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lyudmila [28]
2 years ago
12

In projectile mtion, what is the x-component of the initial velocity? if V= Vi = 100 m/s and the angle with horizontal axis Θ =

60 degrees

Physics
2 answers:
AleksandrR [38]2 years ago
3 0

Answer:

???????????????????????

torisob [31]2 years ago
3 0

Explanation:

The x - component of the initial velocity is given as:

Ux = Vi cos theta

Ux = 100cos 60

Ux = 50m/s

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AveGali [126]

Answer:

b, a

Explanation:

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3 years ago
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The diagram below shows a person using a device called a jetpack. Water is forced downwards from the jetpack and produces an upw
klasskru [66]

Answer:

The condition necessary for the person to be able to remain stationary in mid-air is that there must be an upward force also referred to as THRUST

Explanation:

Thrust in Physics is defined as the force that propels, forces or pushes the mass of a certain object in a specified or particular direction.

Thrust can defined in the terms of a Jet pack can be defined as the force that is required to propel the mass of a person in an upward direction.

Based on the diagram that we have been shown the question, we can see that water is the fuel used to proper the user or person upwards hence, the jet pack used in the question is an HydroJet pack

The condition necessary for the person to be able to remain stationary in mid-air is that the propelling force called the thrust must be upwards such that it can overcome these 4 things:

a) Gravitational force

b) Mass of the jet pack user

c) Mass of the jet pack itself

d) The water which serves as fuel for the jet pack.

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3 years ago
What is the speed of light?
Sergeu [11.5K]
The speed of light : 299 792 458 m / s
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2 years ago
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Diffraction is observable when the opening is ______ than the wavelength.
masya89 [10]
Diffraction is observable when the is smaller than the wavelength 
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I shared a picture of the problem. It’s a basic Physics question and an Algebra question.
julia-pushkina [17]

Hence the expression of ω in terms of m and k is

\omega = \sqrt{\frac{k}{m}

Given the expressions;

T_s = 2 \pi \sqrt{\frac{m}{k} } \ and \ T_s = \frac{2 \pi}{\omega}

Equating both expressions we will have;

2 \pi \sqrt{\frac{m}{k} }  = \frac{2 \pi}{\omega}

Divide both equations by 2π

\frac{2 \pi\sqrt{\frac{m}{2 \pi} } }{2 \pi}=\frac{\frac{2 \pi}{\omega} }{2\pi}\\\sqrt{\frac{m}{2 \pi} } = \frac{1}{\omega}\\

Square both sides

(\sqrt{\frac{m}{k} } )^2 = (\frac{1}{\omega} )^2\\\frac{m}{k} = \frac{1}{\omega ^2} \\\omega ^2 = \frac{k}{m}

Take the square root of both sides

\sqrt{\omega ^2} =\sqrt{\frac{k}{m} } \\\omega = \sqrt{\frac{k}{m}

Hence the expression of ω in terms of m and k is

\omega = \sqrt{\frac{k}{m}

3 0
3 years ago
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