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maw [93]
4 years ago
8

The voltage V in a simple electrical circuit is slowly decreasing as the battery wears out. The resistance R is slowly increasin

g as the resistor heats up. Use Ohm's Law, V = IR, to find how the current.
Physics
1 answer:
Ganezh [65]4 years ago
3 0

Answer:

The current in the circuit decrease slowly .

Explanation:

Given as :

For the electrical circuit

The voltage V in the circuit is slowly decreasing

The resistance R of the resistor slowly increasing after heating

Now, From Ohm's Law

Voltage is directly proportional to the flow of current through circuit

I.e V ∝ I

Or. V = R × I

where R is the proportionate constant and this is the resistance of the resistor

whose property is to opposes the flow of current in the circuit

So, If R value more then current I reduces in the circuit

∵ Here in the circuit ,  The resistance is slowly increasing, so, current I is slowly decreasing .

Hence The current in the circuit decrease slowly . answer

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A ball of mass m= 450.0 g traveling at a speed of 8.00 m/s impacts a vertical wall at an angle of θi =45.00 below the horizontal
Mkey [24]

Answer:

   F = 20.4 i ^

Explanation:

This exercise can be solved using the ratio of momentum and amount of movement.

     I = F t = Dp

Since force and amount of movement are vector quantities, each axis must be worked separately.

X axis

Let's look for speed

      cos 45 = vₓ / v

      vₓ = v cos 45

      vₓ = 8 cos 45

      vₓ = 5,657 m / s

We write the moment

Before the crash                          p₀ = m vₓ

After the shock                            p_{f} = -m vₓ

The variation of the moment      Δp = mvₓ - (-mvₓ) = 2 m vₓ

The impulse on the x axis           Fₓ t = Δp

       

        Fₓ = 2 m vₓ / t

        Fx = 2 0.450 5.657 / 0.250

        Fx = 20.4 N

We perform the same calculation on the y axis

       sin  45 = vy / v

       vy = v sin 45

       vy = 8 sin 45

       vy = 5,657 m / s

We calculate the initial momentum   po = m v_{y}

Final moment                                      p_{f} = m v_{y}

Variations moment                             Δp = mv_{y} - mv_{y} = 0

Force in the Y-axis                             F_{y} = 0

Therefore the total force is

       F = fx i ^ + Fyj ^

       F = Fx i ^

       F = 20.4 i ^

3 0
4 years ago
Two metal disks, one with radius R1 = 2.45 cm and mass M1 = 0.900 kg and the other with radius R2 = 5.00 cm and mass M2 = 1.60 k
larisa86 [58]

Answer:

part (a) a_1\ =\ 2.9\ kg

Part (b) a_2\ =\ 6.25\ kg

Explanation:

Given,

  • Mass of the larger disk = M_2\ =\ 1.60\ kg
  • Mass of the smaller disk = M_1\ =\ 0.900\ kg
  • Radius of the larger disk = R_2\ =\ 5.00\ cm\ =\ 0.05\ m
  • Radius of the smaller disk = R_1\ =\ 2.45\ cm\ =\ 0.0245\ m
  • Mass of the block = M = 1.60 kg

Both the disks are welded together, therefore total moment of inertia of the both disks are the summation of the individual moment of inertia of the disks.

\therefore I\ =\ I_1\ +\ I_2\\\Rightarrow I\ =\ \dfrac{1}{2}M_1R_1^2\ +\ \dfrac{1}{2}M_2R_2^2\\\Rightarrow I\ =\ \dfrac{1}{2} (0.9\times 0.0245^2\ +\ 1.60\times 0.05^2)\\\Rightarrow I\ =\ 2.27\times 10^{-3}\ kgm^2

part (a)

Given that a block of mass m which is hanging with the smaller disk,

Let 'T' be 'a' be the tension in the string and acceleration of the block.

