proton is held stationary at the positive p
aliina [53]

Answer:

Answer is explained in the explanation section below.

Explanation:

Solution:

Data Given:

Electric Field between two parallel plates = 628 N/C

Separation = 4.22 cm

a) In this part, we are asked to calculate the distance from positive plate at which the electron and proton pass each other.

Solution:

First of all:

Force on proton due to the Electric field between the plates is:

F_{p} = q_{p}E

and, we know that, F = ma

So,

m_{p}a = q_{p}E

a = \frac{q_{p}.E }{m_{p} }      Equation 1

So,

The distance covered by the electron is:

S = ut + 1/2at^{2}

Here, u = 0.

S = 1/2at^{2}

Put equation 1 into the above equation:

S = 1/2 x (\frac{q_{p}.E }{m_{p} }  )t^{2}      Equation 2

So,  

Similarly, the distance covered by electron will be:

(D-S) = 1/2 x (\frac{q_{e}.E }{m_{e} }  )t^{2}    Equation 3

We know that the charge of electron is equal to the charge of proton so,

q_{p} = q_{e} = q

By dividing the equation 2 by equation 3, we get:

\frac{S}{D-S} = \frac{m_{e} }{m_{p} }

Solve the above equation for S,

Sm_{p} = m_{e}D - m_{e}S

So,

S = \frac{m_{e}.D }{(m_{e} + m_{p})  }

Plugging in the values,

As we know the mass of electron is 9.1 x 10^{-31} and the mass of proton is 1.67 x 10^{-27}

S = \frac{9.1 . 10^{-31} . 4.22 }{(9.1 . 10^{-31} + 1.67 . 10^{-27}  }

S = 0.002298 cm (Distance from the positive plate at which the two pass each other)

b) In this part, we to calculate distance for Sodium ion and chloride ion as above.

So,

we already have the equation, we need to put the values in it.

So,

S = \frac{m_{Cl}.D }{(m_{Cl} + m_{Na})  }

As we know the mass of chlorine is 35.5 and of sodium is 23

S = \frac{35.5 . 4.22}{(35.5 + 23)}

S = 2.56 cm

7 0
3 years ago
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