In the diagram, R1 = 40.0 ohm, R2 = 25.4 ohm, and R3 = 70.8 ohm. What is the equivalent resistance of the group?​

Answer:

When a polythene rod is rubbed with a duster, the friction causes electrons to gain energy. Electrons gain enough energy to leave the atom and 'rub off' onto the polythene rod. The polythene rod has gained electrons, giving it a negative charge.

Explanation:

Hope it helps

The question is: Two masses with values ​​of 8kg and 12kg each are separated by a distance of 1m. Determine the force of gravitational attraction between them if compared with the value of their weights. What is this value?

Answer: The force of gravitational attraction between them if compared with the value of their weights is [tex]640.704 \times 10^{-11} N[/tex].

Explanation:

Given: [tex]m_{1}[/tex] = 8 kg,       [tex]m_{2}[/tex] = 12 kg

Distance = 1 m

Formula used is as follows.

[tex]F = \frac{G \times m_{1} \times m_{2}}{r^{2}}[/tex]

where,

F = force of gravitational attraction

G = gravitational constant = [tex]6.674 \times 10^{-11} N m^{2}/kg^{2}[/tex]

[tex]m_{1}[/tex] = mass of object 1

[tex]m_{2}[/tex] = mass of object 2

r = distance between the centers of these two objects

Substitute the values into above formula as follows.

[tex]F = \frac{G \times m_{1} \times m_{2}}{r^{2}}\\= \frac{6.674 \times 10^{-11} Nm^{2}/kg^{2} \times 8 kg \times 12 kg}{(1 m)^{2}}\\= 640.704 \times 10^{-11} N[/tex]

Thus, we can conclude that the force of gravitational attraction between them if compared with the value of their weights is [tex]640.704 \times 10^{-11} N[/tex].

Answer:

do you have the answer?

Explanation:

if so whats the answer?

Answer:

R = 0.2 Ω

Explanation:

Para aplicar el concepto de resistencia de un conductor electrico, consideramos la siguiente formula:

R = ρ . l / S

Donde l es la longitud del conductor

S la superficie

y ρ el coeficiente de resistividad

Primero necestiamos hacer ciertas conversiones:

ρ = 0.10 ×10⁻⁶ (Ω . m²/m)

S = 2 mm² → 2 ×10⁻⁶ m²

R = 0.10 ×10⁻⁶ (Ω . m²/m) . 4m / 2 ×10⁻⁶ m²

Dentro de la formula simplificamos las unidades, los metros de ρ con la longitud y ambos m². De esa forma la unidad de R queda en Ω (ohmios).

Resolvemos:

R = 0.10 ×10⁻⁶ Ω . 4 / 2 ×10⁻⁶

R = 0.2 Ω

Answer:

Alpha decay

Explanation:

Answer:

C

Explanation:

Alpha Decay

Answer: I believe that the answer is D.

Explanation:

Hope this helps and have a great day! :)

Answer:

D) all of the above

Explanation:

Answer:

immma send you the link friend me on hereExplanation:

Question:

Find the magnitude and direction of the electric field at points that are at the following distances from the center of the aluminum rod:

(a) 0.5 cm, (b) 1.5 cm, (c) 2.5 cm, (d) 3.5 cm, (e) 7.0 cm

Answer:

(a) 0

(b) 4.795 N/C

(c) 2.877 N/C

(d) 2.055 N/C

(e) 1.027 N/C

Explanation:

The given parameters are;

The inner radius of the copper = 2 cm

The outer radius of the copper - 3 cm

The radius of the aluminum rod, r = 1 cm

The radius of the aluminum rod = 4 pC/m

a) The charge at the 0.5 cm from the center of the aluminum rod, E = 0

There is no charge inside the rod given that the aluminum road is a conductor of electricity, having all the charge on its surface

b) At 1.5 cm, we have;

