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s344n2d4d5 [400]

4 years ago

6

In a TV set, an electron beam moves with horizontal velocity of 4.8 x 10^7 m/s across the cathode ray tube and strikes the scree

n, 50cm away. How far does the electron beam fall while traversing this distance?

1 answer:

devlian [24]4 years ago

7 0

Answer:

$5.32\times 10^{-16} m$

Explanation:

$Speed=\frac {distance}{time}$ hence making time the subject then

$Time=\frac {Distance}{Speed}$

Substituting 50 cm which is equivalent to 0.5 m for distance and velocity as given as $4.8*10^{7}$ we obtain

$Time=\frac {0.5}{4.8*10^{7}}=1.04167\times 10^{-8} s$

From kinematic equation

$s=ut+0.5gt^{2}$ and taking g as 9.81 then

$s=(0*1.04167\times 10^{-8} s)+(0.5*9.81*(1.04167\times 10^{-8} s)^{2})=5.32227\times 10^{-16} m$

$s\approx 5.32\times 10^{-16} m$

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What happens when white light shines through a translucent, red, glass window? a) All colors of light except red are transmitted

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Answer:

b. Red light is transmitted through and reflected by the glass

Explanation:

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3 years ago

A student uses a microwave oven to heat a meal. The wavelength of the radiation is 8.97 cm. What is the energy of one photon of

krek1111 [17]

Answer:

The energy of one photon is 2.21x10⁻²⁴ J. Multiplied by 10²⁵ is 22.10 J.

Explanation:

The energy (E) of a photon is:

$E = h\frac{c}{\lambda}$

Where:

h: is the Planck's constant = 6.62x10⁻³⁴ J.s

λ: is the wavelength of the radiation = 8.97 cm

c: is the speed of light = 3.00x10⁸ m/s

$E = h\frac{c}{\lambda} = 6.62 \cdot 10^{-34} J.s\frac{3.00\cdot 10^{8} m/s}{8.97 \cdot 10^{-2} m} = 2.21 \cdot 10^{-24} J$

Hence, the energy of one photon is 2.21x10⁻²⁴ J.

Now, if we multiply the answer by 10²⁵ we have:

$E = 2.21 \cdot 10^{-24} J \cdot 10^{25} = 22.10 J$

I hope it helps you!

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3 years ago

Which task would a colorectal surgeon perform?

topjm [15]

Answer: 4. removing tumors in the large intestine

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3 years ago

Give an example of something moving that is not accelerating

Anastaziya [24]

Anything that doesnt have a change in magnitude or direction is something thats not accelerating. so, for example, a car moving at a constant speed of 5mph on a flat road is an example of an object moving but not accelerating

6 0

3 years ago

A student measures the diameter of a small cylindrical object and gets the following readings: 4.32, 4.35, 4.31, 4.36, 4.37, 4.3

Zinaida [17]

Answer:

a. $\bar{d}=4.34 cm$

b. $\sigma=0.023 cm$

c. $\rho=(0.0089\pm 0.00058) kg/cm^{3}$

Explanation:

a) The average of this values is the sum each number divided by the total number of values.

$\bar{d}=\frac{\Sigma_{i=1}^(N)x_{i}}{N}$

$x_{i}$ is values of each diameter
N is the total number of values. N=6

$\bar{d}=\frac{4.32+4.35+4.31+4.36+4.37+4.34}{6}$

$\bar{d}=4.34 cm$

b) The standard deviation equations is:

$\sigma=\sqrt{\frac{1}{N}\Sigma^{N}_{i=1}(x_{i}-\bar{d})^{2}}$

If we put all this values in that equation we will get:

$\sigma=0.023 cm$

Then the mean diameter will be:

$\bar{d}=(4.34\pm 0.023)cm$

c) We know that the density is the mass divided by the volume (ρ = m/V)

and we know that the volume of a cylinder is: $V=\pi R^{2}h$

Then:

$\rho=\frac{m}{\pi R^{2}h}$

Using the values that we have, we can calculate the value of density:

$\rho=\frac{1.66}{3.14*(4.34/2)^{2}*12.6}=0.0089 kg/cm^{3}$

We need to use propagation of error to find the error of the density.

$\delta\rho=\sqrt{\left(\frac{\partial\rho}{\partial m}\right)^{2}\delta m^{2}+\left(\frac{\partial\rho}{\partial d}\right)^{2}\delta d^{2}+\left(\frac{\partial\rho}{\partial h}\right)^{2}\delta h^{2}}$

δm is the error of the mass value.
δd is the error of the diameter value.
δh is the error of the length value.

Let's find each partial derivative:

1. $\frac{\partial\rho}{\partial m}=\frac{4m}{\pi d^{2}h}=\frac{4*1.66}{\pi 4.34^{2}*12.6}=0.0089$

2. $\frac{\partial\rho}{\partial d}=-\frac{8m}{\pi d^{3}h}=-\frac{8*1.66}{\pi 4.34^{3}*12.6}=-0.004$

3. $\frac{\partial\rho}{\partial h}=-\frac{4m}{\pi d^{2}h^{2}}=-\frac{4*1.66}{\pi 4.34^{2}*12.6^{2}}=-0.00071$

Therefore:

$\delta\rho=\sqrt{\left(0.0089)^{2}*0.05^{2}+\left(-0.004)^{2}*0.023^{2}+\left(-0.00071)^{2}*0.5^{2}}$

$\delta\rho=0.00058$

So the density is:

$\rho=(0.0089\pm 0.00058) kg/cm^{3}$

I hope it helps you!

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3 years ago