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miv72 [106K]
3 years ago
10

a truck drives up a hill with a 15 incline. if the truck has a constant speed of 22m/s, what are the horizontal and vertical com

ponents of the trucks velocity

Physics
1 answer:
Kamila [148]3 years ago
3 0

Answer:

speed of truck (v) =  22 m/s ,

angle of hill (Θ) =15°

Find

Vertical component (Fv) = ?

Harizontal component (Fh) =?

               Vertical component (Fh) = V cosΘ

                                                        = 22. cos 15

                                                        = 21.25 m/s.

               Harizontal component (Fv) = V sinΘ

                                                            = 22. sin 15

                                                            = 5.69 m/s.

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What separates the terrestrial planets from the jovian or gas planets in space?
Flura [38]
In our solar system, terrestrial planets are separated from the gas giants by the asteroid belt. The asteroid belt is a region in the solar system between Mars and Jupiter where asteroids are located. Gas giants do not have a solid surface and possible a small rocky core. The gas giants are Jupiter, Saturn, Uranus and Neptune. The first four planets, Mercury, Venus, Earth and Mars.  
8 0
3 years ago
A 2 m tall, 0.5 m inside diameter tank is filled with water. A 10 cm hole is opened 0.75 m from the bottom of the tank. What is
valina [46]

Answer:

4.75 m/s

Explanation:

The computation of the velocity of the existing water is shown below:

Data provided in the question

Tall = 2 m

Inside diameter tank = 2m

Hole opened = 10 cm

Bottom of the tank = 0.75 m

Based on the above information, first we have to determine the height which is

= 2 - 0.75 - 0.10

= 2 - 0.85

= 1.15 m

We assume the following things

1. Compressible flow

2. Stream line followed

Now applied the Bernoulli equation to section 1 and 2

So we get

\frac{P_1}{p_g} + \frac{v_1^2}{2g} + z_1 = \frac{P_2}{p_g} + \frac{v_2^2}{2g} + z_2

where,

P_1 = P_2 = hydrostatic

z_1 = 0

z_2 = h

Now

\frac{v_1^2}{2g} + 0 = \frac{v_2^2}{2g} + h\\\\V_2 < < V_1 or V_2 = 0\\\\Therefore\  \frac{v_1^2}{2g} = h\\\\v_1^2 = 2gh\\\\ v_1 = \sqrt{2gh} \\\\v_1 = \sqrt{2\times 9.8\times 1.15}

= 4.7476 m/sec

= 4.75 m/s

6 0
3 years ago
2.0 mol of monatomic gas A initially has 5000 J of thermal energy. It interacts with 3.0 mol of monatomic gas B, which initially
Naddika [18.5K]

Answer:

 E_particle = 1,129 10⁻²⁰ J / particle

  T= 817.5 K

Explanation:

Energy is a scalar quantity so it is additive, let's look for the total energy of each gas

Gas a

         E_a = 2 5000 = 10000 J

Gas b

         E_b = 3 8000 = 24000 J

When the total system energy is mixed it is

          E_total = E_a + E_b

          E_total = 10000 + 24000 = 34000

The total mass is

           M = m_a + m_b

           M = 2 +3 = 5

The average energy among the entire mass is

           E_averge = E_total / M

            E_averago = 34000/5

            E_average = 6800 J

One mole of matter has Avogadro's number of atoms 6,022 10²³ particles

Therefore, each particle has an energy of

                E_particle = E_averag / 6.022 10²³ = 6800 /6.022 10²³

                E_particle = 1,129 10⁻²⁰ J / particle

For  find the temperature let's use equation

               E = kT

               T = E / k

     

               T = 1,129 10⁻²⁰ / 1,381 10⁻²³

               T = 8.175 102 K

               T= 817.5 K

5 0
3 years ago
A train starts at v=10m/s and accelerates at 1m/s2 for 5s. Its final speed is *
ad-work [718]

Answer:

D) 15 m/s

Explanation:

v = u + at

v = 10 m/s + 1 m/s²(5 s)

v = 15 m/s

3 0
2 years ago
A litre of a gas weigh 2 gram at 300 kelvin temperature and 1 atm pressure,if the pressure is made 75 atm then at which temperat
liberstina [14]

Answer:

45000 K .

Explanation:

Given :

A liter of a gas weigh 2 gram at 300 kelvin temperature and 1 atm pressure

We need to find the temperature in which 1 litre of the same gas weigh 1 gram

in pressure 75 atm.

We know, by ideal gas equation :

PV=nRT

Here , n is no of moles , n=\dfrac{Given \ Weight }{Molecular\ Mass}=\dfrac{w}{M}

Putting initial and final values and dividing them :

\dfrac{P_1V_1}{P_2V_2}=\dfrac{\dfrac{w_1}{M}T_1}{\dfrac{w_2}{M}T_2}

\dfrac{1\times 1}{75\times 1}=\dfrac{\dfrac{2}{M}\times 300}{\dfrac{1}{M}\times T_2}\\ \\T_2=45000\ K.

Hence , this is the required solution.

7 0
3 years ago
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