[tex]E = \dfrac{2 \cdot \lambda}{4 \cdot \pi \cdot r \cdot \epsilon_0}[/tex]

Where;

[tex]\dfrac{ \lambda}{4 \cdot \pi \cdot \epsilon_0} = 8.99 \times 10^9 \times 4 \times 10^{-12}[/tex]

r = 1.5 cm = 0.015 m

∴ E = (2 × 8.99 × 10⁹ × 4 × 10⁻¹²)/0.015 ≈ 4.795

The magnitude of the electric field, E = 4.795 N/C

The direction of the charge depends on the charge sign on the rod

c) At 2.5 cm, we have;

r = 2.5 cm = 0.025 m

E = (2 × 8.99 × 10⁹ × 4 × 10⁻¹²)/0.025 ≈ 2.877

The magnitude of the electric field, E ≈ 2.877 N/C

d) At r = 3.5 cm, we have;

r = 3.5 cm =  0.035 m

E = (2 × 8.99 × 10⁹ × 4 × 10⁻¹²)/0.035 ≈ 2.055

The magnitude of the electric field, E ≈ 2.055 N/C

e) At

r = 7.0 cm = 0.070 m

E = (2 × 8.99 × 10⁹ × 4 × 10⁻¹²)/0.07 ≈ 1.027

The magnitude of the electric field, E ≈ 1.027 N/C

-----------------------------------------------------------------------------------------------------------------

When light strikes a black surface, it is reflected by the surface and

nothing is absorbed    This would be False because black absorbs the light not reflect

Mirrors are used in telescopes because they have the ability to reflect

True? Telescopes designed with mirrors avoid the problems of refracting telescopes. Because the light is reflected from the front surface only, flaws and bubbles within the glass do not affect the path of the light

Transmission in the passing of light through some materials​

False? Some materials allow much of the light that falls on them to be transmitted through the material without being reflected. Materials that allow the transmission of light waves through them are called optically transparent. ... Materials which do not allow the transmission of any light wave frequencies are called opaque.

Answer:

They don’t ‘represent’ anything, they are properties of the wave.

Depending on the type of wave, we experience them as various phenomena. For example, with a sound wave we experience frequency (or wavelength, which is just another way to describe the same property) as the pitch of the sound. We experience amplitude as the loudness of the sound, although due to the characteristics of the ear, frequency also effects perceived loudness.

If the wave is a light wave, we experience the frequency (wavelength) as the colour of the light, and the amplitude as the brightness of the light.

For many waves, we don’t perceive them at all (e.g. radio waves).

For ocean waves, frequency is the time for each peak or trough to reach us, and amplitude is how tall the wave is.

Answer:

A battery is a device that stores chemical energy and converts it to electrical energy. The chemical reactions in a battery involve the flow of electrons from one material (electrode) to another, through an external circuit. The flow of electrons provides an electric current that can be used to do work.

Answer:

v_f = -7.5 m / s

Explanation:

Let's analyze this exercise a little, two forces that act on a body for different time intervals are indicated, each force creates an impulse and since this is a vector quantity we must add in the form of vectors. The net momentum is

we assume that the direction to the right is positive

             I = I₁ + I₂

             I = F₁ Δt₁ - F₂ Δt₂

             I = 3  1.5 - 4  3

             I = -7.5   N s

now let's use the relationship between momentum and momentum, suppose the object starts from rest (vo = 0)

             I = Δp

             I = m (v_f - v₀)

             v_f = I / m

              v_f = -7.5 / m

to finish the calculation we must assume a mass m = 1 kg

              v_f = -7.5 m / s

the negative sign in the body is moving to the left

Answer:

true is the answer of the question

Answer:

it depends on a person's own weight

Answer:

The gravity is roughly the same. The inverse square law applies to gravity and that means that being 60 miles above Earth is about the same as being on Earth. The thing is, they are going really fast in an orbit and thus falling around the Earth. The weightlessness comes from a sort of cancellation of gravity pulling them down and their going perpendicular to that force. The forces are cancelling and making them relatively “weightless”. It’s not really weightlessness but just an equilibrium reached between the two accelerations of gravity and their motion against the pull of the Earth.

