A tennis ball moves 16 meters northward, then 22 meters southward, then 12 meters northward, and finally 32 meters southward.

Answers

Answer 1
U didn’t ask a question so I’ll just say that’s cool that it moves like that

Related Questions

On Venus, the atmospheric temperature is a hot 720 K due to the greenhouse effect. It consists mostly of carbon dioxide (molar mass 44 g/mol) and the pressure is 92 atm. What is the total translational kinetic energy of 3 moles of carbon dioxide molecules?

Answers

Answer:

The value is   [tex]E_t = 17958.2 \ J[/tex]

Explanation:

From the question we are told that

     The atmospheric temperature is [tex]T_a = 720 \ K[/tex]

       The molar mass of carbon dioxide is  [tex]Z = 44 \ g/mol[/tex]

        The pressure is [tex]P = 92 \ atm =[/tex]

      The number of moles is [tex]n = 3 \ moles[/tex]

Generally the translational kinetic energy is mathematically represented as

        [tex]E_t = \frac{f}{2} * n * R T[/tex]

       Here  R is the gas constant with value  [tex]R = 8.314 J\cdot K^{-1}\cdot mol^{-1}[/tex]

Generally the degree of freedom of carbon dioxide in terms of  translational motion is  f =  3

     So

           [tex]E_t = \frac{ 3}{2} * 2 * 8.314 * 720[/tex]

=>         [tex]E_t = 17958.2 \ J[/tex]

The speed of revolution of particle going around a circlr is doubled and its angular speed is havled. What happen to the centripetal acceleration?
a) unchanged
b) doubles
c) halves
d) becomes four times​

Answers

Answer: The correct answer is C

Explanation:

A 1 200-kg automobile moving at 25 m/s has the brakes applied with a deceleration of 8.0 m/s2. How far does the car travel before it stops?

Answers

Answer:

Δx = 39.1 m

Explanation:

Assuming that deceleration keeps constant during the braking process, we can use one of the kinematics equations, as follows:

        [tex]v_{f} ^{2} - v_{o} ^{2} = 2* a * \Delta x (1)[/tex]

        where  vf is the final velocity (0 in our case), v₀ is the initial velocity

        (25 m/s), a is the acceleration (-8.0 m/s²), and Δx is the distance

        traveled since the brakes are applied.

Solving (1) for Δx, we have:

        [tex]\Delta x = \frac{-v_{o} ^{2} }{2*a} = \frac{-(25m/s)^{2}}{2*(-8.0m/s2} = 39.1 m (2)[/tex]        

The car will travel a distance of 39.1 m before its stops.

To solve the problem above, use the equations of motion below.

Equation:

v² = u²+2as................... Equation 1

Where:

v = final velocity of the automobileu = initial velocity of the automobilea = accelerations = distance covered

From the question,

Given:

v = 0 m/s (before its stops)u = 25 m/sa = -8 m/s² (decelerating)

Substitute these values into equation 1

⇒ 0² = 25²+2(-8)(s)

Solve for s

⇒ 0²-25² = -16s⇒ -16s = -625⇒ s = -625/16⇒ s = 39.1 m

Hence, The car will travel a distance of 39.1 m before its stops.

Learn more about acceleration here: https://brainly.com/question/605631

A balloon contains 0.075 m^3 of
gas. The pressure is reduced to
100kPa and fills a box of 0.45 m^3.
What is the initial pressure inside the
balloon if the temperature remains
constant?

Answers

Answer:

600 KPa.

Explanation:

From the question given above, the following data were obtained:

Initial volume (V1) = 0.075 m³

Final volume (V2) = 0.45 m³

Final pressure (P2) = 100 KPa

Initial pressure (P1) =?

Temperature = constant

The initial pressure can be obtained by using the Boyle's law equation as shown below:

P1V1 = P2V2

P1 × 0.075 = 100 × 0.45

P1 × 0.075 = 45

Divide both side by 0.075

P1 = 45 / 0.075

P1 = 600 KPa.

Thus, the initial pressure in the balloon is 600 KPa.

g You heard the sound of a distant explosion (3.50 A/10) seconds after you saw it happen. If the temperature of the air is (15.0 B) oC, how far were you from the site of the explosion

Answers

Answer:

The answer is "1557 meters".

Explanation:

speed of sound in ([tex]\frac{m}{s}[/tex]) [tex]= 331.5 + 0.60 \ T^{\circ}\ C\\\\[/tex]

[tex]\to V = 331.5 + 0.6 \times 24 = 346 \frac{m}{s}\\\\\to t = 4.5 \ seconds \\\\\to S = vt = 346 \times 4.5 = 1557 \ meters[/tex]

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