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gladu [14]
2 years ago
8

While working on her science fair project Venus connected a battery to a circuit that contained a light bulb. Venus decided to c

hange the light bulb to a higher resistance, but she wanted to keep the current the same. What will Venus need to do?
Physics
1 answer:
inessss [21]2 years ago
3 0

Answer:

<em>Venus will need to use more voltage to have the same current</em>

Explanation:

<u>Ohm's Law</u>

It states that the current through a conductor is directly proportional to the voltage applied to it and inversely proportional to the resistance of the conductor. This relationship can be written as

\displaystyle I=\frac{V}{R}

Solving for V

V=IR

Venus originally connected a resistance R_1 to a voltage V_1 to get some current I

\displaystyle I=\frac{V_1}{R_1}

The voltage needed to get that current is

V_1=IR_1

If Venus changes the light bulb to another with higher resistance, say, R_2, then the current would change, unless she uses another voltage

V_2=IR_2

Since\ R_2>R_1,\ then\ V_2>V_1

Venus will need to use more voltage to have the same current

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The electric potential in a particular region of space varies only as a function of y-position and is given by the function V(y)
nikdorinn [45]

Answer:

E = 55.9583\ Volts/meter

Explanation:

First let's find the electric potential using y = 22.5:

V(y) = 1.69y^2 +15.6y+52.5

V(22.5) = 1.69(22.5)^2 + 15.6*22.5 + 52.5

V(22.5) = 1259.0625\ Volts

Then, to find the magnitude of the electric field, we just need to divide the electric potential by the distance y:

E = V/d

E = 1259.0625/22.5

E = 55.9583\ Volts/meter

3 0
2 years ago
A hummingbird can a flutter its wings 4800 times per minute what is the frequency of wing flutters per second
Scilla [17]

the answer i got is 80 i hope this helpss!!!!

7 0
2 years ago
Read 2 more answers
Derive an expression for the total mechanical energy of the system as the monkey reaches the top of the motion, Etop, in terms o
ipn [44]

Answer:

U =  0.5 * k *(x + d - h_max)^2 + m*g*h_max

Explanation:

Given:

- The extension in spring @ equilibrium = x m

- The spring constant = k

- The amount of distance pulled down = d

- mass of the toy = m

Find:

- The total mechanical energy E_top at the top position h_max in terms of the available variables.

Solution:

- First we need to determine the types of Energy that are in play:

- The Elastic potential Energy E_p in a spring is given:

                              E_p: 0.5 * k * (ext)

- In our case when the toy at the top most position h_max will have a net extension ext, by summing displacement of spring:

             ext = Equilibrium + distance pulled - h_max = (x + d - h_max)

Hence, the elastic potential energy will be:

                              E_p = 0.5 * k *(x + d - h_max)^2

- The gravitational potential energy E_g is given by:

                              E_g = m*g*h_max

Where, bottom most position is taken as reference (datum).

- The kinetic Energy E_k is given by:

                              E_k = 0.5*m*v_top^2

- Since we know that the maximum height is reached when velocity is zero

Hence,                   E_k = 0.5*m*0^2 = 0.

The total Energy of the system U is sum of all energies and play:

                               U = E_p + E_k + E_g

                               U =  0.5 * k *(x + d - h_max)^2 + m*g*h_max

8 0
2 years ago
1. What is the momentum of a golf ball with a mass of 62 g moving at 73 m/s?
Anit [1.1K]

Answer:

<h3>The answer is 4.53 kgm/s</h3>

Explanation:

The momentum of an object can be found by using the formula

<h3>momentum = mass × velocity</h3>

From the question

mass = 62 g = 0.062 kg

velocity = 73 m/s

We have

momentum = 0.062 × 73 = 4.526

We have the final answer as

<h3>4.53 kgm/s</h3>

Hope this helps you

4 0
2 years ago
Find the object's speeds v1, v2, and v3 at times t1=2.0s, t2=4.0s, and t3=13s.
Burka [1]
Since this is a distance/time graph, the speed at any time is the slope
of the part of the graph that's directly over that time on the x-axis.

At time  t1 = 2.0 s
That's in the middle of the first segment of the graph,
that extends from zero to 3 seconds.
Its slope is  7/3 .              v1 = 7/3 m/s .

At time  t2 = 4.0 s
That's in the middle of the horizontal part of the graph
that runs from 3 to 6 seconds.
Its slope is zero.
                                     v2 = zero .

At time  t3 = 13 s.
That's in the middle of the part of the graph that's sloping down,
between 11 and 16 seconds.
Its slope is  -3/5 .            v3 = -0.6 m/s .              
7 0
2 years ago
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