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Marta_Voda [28]
2 years ago
9

Find the x-component of this

Physics
1 answer:
valentina_108 [34]2 years ago
4 0

The x-component of this vector inclined from the horizontal axis is  15m.

<h3>What is vector?</h3>

A vector is the representation of a physical quantity in magnitude and direction.

A vector incline at an angle in has two components. On in x direction and other is in y direction.

For the given length of vector = 45.6 m and the angle of inclination from the +x axis θ = 70.8°

The x component of the given vector is

= 45.6cos 70.8°

= 14.99 m

Thus, the x- component of this vector is 15m.

Learn  more about vector.

brainly.com/question/13322477

#SPJ1

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lisov135 [29]

Answer:

Glow

Explanation:

Actually, it is the air in front of the meteoroid that heats up. The particle is traveling at speeds between 20 and 30 kilometers per second. It compresses the air in front, causing the air to get hot. The air is so hot it begins to glow — creating a meteor - the streak of light observed from Earth.

Hope this helped!

7 0
3 years ago
I need help please...
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6 0
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Read 2 more answers
Set the radius to 2.0 m and the velocity to 1.0 m/s. Keeping the radius the same, record the magnitude of centripetal accelerati
jek_recluse [69]

Answer:

a=4\ m/s^2

Explanation:

Given that,

Radius, r = 2 m

Velocity, v = 1 m/s

We need to find the magnitude of the centripetal acceleration. The formula for the centripetal acceleration is given by :

a=\dfrac{v^2}{r}\\\\a=\dfrac{(2)^2}{1}\\\\=4\ m/s^2

So, the magnitude of centripetal acceleration is 4\ m/s^2.

5 0
2 years ago
Who much force is needed to accelerate a 68 kilogram-skier at a Rate of 1.2 m/sec
liberstina [14]
If F =m*a
and the question says how much force the s needed to accelerate a 68kg skier to a rate of 1.2ms^-2
Then F = 68*1.2
7 0
3 years ago
A mole of ideal gas expands at T=27 °C. The pressure changes from 20 atm to 1 atm. What’s the work that the gas has done and wha
Airida [17]

Answer:

  • The work made by the gas is 7475.69 joules
  • The heat absorbed is 7475.69 joules

Explanation:

<h3>Work</h3>

We know that the differential work made by the gas  its defined as:

dW =  P \ dv

We can solve this by integration:

\Delta W = \int\limits_{s_1}^{s_2}\,dW = \int\limits_{v_1}^{v_2} P \ dv

but, first, we need to find the dependence of Pressure with Volume. For this, we can use the ideal gas law

P \ V = \ n \ R \ T

P = \frac{\ n \ R \ T}{V}

This give us

\int\limits_{v_1}^{v_2} P \ dv = \int\limits_{v_1}^{v_2} \frac{\ n \ R \ T}{V} \ dv

As n, R and T are constants

\int\limits_{v_1}^{v_2} P \ dv = \ n \ R \ T \int\limits_{v_1}^{v_2} \frac{1}{V} \ dv

\Delta W= \ n \ R \ T  \left [ ln (V) \right ]^{v_2}_{v_1}

\Delta W = \ n \ R \ T  ( ln (v_2) - ln (v_1 )

\Delta W = \ n \ R \ T  ( ln (v_2) - ln (v_1 )

\Delta W = \ n \ R \ T  ln (\frac{v_2}{v_1})

But the volume is:

V = \frac{\ n \ R \ T}{P}

\Delta W = \ n \ R \ T  ln(\frac{\frac{\ n \ R \ T}{P_2}}{\frac{\ n \ R \ T}{P_1}} )

\Delta W = \ n \ R \ T  ln(\frac{P_1}{P_2})

Now, lets use the value from the problem.

The temperature its:

T = 27 \° C = 300.15 \ K

The ideal gas constant:

R = 8.314 \frac{m^3 \ Pa}{K \ mol}

So:

\Delta W = \ 1 mol \ 8.314 \frac{m^3 \ Pa}{K \ mol} \ 300.15 \ K  ln (\frac{20 atm}{1 atm})

\Delta W = 7475.69 joules

<h3>Heat</h3>

We know that, for an ideal gas, the energy is:

E= c_v n R T

where c_v its the internal energy of the gas. As the temperature its constant, we know that the gas must have the energy is constant.

By the first law of thermodynamics, we know

\Delta E = \Delta Q - \Delta W

where \Delta W is the Work made by the gas (please, be careful with this sign convention, its not always the same.)

So:

\Delta E = 0

\Delta Q = \Delta W

7 0
3 years ago
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