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3 years ago

Calculate the speed of a gamma ray with a frequency of 3.0 x 10^19 Hz and a wavelength of 1.0 x 10^-11 m.

Physics

1 answer:

Ulleksa [173]3 years ago

6 0

Answer:

Speed of gamma rays = 3 x 10⁸ m/s

Explanation:

Given:

Frequency of gamma ray = 3 x 10¹⁹ Hz

Wavelength of gamma rays = 1 x 10⁻¹¹ meter

Find:

Speed of gamma rays

Computation:

Velocity = Frequency x wavelength

Speed of gamma rays = Frequency of gamma ray x Wavelength of gamma rays

Speed of gamma rays = [3 x 10¹⁹][1 x 10⁻¹¹]

Speed of gamma rays = 3 x [10¹⁹⁻¹¹]

Speed of gamma rays = 3 x [10⁸]

Speed of gamma rays = 3 x 10⁸ m/s

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Solve for work when

BlackZzzverrR [31]

So, <u>the value of the work is approximately 84.65 J</u>.

<h2>Introduction</h2>

Hi ! Here I will help you to discuss the subject about work that caused by force in amount value of angle. Work is affected by the force and displacement.

If related to the magnitude of the force, the amount of work will be proportional to the magnitude of the applied force. Thats mean, if the value of the force that applied on it is greater, then the value of the work will be greater.
If related to the magnitude of shift, the amount of work will be proportional to the magnitude of shift of object. Thats mean, if the value of the shift on it is greater, then the value of the work will be greater.

<h3>Formula Used</h3>

The work done by a moving object can be expressed in the equation:

If the Angle Is Ignored

$\boxed{\sf{\bold{W = F \times s}}}$

If the Angle Effect on Work

$\boxed{\sf{\bold{W = F \times s \times \cos(\theta)}}}$

With the following condition:

W = work that done by object (J)
F = force that applied (N)
s = shift or distance (m)
$\sf{\theta}$ = angle of elevation (°)

<h3>Solution</h3>

We know that :

F = force that applied = $\sf{1.41 \times 10^4}$ N
s = shift or distance = 84.9 m
$\sf{\theta}$ = angle of elevation = 45°

What was asked ?

W = work that done by object = ... J

Step by step :

$\sf{W = F \times s \times \cos(\theta)}$

$\sf{W = (1.41 \cdot 10^4) \times 84.9 \times \cos(45^o)}$

$\sf{W = (1.41 \cdot 10^4) \times 84.9 \times \frac{\sqrt{2}}{2}}$

$\sf{W = 119.709 \times \frac{\sqrt{2}}{2}}$

$\sf{W = 59.8545 \sqrt{2}}$

$\boxed{\sf{W \approx 84.65 \: J}}$

<h3>Conclusion</h3>

So, the value of the work is approximately 84.65 J.

3 0

1 year ago

You push on a box with a force of 300 N directly north. Another person pushes the box with a

tangare [24]

Answer:

$resultant \\ \: F = \sqrt{ {300}^{2} + {600}^{2} } \\ = \sqrt{450000} \\ = 670.82 \: newtons$

7 0

3 years ago

While standing on the edge of the grand canyon, you see a rattlesnake and scream. You hear the echo of the scream off canyon flo

bezimeni [28]

Answer:

1904 m

Explanation:

$\pink{\frak{Given}}\begin{cases} \textsf{ You scream after seeing a rattle snake .}\\\textsf{Scream is heard after 11.2 s .} \end{cases}$

Here we need to find out the depth of the canyon . When you will scream after seeing a snake , the sound produced will travel till the end of the canyon and after hitting the end , it will travel back to you .

So if the depth of the canyon is d (say) , then the total distance travelled by the sound wave will be d + d = 2d .

And we know that the speed of the sound is approximately 340m/s in air , so we can use the formula distance = speed * time to calculate the depth of the canyon . So ,

$\sf \longrightarrow Distance = (Speed)(Time) \\$

$\sf \longrightarrow 2d = 340m/s * 11.2 s\\$

$\sf \longrightarrow d = \dfrac{340m/s * 11.2s}{2} \\$

$\sf \longrightarrow \boxed{\bf Depth_{canyon}= 1904m }\\$

6 0

2 years ago

A block of mass 14.9 kg is pulled to the right by an applied force of 39.4 N. If it moves with constant velocity, how much frict

lakkis [162]

The frictional force is 39.4 N

Explanation:

We can solve this problem by applying Newton's 2nd law of motion: in fact, the net force acting on the block is equal to the product between its mass and its acceleration. So we can write

$\sum F = ma$

where

$\sum F$ is the net force

m is the mass

a is the acceleration

Here we know that the box is moving with constant velocity, so its acceleration is zero:

$a=0$

This means that the net force is also zero:

$\sum F=0$

The net force on the block is given by the applied force, forward, and the frictional force, backward:

$\sum F = F_a-F_f=0$

where

$F_a=39.4 N$ is the applied force

$F_f$ is the frictional force

Therefore, solving for $F_f$ ,

$F_f=F_a=39.4 N$

Learn more about friction:

brainly.com/question/6217246

brainly.com/question/5884009

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#LearnwithBrainly

8 0

2 years ago

A liquid is used to make a mercury-type barometer. The barometer is intended for space-faring astronauts. At the surface of the

Anarel [89]

Answer:

Density of liquid = 4730 kg/m³

Atmospheric pressure on planet X = 8401.7 N/m²

Explanation:

Pressure, P = ρgh where ρ = density of liquid, g =9.8 m/s² and h = height of column at earth's surface = 2185 mm. Since P = atmospheric pressure, for mercury, P = ρ₁gh₁ where ρ₁ = 13.6 g/cm³ and h₁ = 760 mm

So, ρgh = ρ₁gh₁

ρ = ρ₁h₁/h = 13.6 g/cm³ × 760/2185 = 4.73 g/cm³ = 4730 kg/m³

The atmospheric pressure on planet X

P = ρg₁h₃ g₁ = g/4 and h₃ = 725 mm = 0.725 m

on planet X

P = ρg₁h₃ = (4730 kg/m³ × 9.8 m/s² × 0.725 m)/4 = 8401.7 N/m²

6 0

3 years ago