An astronaut stands on the surface of an asteroid. The astronaut then jumps such that the astronaut is no longer in contact with the surface. The astronaut falls back down to the surface after a short time interval. Which of the following forces CANNOT be neglected when analyzing the motion of the astronaut?
Answers
Asteroids are known through the help of artificial gravity, to have small gravity. The forces that cannot be neglected when analyzing the motion of the astronaut is that the gravitational force between the astronaut and the asteroid.
The gravitational force between two objects is said to be inversely proportional to the distance between them when squared. Therefore, when an individual halve the distance then the force increases by four times.
Unbalanced forces are simply known to be brought about due to a change in motion, speed, and/or direction. If two forces act in the same direction on an object, the net force is said to be equal to the sum of the two forces.
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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 1Where:
v = final velocity of the automobileu = initial velocity of the automobilea = accelerations = distance coveredFrom 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 mHence, The car will travel a distance of 39.1 m before its stops.
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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.
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]
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]
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: