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QveST [7]
2 years ago
14

Which year was pluto no longer considered a planet?.

Physics
1 answer:
katrin2010 [14]2 years ago
6 0
2006, i hope this helps
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How much power is used by a hair dryer if it does 40,000 J of work in 40 seconds?
bearhunter [10]

Answer:

40000÷40=1000 joules is required to work in 40 seconds

6 0
3 years ago
A 5kg object moving horizontally at 3m/s collides with a stationary 3kg object. After the collision, the 5kg object is deflected
gavmur [86]

Answer:

The velocity of each ball after the collision are 2.19 m/s and 2.58 m/s.

Explanation:

Given that,

Mass of object = 5 kg

Speed = 3 m/s

Mass of stationary object = 3 kg

Moving object deflected  = 30°

Stationary object deflected = 31°

We need to calculate the velocity of each ball after collision

Using conservation of momentum

Along x-axis

m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}\cos\theta+m_{2}v_{2}\cos\theta

Put the value into the fomrula

5\times3+0=5\times v_{1}\cos30+3\times v_{2}\cos45

15=5v_{1}\times\dfrac{\sqrt{3}}{2}+3v_{2}\times\dfrac{1}{\sqrt{2}}

15=\dfrac{5\sqrt{3}}{2}v_{1}+\dfrac{3}{\sqrt{2}}v_{2}....(I)

Along y -axis

m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}\sin\theta+m_{2}v_{2}\sin\theta

Put the value into the formula

0+0=5\times v_{1}\sin30-3\times v_{2}\sin45

\dfrac{5}{2}v_{1}-\dfrac{3}{\sqrt{2}}v_{2}=0...(II)

From equation (I) and (II)

v_{1}=\dfrac{15\times2}{5\sqrt{3}+5}

v_{1}=2.19\ m/s

Put the value of v₁ in equation (I)

\dfrac{5}{2}\times2.19-\dfrac{3}{\sqrt{2}}v_{2}=0

v_{2}=\dfrac{5.475\times\sqrt{2}}{3}

v_{2}=2.58\ m/s

Hence, The velocity of each ball after the collision are 2.19 m/s and 2.58 m/s.

3 0
3 years ago
Consider a sound wave moving through the air modeled with the equation s(x, t) = 5.00 nm cos(60.00 m−1x − 18.00 ✕ 103 s−1t). Wha
GaryK [48]

Answer:

Shortest time = 58.18 × 10^(-6) s

Explanation:

We are given;

s(x,t) = 5.00 nm cos((60.00 m^(−1)x) − (18.00 X 10³ s^(−1)t))

Let us set x = 0 as origin.

Now, for us to find the time difference, we need to solve 2 equations which are;

s(x,t) = 5.00 nm cos((60.00 m^(−1)x) − (18.00 X 10³ s^(−1)t1))

And

s(x,t) = 5.00 nm cos((60.00 m^(−1)x) − (18.00 X 10³ s^(−1)t2))

Now, since the wave starts from maxima at time at t = 0, the required time would be the difference (t2 - t1)

Thus, the solutions are;

t1 = (1/(18 × 10³)) cos^(-1) (2.5/5)

And

t2 = (1/(18 × 10³)) cos^(-1) (-2.5/5)

Angle of the cos function is in radians, thus;

t1 = 58.18 × 10^(-6) s

t2 = 116.36 × 10^(-6) s

So,

Required time = t2 - t1 = (116.36 × 10^(-6) s) - (58.18 × 10^(-6) s) = 58.18 × 10^(-6) s

4 0
3 years ago
Measurements show that the enthalpy of a mixture of gaseous reactants decreases by during a certain chemical reaction, which is
snow_tiger [21]

The given question is incomplete. The complete question is as follows.

Measurements show that the enthalpy of a mixture of gaseous reactants decreases by 338 kJ during a certain chemical reaction, which is carried out at a constant pressure. Furthermore, by carefully monitoring the volume change it is determined that 187 kJ of work is done on the mixture during the reaction. Calculate the change in energy of the gas mixture during the reaction. Be sure your answer has the correct number of significant digits. Is the reaction exothermic or endothermic ?

Explanation:

The given data is as follows.

    Change in enthalpy (\Delta H) = -338 kJ (as it is a decrease)

    Work done = 187 kJ,

    Change in energy (\Delta E) = ?

Now, according to the first law of thermodynamics the formula is as follows.

          \Delta H = \Delta E + P \Delta V

Hence, putting the given values into the above formula as follows.

        \Delta H = \Delta E + P \Delta V

Also, we know that W = P \Delta V

so,           \Delta H = \Delta E + W      

              -338 kJ = \Delta E + 187

              \Delta E = -151 kJ

Thus, we can conclude that the change in energy of the gas mixture during the reaction is -151 kJ.

5 0
3 years ago
1. You push a deak across the room. A force was applied. The object moves as a result of a force.​
FinnZ [79.3K]

Answer:

what is the question?

Explanation:

I'm not sure if this was a question because you gave the answer lol

6 0
3 years ago
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