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3 years ago

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Two tractors pull against a 1000 kg log. If the angle of the tractors' chains in relation to each other is 18.0° (each 9.0° from

being parallel with the log) , and each tractor pulls with a force of 800 N, what forces will they be able to exert?
A. 250 N
B. 1520 N
C. 1580 N
D. 1600 N

I'm 80% sure it's C but I don't know how to go about calculating the answer. Any help is appreciated.

1 answer:

Morgarella [4.7K]3 years ago

7 0

Pie ?????????????? π pizza ??? wut

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28. Which of the following correctly shows the order of highest amount of friction to the lowest amount of

Bad White [126]

Answer:

$\mathrm{d.\:Static,\: Sliding,\:Rolling}$

Explanation:

Static friction occurs when an object initially starts at rest. When the surfaces of the materials touch, the microscopic unevenness interlock greatest with each other, causing the most friction out of the three.

During sliding friction, an object is already moving or in motion. The microscopic surfaces still interlock, but because the object is in motion, it has a momentum. Therefore, the magnitude of sliding friction is less than that of static friction.

Rolling friction occurs when an object rolls across some surface. Rather than surfaces interlocking, rolling friction is caused by the constant distortion of surfaces. As it rolls, the surfaces of the object are constantly wrapping and changing. This distortion causes the rolling friction. However, it is much less in magnitude when compared to static or sliding friction.

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2 years ago

Aloop of wire of area 71 cm^2 is placed with its plane parallel to a 16 mt magnetic field. the loop is then rotated so that its

kkurt [141]

Answer:

Approximately 1.62 × 10⁻⁴ V.

Explanation:

The average EMF in the coil is equal to

$\displaystyle \frac{\text{Final Magnetic Flux} - \text{Initial Magnetic Flux}}{2}$ ,

Why does this formula work?

By Faraday's Law of Induction, the EMF $\epsilon$ induced in a coil (one loop) is equal to the rate of change in the magnetic flux $\Phi$ through the coil.

$\displaystyle \epsilon(t) = \frac{d}{dt}(\Phi(t))$ .

Finding the average EMF in the coil is similar to finding the average velocity.

$\displaystyle \text{Average}\; \epsilon = \frac{1}{t}\int_0^t \epsilon(t)\cdot dt$ .

However, by the Fundamental Theorem of Calculus, integration reverts the action of differentiation. That is:

$\displaystyle \int_0^{t} \epsilon(t)\cdot dt = \int_0^{t} \frac{d}{dt}\Phi(t)\cdot dt = \Phi(t) - \Phi(0)$ .

Hence the equation

$\displaystyle \text{Average}\; \epsilon = \frac{1}{t}\int_0^t \epsilon(t)\cdot dt = \frac{\Phi(t)- \Phi(0)}{t}$ .

Note that information about the constant term in the original function will be lost. However, since this integral is a definite one, the constant term in $\Phi(t)$ won't matter.

Apply this formula to this question. Note that $\Phi$ , the magnetic flux through the coil, can be calculated with the equation

$\Phi = B \cdot A \cdot N \; \sin{\theta}$ .

For this question,

$B = \rm 16\; mT = 16\times 10^{-3}\; T$ is the strength of the magnetic field.
$A = \rm 71\; cm^{2} = 71\times \left(10^{-2}\right)^2 \; m^{2}$ is the area of the coil.
$N = 1$ is the number of loops in the coil.
$\theta$ is the angle between the field lines and the coil.
At $\rm 0\;s$ , the field lines are parallel to the coil, $\theta = 0^{\circ}$ .
At $\rm 0.7\; s$ , the field lines are perpendicular to the coil, $\displaystyle \theta = 90^{\circ}$ .

Initial flux: $\Phi(0)= 0$ .

Final flux: $\Phi(0.7) = \rm 1.1136\times 10^{-4}\; Wb$ .

Average EMF, which is the same as the average rate of change in flux:

$\displaystyle \frac{\Phi(0.7) - \Phi(0)}{0.7} \approx\rm 1.62\times 10^{-4}\; V$ .

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3 years ago

A severe storm has an average peak wave height of 16.4 feet for waves hitting the shore. suppose that a storm is in progress wit

Agata [3.3K]

Before we answer this question, let us first understand what alternate hypothesis is.

The alternative hypothesis is the hypothesis which is used in the hypothesis testing and this is opposite to the null hypothesis. This is the test hypothesis which is usually taken to be that the observations are the result of a real effect in an experiment.

In this case since what we want to set up is the statistical test to see if the waves are dying down, then this means we are trying to determine if the wave height are decreasing, so lesser than 16.4 feet. Therefore:

The alternative hypothesis would state (ANSWER)

Ha: μ less than 16.4 feet and P-value area is on the left of the mean.

While the null hypothesis is the opposite and would state

H0: mu equals 16.4 feet

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3 years ago

Determine the amount of work done on an ideal gas as it is heated in an enclosed thermally insulated cylinder topped with a free

sp2606 [1]

Answer:

W = 3/2 n (T₁- T₂)

Explanation:

Let's use the first law of thermodynamics

ΔE = Q + W

in this case the cylinder is insulated, so there is no heat transfer

ΔE = W

internal energy can be related to the change in temperature

ΔE = 3/2 n K ΔT

we substitute

3/2 n (T₂-T₁) = W

as the work is on the gas it is negative

W = 3/2 n (T₁- T₂)

3 0

3 years ago

Three charges are located at a different position in a plane: q1= 10μC at →r1=(5,6)cm q2=−27μC at →r2=(−6,10)cm and q3=−12μC at

sasho [114]

Answer:

E = (2.29 i ^ - 0.917 j ^) 10⁶ N / C

E_{total} = 2,467 10⁶ N / A θ = -21.8

Explanation:

For this exercise we will use that the electric field is a vector quantity, so the total field is

E_total = E₁₃ + E₂₃

bold font vectors . We can work with the components of the electric field in each axis

X- axis

E_ total x = E₁₃ₓ + E_{23x}

y-axis

E_{total y} = E_{13y} + E_{23y}

the expression for the electric field is

E = k q / r²

where r is the distance between the charge and the positive test charge

in this exercise

Let's find the field created by charge 1

q₁ = 10 μC = 10 10⁻⁶ C

x₁ = 5 cm = 0.05 m

x₃ = 21 cm = 0.21 m

E_{13x} = 9 10⁹ 10 10⁻⁶ / (0.21 -0.05)²

E_{13x} = 3.516 10⁶ N / C

y₁ = 6 cm = 0.06 cm

y₃ = -12 cm = -0.12 m

E_{13y} = 9 10⁹ 10 10⁻⁶ / (-0.12 - 0.06)²

E_{13y} = 2,777 10⁶ N / C

let's find the field produced by charge 2

q₂ = -27 μC = - 27 10⁻⁶ C

x₂ = -6 cm = -0.06 m

x₃ = 0.21 m

E_{23x} = 9 10⁹ 27 10⁻⁶ / (0.21 + 0.06)²

E_{23x} = 1.23 10⁶ N / A

y₂ = 10 cm = 0.10 m

y₃ = -0.12 m

E_{23y} = 9 10⁹ 27 10⁻⁶ / (-0.12 - 0.10)²

E_{23y} = 1.86 10⁶ N / C

Taking the components we can calculate the total electric field, we must use that charge of the same sign repel and attract the opposite sign, remember that the test charge is always considered positive.

E_{total x} = E_{13x} - E_{23x}

E_{total x} = (3.516 - 1.23) 10⁶

E_{total x} = 2.29 10⁶ N / A

E_{total y} = -E_{13y} + E_{23y}

E_{total y} = (-2.777 +1.86) 10⁶ N / A

E_{total y} = -0.917 10⁶ N / A

we can give the result in two ways

E = (2.29 i ^ - 0.917 j ^) 10⁶ N / C

or in the form of modulus and angle, let's use the Pythagorean theorem to find the modulus

E_{total} = √ (E_{total x}^2 + E_{total y}^ 2)

E_{total} = √ (2.29² + 0.917²) 10⁶

E_{total} = 2,467 10⁶ N / A

let's use trigonometry for the angle

tan θ = E_total and / E_totalx

θ = tan⁻¹ E_{total y} / E_{total x}

θ = tan⁻¹ (-0.917 / 2.29)

θ = -21.8

The negative sign indicates that the angle is measured with respect to the x-axis in a clockwise direction.

7 0

3 years ago