From the free body diagram of the smaller block,

mg\ -\ T\ =\ ma\\\Rightarrow T\ =\ mg\ -\ ma\,\,\,\,eqn (1)

From the pulley,

\sum \tau\ =\ I\alpha\\\Rightarow T\times R_1\ =\ I\alpha\ =\ \dfrac{Ia}{R_1}\\\Rightarrow T\ =\ \dfrac{I\alpha}{R_1^2}\,\,\,eqn(2)

From the equation (1) and (2),

mg\ -\ ma\ =\ \dfrac{Ia}{R_1^2}\\\Rightarrow a\ =\ \dfrac{mg}{\dfrac{I}{R_1^2}\ +\ m}\\\Rightarrow a\ =\ \dfrac{1.60\times 9.81}{\dfrac{2.27\times 10^{-3}{0.0245^2}}\ +\ 1.60}\\\Rightarow a\ =\ 2.91\ m/s^2

part (b)

Above expression for the acceleration of the block is only depended on the radius of the pulley.

Radius of the larger pulley = R_2\ =\ 0.05\ m

Let a_2 be the acceleration of the block while connecting to the larger pulley.\therefore a\ =\ \dfrac{mg}{\dfrac{I}{R_2^2}\ +\ m}\\\Rightarrow a\ =\ \dfrac{1.60\times 9.81}{\dfrac{2.27\times 10^{-3}{0.05^2}\ +\ 1.60}}\\\Rightarow a\ =\ 6.25\ m/s^2

4 0
3 years ago
If a person (weighs 70kg) jumped of a moving car (at 100km/h) and fell on asphalt what is the amount of force applyied to his bo
jok3333 [9.3K]

Answer:

The amount of force applied to his body is 1944.44 N

<em>The chances of the person dying is very high owing to the high impact force with which the person would experience when he or she lands on the asphalt road due to the jump out of the moving car.</em>

Explanation:

We all know that,

F = Ma where,

F = Force

M = weight of the person

a = acceleration or velocity of the moving car

Therefore;

F = { 70 x (100 x 1000) } / [3600]

= [7 000 000] / 3600

= <u>1944.44 N</u>

6 0
3 years ago
What is density? Write a formula by showing the relation of mass density and volume​
slava [35]

Answer: below

Explanation: Density, mass of a unit volume of a material substance. The formula for density is d = M/V, where d is density, M is mass, and V is volume

7 0
3 years ago
An early model of the atom, proposed by Rutherford after his discovery of the atomicnucleus, had a positive point charge +Ze(the
Amanda [17]

Answer:

a)  E = k Ze (1- r³ / R³)  1/r², b) E=0, c)   E = -6.62 10¹⁰  N / C

Explanation:

a) For this we can use the law of Gauus

         Ф = E- dA = q_{int} / ε₀

where we take a sphere as a Gaussian surface, so that the electric field lines and the radii of the sphere are parallel, consequently the dot product is reduced to the algebraic product

       E A =q_{int} / ε₀

the area of ​​a sphere  

      A = 4π r²

      E 4π r² = q_{int} / ε₀

      E = 1 / 4πε₀   q_{int} / r²

       k = 1 /4π ε₀

       E = k q_{int} / r²       (1)

       

let's analyze the charge inside the gaussian sphere,

let's use the concept of density for electrons, since they indicate that the charge is evenly distributed

     ρ = Q / V

where the volume of the sphere is

    V = 4/3 πr³

     Qe = ρ V

     Qe = ρ 4 / 3π r³

the density of the electrons is

     ρ = Ze 3 / (4π R³)

where R is the atomic radius

we substitute

       Qe = Ze r³/ R³

for protons they are in a very small space, the atomic nucleus, so we can superno that they are a point charge.

The net charge inside our Gaussian surface, the charge of the protoens plus the charge of the electroens (Qe)

     q_{int} = q_proton + Q_electron

     q_{int} = + Ze - Qe

     q_{int} = + Ze - Ze r³ / R³

     q_{int} = Ze (1- r³ / R³)

   

  we substitute in equation 1

     E = k Ze (1- r³ / R³)  1/r²

b) on the surface of the atom r = R

therefore the electric field is zero

      E = 0

c) Calculate the electric field for the Uranium for

       r = R / 2 = 0.10 10⁻⁹ / 2 = 0.05 10⁻⁹ = 5 10⁻¹¹ m

     

       E = 8.99 10⁹ 92 1.6 10⁻¹⁹ (1-1/2)³   1/ (5 10⁻¹¹)²

       E = -6.62 10¹⁰  N / C

6 0
4 years ago
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