Explanation:

Answer:

Explanation:

2) From F=ma

Force =15×40=600N or kgm/s2

3)From the same equation making acceleration the subject of the formula will give

a=f÷m

=24÷4=6m/s2

4)m=f÷a

=45÷15=3kg

Answer: 75 N

Because we're finding the force, we need to use F = ma. M (Mass) = 25 and A (Acceleration) = 3. Multiply 25(3) to get 75. So, you're answer should be 75 N.

Answer:

(i) Please find attached the required velocity time graphs plotted with MS Excel

(ii) The velocity of vehicle A at the 18th second = 20 m/s

The velocity of vehicle B at the 18th second = 0 m/s

(iii) The distance between the two vehicles at the moment in (ii) above is 60 meters

Explanation:

The given parameters of the motion of vehicles A and B are;

The acceleration of vehicles A and B = Uniform acceleration starting from rest

The maximum velocity attained by vehicle A = 30 m/s

The time it takes vehicle A to attain maximum velocity = 10 s

The maximum velocity attained by vehicle B = 30 m/s

The time it takes vehicle B to attain maximum velocity = The time it takes vehicle A to attain maximum velocity = 10 s

The time duration vehicle A maintains its maximum velocity = 6 s

The time duration vehicle B maintains its maximum velocity = 4 s

(i) From the question, we get the following table;

[tex]\begin{array}{ccc}Time &V_A&V_B\\0&0&0\\10&30&40\\14&30&40\\16&30&20\\18&20&0\\22&0&\end{array}[/tex]

From the above table the velocity time graphs of vehicles A and B is created with MS Excel and can included here

(ii) The velocity of vehicle A at the start = 0 m/s

After accelerating for 10 seconds, the velocity of vehicle A = The maximum velocity of vehicle A = 30 m/s

The maximum velocity is maintained for 6 seconds which gives;

At 10 s + 6 s = 16 s, the velocity of vehicle A = 30 m/s

The time it takes vehicle A to decelerate to rest = 6 s

The deceleration of vehicle A, [tex]a_A[/tex] = (30 m/s - 0 m/s)/(6 s) = 5 m/s²

Therefore, we get;

v = u - [tex]a_A[/tex]·t

At the 18th second, the deceleration time, t = 18 s - 16 s = 2 s

u = 30 m/s

∴ v₁₈ = 30 - 5 × 2 = 20

The velocity of vehicle A at the 18th second, [tex]V_{18A}[/tex] = 20 m/s

For vehicle B, we have;

At the 14th second, the velocity of vehicle B = 40 m/s

Vehicle B decelerates to rest in, t = 4 s

The deceleration of vehicle B, [tex]a_B[/tex] = (40 m/s - 0 m/s)/(4 s) = 10 m/s²

For vehicle B, at the 18th second, t = 18 s - 14 s = 4 s

∴ [tex]v_{18B}[/tex] = 40 m/s - 10 m/s² × 4 s = 0 m/s

The velocity of the vehicle B at 18th second, [tex]v_{18B}[/tex] = 0 m/s

(iii) The distance covered by vehicle A up to the 18th second is given by the area under the velocity-time graph as follows;

The area triangle A₁ = (1/2) × 10 × 30 = 150

Area of rectangle, A₂ = 6 × 30 = 180

Area of trapezoid, A₃ = (1/2) × (30 + 20) × 2 = 50

The distance covered in the 18th second by vehicle [tex]S_A[/tex] = A₁ + A₂ + A₃

∴ [tex]S_A[/tex] = 150 + 180 + 50 = 380

The distance covered in the 18th second by vehicle [tex]S_A[/tex] = 380 m

The distance covered by the vehicle B in the 18th second is given by the area under the velocity time graph of vehicle B as follows;

Area of trapezoid, A₅ = (1/2) × (18 + 4) × 40 = 440

The distance covered by the trapezoid, [tex]S_B[/tex] = 440 m

The distance of the two vehicles apart at the 18t second, [tex]S_{AB}[/tex] = [tex]S_B[/tex] - [tex]S_A[/tex]

∴ [tex]S_{AB}[/tex] = 440 m - 380 m = 60 m

The distance of the two vehicles from one another at the 18th second, [tex]S_{AB}[/tex] = 60 m.

Answer:

hope it is helpful to you ☺️☺️✌️

Answer:

The voltage across light bulb 1 and light bulb 2 is the the same i.e V

Explanation:

In a parallel circuit, the Voltage is same across all the components of the circuit and the current flowing through each component is added to get the total current across the circuit.

Let us say, the voltage across the circuit is V. The voltage across light bulb 1 and light bulb 2 is the the same i.e V

Answer:

c and a

Explanation:

ape x

Answer:Shape

I took the quiz

Answer:shape

Explanation:

I took the quiz

Answer:

Solid objects will deform when adequate loads are applied to them; if the material is elastic, the object will return to its initial shape and size after removal. This is in contrast to plasticity, in which the object fails to do so and instead remains in its deformed state.

Explanation:

Answer:

It is false

Explanation:

because when it like a rolling ball it can deform no matter what unlesss you throw it at the ground really hard then it breaks but a soft or sometimes metal yeah it can deform but not all things solid deform.

Answer:

conductor

Does not easily transfer electricity

Answer:

W = 439998 J = 439.99 KJ

Explanation:

First, we will calculate the acceleration of the car by using the first equation of motion:

[tex]v_f = v_i + at\\\\a = \frac{v_f-v_i}{t}[/tex]

where,

a = acceleration = ?

vf = final speed = [tex]108(\frac{km}{h})(\frac{1000\ m}{1\ km})(\frac{1\ h}{3600\ s})[/tex] = 30 m/s

vi = initial speed = [tex]36(\frac{km}{h})(\frac{1000\ m}{1\ km})(\frac{1\ h}{3600\ s})[/tex] = 10 m/s

t = time = 30 s

Therefore,

[tex]a = \frac{30\ m/s - 10\ m/s}{30\ s}[/tex]

a = 0.67 m/s²

Now, we will calculate the force applied by the engine:

F = ma

where,

F = force = ?

m = mass = 1100 kg

Therefore,

F = (1100 kg)(0.67 m/s²)

F = 733.3 N

Now, we will calculate the distance covered by the car by using the second equation of motion:

[tex]s = v_it+\frac{1}{2}at^2\\\\s = (10\ m/s)(30\ s)+\frac{1}{2} (0.67\ m/s^2)(30\ s)^2[/tex]

s = 600 m

Now, the work done (W) by engine can be calculated as follows:

W = Fs

W = (733.3 N)(600 m)

W = 439998 J = 439.99 KJ

Answer:

Halos, blurriness, and night blindness causes night vision (symptoms) while vitamin-A,corrective lenses, such as eyeglasses or contacts, can improve or help night vision.

Explanation:

I bet you will get full Marks if you write it!!

(please note that you will have to mark me as brainliest if it helped you)£

Answer:

120N

Explanation:

Weight=mass x acceleration due to gravity (on the planet)

Mass of an astronaut is 75kg and value of acceleration due to gravity on moon is 1.6m/s2

Weight of astronaut on moon=mass of astronaut x value of acceleration due to gravity on moon

75kg x 1.6m/s2

=120N

Answer:

Acceleration is the name we give to any process where the velocity changes. Since velocity is a speed and a direction, there are only two ways for you to accelerate: change your speed or change your direction—or change both.

Answer: A Graduated Cylinder or A Beaker ?

Explanation: