Which one of the following formulas is used to find the voltage of a circuit?

Trial 1: Get a textbook and put a sheet of paper on top of it. Fold the paper as needed to keep the paper from sticking over the

siniylev [52]

Answer:

1) the two objects reach the floor at the same time.

2)the book reaches the floor much earlier than the foil

In conclusion, the difference in motion between the two systems subjected to the same acceleration depends on the weight of the body and friction force, when the body has less weight, the friction of the air affects it more

Explanation:

This interesting experiment has the following results

1) first case. Sheet on top of book

In this case the two objects reach the floor at the same time.

This shows that the acceleration in the two objects is the same and we call it the acceleration of gravity.

The speed of the body increases as it goes down linearly.

This occurs because the book that receives air resistance is much heavier, so the resistance has almost no effect on its movement, the sheet does not have the air resistance because it goes down next to the book.

2) second case. Book and sheet next to each other.

In this case the book reaches the floor much earlier than the foil.

This is because the resisting force of the air has almost no effect on the book and its movement is little affected by this force.

In the case of the blade, it has very little weight, therefore as its speed increases, the resistance force of the air rapidly equals the weight of the blade.

W_sheet - fr = 0

so after this, since the acceleration is zero, it goes down at constant speed, this speed is called the terminal velocity.

In conclusion, the difference in motion between the two systems subjected to the same acceleration depends on the weight of the body and friction force, when the body has less weight, the friction of the air affects it more.

3 0

2 years ago

A bowler throws a bowling ball of radius R = 11 cm down the lane with initial speed v₀ = 7.5 m/s. The ball is thrown in such a w

salantis [7]

Answer:

a) t = 0.74s

b) D = 4.76m

c) Vf = 5.35m/s

Explanation:

The ball starts rolling when Vf = ωf*R.

We know that:

Vf = Vo - a*t

ωf = ωo + α*t

With a sum of forces on the ball:

$Ff = m*a$

$\mu*N = m*a$

$\mu*m*g = m*a$

$a=\mu*g=2.9m/s^2$

With a sum of torque on the ball:

$Ff*R = I*\alpha$

$\mu*m*g*R = 2/5*m*R^2*\alpha$

$\alpha=5/2*\mu*g/R=65.9rad/s^2$

Replacing both accelerations:

$Vo - a*t=\alpha*t*R$

$7.5 - 2.9*t=65.9*t*0.11$

t=0.74s

The distance will be:

$D = Vo*t-1/2*a*t^2$

$D = 4.76m$

Final velocity:

$Vf=Vo-a*t$

Vf=5.35m/s

7 0

2 years ago

Let A^=6i^+4j^_2k^ and B= 2i^_2j^+3k^. find the sum and difference of A and B

andreyandreev [35.5K]

Explanation:

Let $\textbf{A} = 6\hat{\textbf{i}} + 4\hat{\textbf{j}} - 2\hat{\textbf{k}}$ and $\textbf{B} = 2\hat{\textbf{i}} - 2\hat{\textbf{j}} + 3\hat{\textbf{k}}$

The sum of the two vectors is

$\textbf{A + B} = (6 + 2)\hat{\textbf{i}} + (4 - 2)\hat{\textbf{j}} + (-2 + 3)\hat{\textbf{k}}$

$= 8\hat{\textbf{i}} + 2\hat{\textbf{j}} + \hat{\textbf{k}}$

The difference between the two vectors can be written as

$\textbf{A - B} = (6 - 2)\hat{\textbf{i}} + (4 - (-2))\hat{\textbf{j}} + (-2 - 3)\hat{\textbf{k}}$

$= 4\hat{\textbf{i}} + 6\hat{\textbf{j}} - 5\hat{\textbf{k}}$

8 0

3 years ago

An 67-kg jogger is heading due east at a speed of 2.3 m/s. A 70-kg jogger is heading 61 ° north of east at a speed of 1.3 m/s. F

ololo11 [35]

Answer

given,

mass of jogger = 67 kg

speed in east direction = 2.3 m/s

mass of jogger 2 = 70 Kg

speed = 1.3 m/s in 61 ° north of east.

jogger one

$P_1 = m_1 v_1 \hat{i}$

$P_1 = 67 \times 2.3\hat{i}$

$P_1 = 154.1 \hat{i}$

$P_2 = m_2 v_2 \hat{i} +m_2 v_2 \hat{j}$

$P_2 = 70\times v cos \theta \hat{i} +70\times v sin \theta \hat{j}$

$P_2 = 70\times 1.3 cos 61^0 \hat{i} +70\times 1.3 sin 61^0\hat{j}$

$P_2 = 44.12\hat{i} +79.59\hat{j}$

now

P = P₁ + P₂

$P = 198.22 \hat{i} +79.59 \hat{j}$

magnitude

$P = \sqrt{198.22^2 + 79.59^2}$

$P =213.60 kg.m/s$

$\theta = tan^{-1}\dfrac{79.59}{198.22}$

$\theta = 21.87$

the angle is $\theta = 21.87$ north of east

7 0

3 years ago

Two parallel 3.0-meter long wires conduct current. The current in the top wire is 12.5 A and flows to the right. The top wire fe

Aleksandr [31]

Complete question:

Two parallel 3.0-meter long wires conduct current. The current in the top wire is 12.5 A and flows to the right. The top wire feels a repulsive force of 2.4 x 10^-4 N created by the interaction of the 12.5 A current and the magnetic field created by the bottom current (I). Find the magnitude and direction of the bottom current, if the distance between the two wires is 40cm.

Answer:

The bottom current is 12.8 A to the right.

Explanation:

Given;

length of the wires, L = 3.0 m

current in the top wire, I₁ = 12.5 A

repulsive force between the two wires, F = 2.4 x 10⁻⁴ N

distance between the two wires, r = 40 cm = 0.4 m

The repulsive force between the two wires is given by;

$F = \frac{\mu_oI_1I_2L}{2\pi r}\\\\I_{2} = \frac{2F\pi r}{\mu_oI_1L}$

Where;

I₂ is the bottom current

The direction of the bottom current must be in the same direction as the top current since the force between the two wires is repulsive.

$I_{2} = \frac{2F\pi r}{\mu_oI_1L}\\\\I_{2} = \frac{2(2.4*10^{-4})(\pi)(0.4)}{(4\pi*10^{-7})(12.5)(3)}\\\\I_{2} = 12.8 \ A$

Therefore, the bottom current is 12.8 A to the right.

3 0

2 years ago