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3 years ago

How did conflict between countries affect trade along the Silk Road?

Physics

2 answers:

Eva8 [605]3 years ago

6 0

Answer:

A: It decreased trade because soldiers no longer protected the oases.

I took it on E2020. I hope I've helped. If you need further explanation, please comment below.

Alborosie3 years ago

5 0

Answer:

Yes it is A

It decreased trade because soldiers no longer protected the oases.

Explanation:

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Calculate the frequency of a wave using an equation

Gemiola [76]

Answer:

1. f=1T. where: f is the frequency of the wave in hertz. T is the period of the wave in seconds.

2. f=vλ where: f is the frequency of the wave in hertz. v is the velocity of the wave in meters per second. λ is the wavelength of the wave in meters

3. f=cλ

Explanation:

8 0

3 years ago

Taking the resistivity of platinoid as 3.3 x 10-7 m, find the resistance of 7.0 m of platinoid wire of average diameter 0.14 cm.

mr_godi [17]

Answer:

$1.5 \Omega$

Explanation:

The resistance of a wire is given by the equation:

$R=\rho \frac{L}{A}$

where

$\rho$ is the resistivity of the material

L is the length of the wire

A is the cross-sectional area of the wire

In this problem, we have a wire of platinoid, whose resistivity is

$\rho = 3.3\cdot 10^{-7} \Omega m$

The length of the wire is

L = 7.0 m

And its radius is

$r=\frac{0.14 cm}{2}=0.07 cm = 7\cdot 10^{-4} m$ , so the cross-sectional area is

$A=\pi r^2=\pi(7\cdot 10^{-4})^2=1.54\cdot 10^{-6}m^2$

Solving for R, we find the resistance of the wire:

$R=(3.3\cdot 10^{-7})\frac{7.0}{1.54\cdot 10^{-6}}=1.5 \Omega$

3 0

3 years ago

The sun is 60° above the horizon. Rays from the sun strike the still surface of a pond and cast a shadow of a stick that is stuc

Kipish [7]

Answer:

shadow length 7.67 cm

Explanation:

given data:

refractive index of water is 1.33

by snell's law we have

$n_{air} sin30 =n_{water} sin\theta$

$1*0.5 = 1.33*sin\theta$

solving for $\theta$

$sin\theta = \frac{3}{8}$

$\theta = sin^{-1}\frac{3}{8}$

$\theta = 22 degree$

from shadow- stick traingle

$tan(90-\theta) = cot\theta = \frac{h}{s}$

$s = \frac{h}{cot\theta} = h tan\theta$

s = 19tan22 = 7.67 cm

s = shadow length

5 0

3 years ago

A large crate with mass m rests on a horizontal floor. The static and kinetic coefficients of friction between the crate and the

Mariana [72]

a) $F=\frac{\mu_k mg}{cos \theta - \mu_k sin \theta}$

Here the crate is moving at constant velocity, so no acceleration:

a = 0

Let's analyze the forces acting along the horizontal and vertical direction.

- Vertical direction: the equation of the forces is

$R-Fsin \theta - mg = 0$ (1)

where

R is the normal reaction of the floor (upward)

$F sin \theta$ is the component of the force F in the vertical direction (downward)

mg is the weight of the crate (downward)

- Horizontal direction: the equation of the forces is

$F cos \theta - \mu_k R = 0$ (2)

where

$F cos \theta$ is the horizontal component of the force F (forward)

$\mu_k R$ is the force of friction (backward)

From (1) we get

$R=Fsin \theta +mg$

And substituting into (2)

$F cos \theta - \mu_k (Fsin \theta +mg) = 0\\F cos \theta -\mu _k F sin \theta = \mu_k mg\\F(cos \theta - \mu_k sin \theta) = \mu_k mg\\F=\frac{\mu_k mg}{cos \theta - \mu_k sin \theta}$

b) $\mu_s=cot(\theta)$

In this second case, the crate is still at rest, so we have to consider the static force of friction, not the kinetic one.

The equations of the forces will be:

$R-Fsin \theta - mg = 0$ (1)

$F cos \theta - \mu_s R = 0$ (2)

In this second case, we want to find the critical value of $\mu_s$ such that the woman cannot start the crate: this means that the force of friction must be at least equal to the component of the force pushing on the horizontal direction, $F cos \theta$ .

Therefore, using the same procedure as before,

$R=Fsin \theta +mg$

$F cos \theta - \mu_s (Fsin \theta +mg) = 0$

And solving for $\mu_s$ ,

$F cos \theta = \mu_s (Fsin \theta +mg) \\\mu_s = \frac{F cos \theta}{F sin \theta + mg}$

Now we analyze the expression that we found. We notice that if the force applied F is very large, $F sin \theta >> mg$ , therefore we can rewrite the expression as

$\mu_s \sim \frac{F cos \theta}{F sin \theta}\\\mu_s=cot(\theta)$

So, this is the critical value of the coefficient of static friction.

8 0

3 years ago

Whenever a musician plays a guitar, they pluck one of the guitar strings to produce a standing wave in the string. Then pinch do

UkoKoshka [18]

Answer:

It's due to the distance from either ends of strings origin...

Explanation:

As we know that waves behave moving in a flow from one side to another side and this gives a prospective of motion. Suppose a wave is pinched from the near one end of a guitar then due to the distortion created by the point of tie of strings the wave super imposes and moves with a velocity v and produces a wave frequency f. as we the pinching go down to the center the wave stabilizes itself to a stationary origin right at the center and the frequency then changes accordingly as moving down on the string.

7 0

3 years ago