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3 years ago

The speed that light travels :

Physics

2 answers:

Monica [59]3 years ago

6 0

The answer has to be <span>c.changes when going through different mediums</span>

OLga [1]3 years ago

4 0

I think it might be C.

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A searchlight is 210 ft from a straight wall. As the beam moves along the wall, the angle between the beam and the perpendicula

Ivenika [448]

Answer:

The length of the beam increasing is 9.64 ft/s.

Explanation:

Given that,

Height = 210 ft

Distance =290 ft

According to figure,

We need to calculate the angle

$\cos\theta=\dfrac{210}{x}$ ....(I)

Put the value of x in the equation

$\cos\theta=\dfrac{210}{290}$

$\cos\theta=\dfrac{21}{29}=0.72$

Now, $\sin\theta=\dfrac{20}{29}$

On differentiate of equation (I)

$-\sin\theta\dfrac{d\theta}{dt}=-\dfrac{-210}{x^2}\dfrac{dx}{dt}$

$\sin\theta=\dfrac{210}{x^2}\dfrac{dx}{dt}$

Put the value in the equation

$\sin\dfrac{20}{29}\times2.0=\dfrac{210}{(290)^2}\dfrac{dx}{dt}$

$\dfrac{dx}{dt}=\sin\dfrac{20}{29}\times2.0\times\dfrac{290^2}{210}$

$\dfrac{dx}{dt}=9.64\ ft/s$

Hence, The length of the beam increasing is 9.64 ft/s.

4 0

3 years ago

How to calculate motion

otez555 [7]

Answer:

Distnace per unit time

Explanation:

3 0

3 years ago

Read 2 more answers

What makes a hypothesis testable

Sergio [31]

<span>A hypothesis is testable when you can create an experiment to study the proposition contained within the hypothesis. For example, the hypothesis ‘Santa travels slower than a unicorn’ is testable in theory by measuring the speeds of both, but it is not truly testable because neither exists in reality.</span>

7 0

3 years ago

Read 2 more answers

t. In the fig. the long straight wires at separatioad-16.0 cm carr currents i1=3.6 mA and i2=3.00 i1 out the page a, were on the

timurjin [86]

hope it will help

if it is wrong let me know

8 0

2 years ago

An ore sample weighs 17.50 N in air. When the sample is suspended by a light cord and totally immersed in water, the tension in

valkas [14]

Answer:

Volume of the sample: approximately $\rm 0.6422 \; L = 6.422 \times 10^{-4} \; m^{3}$ .

Average density of the sample: approximately $\rm 2.77\; g \cdot cm^{3} = 2.778 \times 10^{3}\; kg \cdot m^{3}$ .

Assumption:

$\rm g = 9.81\; N \cdot kg^{-1}$ .
$\rho(\text{water}) = \rm 1.000\times 10^{3}\; kg \cdot m^{-3}$ .
Volume of the cord is negligible.

Explanation:

<h3>Total volume of the sample</h3>

The size of the buoyant force is equal to $\rm 17.50 - 11.20 = 6.30\; N$ .

That's also equal to the weight (weight, $m \cdot g$ ) of water that the object displaces. To find the mass of water displaced from its weight, divide weight with $g$ .

$\displaystyle m = \frac{m\cdot g}{g} = \rm \frac{6.30\; N}{9.81\; N \cdot kg^{-1}} \approx 0.642\; kg$ .

Assume that the density of water is $\rho(\text{water}) = \rm 1.000\times 10^{3}\; kg \cdot m^{-3}$ . To the volume of water displaced from its mass, divide mass with density $\rho(\text{water})$ .

$\displaystyle V(\text{water displaced}) = \frac{m}{\rho} = \rm \frac{0.642\; kg}{1.000\times 10^{3}\; kg \cdot m^{-3}} \approx 6.42201 \times 10^{-4}\; m^{3}$ .

Assume that the volume of the cord is negligible. Since the sample is fully-immersed in water, its volume should be the same as the volume of water it displaces.

$V(\text{sample}) = V(\text{water displaced}) \approx \rm 6.422\times 10^{-4}\; m^{3}$ .

<h3>Average Density of the sample</h3>

Average density is equal to mass over volume.

To find the mass of the sample from its weight, divide with $g$ .

$\displaystyle m = \frac{m \cdot g}{g} = \rm \frac{17.50\; N}{9.81\; N \cdot kg^{-1}} \approx 1.78389 \; kg$ .

The volume of the sample is found in the previous part.

Divide mass with volume to find the average density.

$\displaystyle \rho(\text{sample, average}) = \frac{m}{V} = \rm \frac{1.78389\; kg}{6.42201 \times 10^{-4}\; m^{3}} \approx 2.778\; kg \cdot m^{-3}$ .

3 0

3 years ago