A 30 kg child on a 2 m long swing is released from rest when the swing supports make an angle of 34 ◦ with the vertical. The acceleration of gravity is 9.8 m/s 2 . If the speed of the child at the lowest position is 2.31547 m/s, what is the mechanical energy dissipated by the various resistive
Answers
Answer:
Energy dissipated = 13.453 Joules
Explanation:
In order to solve this problem, we first compute the gravitational potential energy the child has, and then find the kinetic energy at the lowest position.
The gravitational potential energy (relative to lowest position) is found as follows:
G.P.E = mass * gravity * height
Where, Height = 2 - 2 * Cos(34°)
Height = 0.3193 meters
G.P.E = 30 * 9.8 * 0.3193
G.P.E = 93.874 J
Kinetic energy:
K.E = 0.5 * mass * velocity^2
K.E = 0.5 * 30 * 2.31547^2
K.E = 80.421 J
Energy dissipated = G.P.E - K.E
Energy dissipated = 93.874 - 80.421
Energy dissipated = 13.453 J
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A car (m = 2000 kg) is going around an unbanked curve at the recommended speed of 11m/s (24.6 MPH). (a) If the radius of the curvature of the path is 25m and the coefficient of static friction between the rubber tires and the road is µs = 0.70, does the car skid as it goes around the curve? (b) What will happen if the driver ignores the highway speed limit sign and travels at 18 m/s (40.3 MPH)? (c) What speed is safe for traveling around the curve if the road surface is wet from a recent rainstorm and the coefficient of static friction between the wet roud and the rubber tires is µs = 0.50?
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Answer:
The amplitude of oscillation for the oscillating mass is 0.28 m.
Explanation:
Given that,
Mass = 0.14 kg
Equation of simple harmonic motion
....(I)
We need to calculate the amplitude
Using general equation of simple harmonic equation
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x = vi(cos )t
x = ayt
x = vxt (RIGHT ANSWER)
Answers
The formula for calculating the horizontal displacement of a horizontally launched projectile is
A projectile launched horizontally with a velocity v, at a height y ,travels a horizontal distance x, while falling through a distance y. The horizontal velocity of a projectile remains constant throughout its motion, in the absence of air resistance. The vertical component of the velocity is under the action of the gravitational force and hence it increases in magnitude as it falls through the height.
The horizontal displacement of the projectile is a uniform motion and it occurs at a constant speed v.
Thus, the horizontal displacement of the projectile is given by the expression.
Answers
Answer:
5.78971 m
Explanation:
= Initial pressure = 0.873 atm
= Final pressure = 0.0282 atm
= Initial volume
= Final volume
= Initial radius = 16.2 m
= Final radius
Volume is given by
From the ideal gas law we have the relation
The radius of balloon at lift off is 5.78971 m
Final answer:
To find the radius of the weather balloon at lift-off, the ideal gas law can be used. Using the equation P1V1 = P2V2, where P1 and V1 are the initial pressure and volume, and P2 and V2 are the pressure and volume at lift-off, the radius at lift-off can be calculated to be approximately 4.99 m.
Explanation:
To find the radius of the weather balloon at lift-off, we can use the ideal gas law, which states that PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.
In this case, we know that the number of moles is constant, as the balloon is filled with the same amount of helium at lift-off and in flight. Therefore, we can write the equation as P1V1 = P2V2, where P1 and V1 are the initial pressure and volume, and P2 and V2 are the pressure and volume at lift-off.
Plugging in the given values, we have (0.873 atm)(V1) = (0.0282 atm)(16.2 m)^3. Solving for V1, we find that the volume at lift-off is approximately 110.9 m^3. The radius can then be calculated using the formula for the volume of a sphere: V = (4/3) * π * r^3, where r is the radius.
Therefore, the radius at lift-off is approximately 4.99 m.
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B. 3.75 × 10–7 N toward C
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D. 1.15 × 10–7 N toward D
Answers
The magnitude and direction of the net gravitationalforce on mass A due to the other masses is 1.15 × 10⁻⁷ N toward D.
The gravitational force between two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. In this case, all of the masses are equal to 3 kg, and the distance between mass A and mass D is 3 m.
The gravitational force between mass A and mass D is therefore:
F = G * m_A * m_D / r²
= 6.674 × 10⁻¹¹ N m² / kg² * 3 kg * 3 kg / 3 m²
= 1.15 × 10⁻⁷ N
The direction of the gravitational force is towards mass D.
Therefore, the net gravitational force on mass A due to the other masses is 1.15 × 10⁻⁷ N toward D.
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Answer:
THE ANSER IS B
Explanation:
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Answer:
kilogram
Explanation:
Answer:
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It is shared electrons.
The following information should be considered:
- In the case when two or more atoms can be together at the time when they share electrons with each other.
- By sharing, they create a covalent bond and that way the atoms can be stable.
learn more: brainly.com/question/2514933?referrer=searchResults
Answer:
try electrons i hope this helps!!
Explanation: