A string is attached to the rear-view mirror of a car. A ball is hanging at the other end of the string. The car is driving around in a circle, at a constant speed. Which of the following lists gives all of the forces directly acting on the ball?a. tension
b. tension and gravity
c. tension, gravity, and the centripetal force
d. tension, gravity, the centripetal force, and friction

Answers

Answer 1

Answer:

Final answer:

The forces directly acting on the ball hanging from a rear-view mirror while a car drives in a circle are tension, gravity, and the centripetal force.

Explanation:

The correct answer is c. tension, gravity, and the centripetal force.

When the car is driving in a circle, the ball experiences both tension and gravity. The tension in the string is what keeps the ball from falling, while gravity pulls the ball downward.

In addition to tension and gravity, the ball also experiences the centripetal force. This force is directed towards the center of the circular motion and keeps the ball moving in a circular path.

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A softball player swings a bat, accelerating it from rest to 2.6 rev/srev/s in a time of 0.20 ss . Approximate the bat as a 0.90-kgkg uniform rod of length 0.95 mm, and compute the torque the player applies to one end of it.

Answers

Answer:

$\tau=22.13Nm$

Explanation:

information we have:

mass: $m=0.9kg$

lenght: $L=0.95m$

frequency: $f=2.6rev/s$

time: $t=0.2s$

and from the information we have we can calculate the angular velocity $\omega$ . which is defined as

$\omega=2\pi f$

$\omega=2\pi (2.6rev/s)\n\omega=16.336 rev/s$

----------------------------

Now, to calculate the torque

We use the formula

$\tau=I \alpha$

where $I$ is the moment of inertia and $\alpha$ is the angular acceleration

moment of inertia of a uniform rod about the end of it:

$I=(1)/(3)mL^2$

substituting known values:

$I=(1)/(3) (0.9kg)(0.95m)^2\nI=0.271kg/m^2$

for the torque we also need the acceleration $\alpha$ which is defined as:

$\alpha=(\omega)/(t)$

susbtituting known values:

$\alpha=(16.336rev/s)/(0.2s) \n\alpha=81.68rev/s^2$

and finally we substitute $I$ and $\alpha$ into the torque equation $\tau=I \alpha$ : $\tau=(0.271kg/m^2)(81.68rev(s^2)\n\tau=22.13Nm$

Final answer:

To calculate the torque, we need to use the formula: Torque = Moment of Inertia * Angular Acceleration. By approximating the bat as a uniform rod and using its length and mass, we can find the moment of inertia. Then, using the given angular velocity, we can calculate the angular acceleration. Finally, we can determine the torque by multiplying the moment of inertia by the angular acceleration.

Explanation:

To compute the torque the player applies to one end of the bat, we need to use the formula:

Torque = Moment of Inertia * Angular Acceleration

Given that the bat is approximated as a uniform rod and we know its length and mass, we can calculate the moment of inertia. Then, using the given angular velocity, we can compute the angular acceleration. Finally, we can find the torque by multiplying the moment of inertia by the angular acceleration.

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A capacitor consists of two closely spaced metal conductors of large area, separated by a thin insulating foil. It has an electrical capacity of 3000.0 μF and is charged to a potential difference of 60.0 V. Calculate the amount of energy stored in the capacitor. Tries 0/20 Calculate the charge on this capacitor when the electrical energy stored in the capacitor is 6.53 J. Tries 0/20 If the two plates of the capacitor have their separation increased by a factor of 5 while the charge on the plates remains constant, by what factor is the energy stored in the capacitor increased?

Answers

Answer:

1 = 5.4 J

2 = 0.1979 C

3 = 5

Explanation:

Energy in a capacitor, E is

E = 1/2 * C * V²

E = 1/2 * 3000*10^-6 * 60²

E = 1/2 * 3000*10^-6 * 3600

E = 1/2 * 10.8

E = 5.4 J

E = Q²/2C = 6.53 J

E * 2C = Q²

Q² = 6.53 * 2 * 3000*10^-6

Q² = 13.06 * 3000*10^-6

Q² = 0.03918

Q = √0.03918

Q = 0.1979 C

The Capacitor, C is inversely proportional to the distance of separation, D. Thus, if D is increased by 5 to be 5D, then C would be C/5. And therefore, our energy stored in the capacitor is increased by a factor of 5.

What is the kinetic energy k of an electron with momentum 1.05×10−24 kilogram meters per second?

Answers

Momentum = mv
where m is the mass of an electron and v is the velocity of the electron.

v = momentum ÷ m
= (1.05×10∧-24)÷(9.1×10∧-31) = 1,153,846.154 m/s

kinetic energy = (mv∧2)÷2
= (9.1×10∧-31 × 1,153,846.154∧2) ÷2
= (1.21154×10∧-18) ÷ 2
= 6.05769×10∧-19 J

Answer:

K = 6.02 × 10⁻¹⁹ J

Explanation:

The momentum (p) of an electron is its mass (m) times its speed (v).

p = m × v

v = p / m = (1.05 × 10⁻²⁴ kg.m/s) / 9.11 × 10⁻³¹ kg = 1.15 × 10⁶ m/s

We can find the kinetic energy (K) using the following expression.

K = 1/2 × m × v²

K = 1/2 × 9.11 × 10⁻³¹ kg × (1.15 × 10⁶ m/s)²

K = 6.02 × 10⁻¹⁹ J

A piston-cylinder device contains 5 kg of refrigerant-134a at 0.7 MPa and 60°C. The refrigerant is now cooled at constant pressure until it exists as a liquid at 24°C. If the surroundings are at 100 kPa and-24°C, determine: (a) the exergy of the refrigerant at the initial and the final states and
(b) the exergy destroyed during this process.

Answers

A) The exergy of the refrigerant at the initial and final states are :

Initial state = - 135.5285 kJ
Final state = -51.96 kJ

B) The exergy destroyed during this process is : - 1048.4397 kJ

Given data :

Mass ( M ) = 5 kg

P1 = 0.7 Mpa = P2

T1 = 60°C = 333 k

To = 24°C = 297 k

P2 = 100 kPa

A) Determine the exergy at initial and final states

At initial state :

U = 274.01 kJ/Kg , V = 0.034875 m³/kg , S = 1.0256 KJ/kg.k

exergy ( Ф ) at initial state = M ( U + P₂V - T₀S )

= 5 ( 274.01 + 100* 10³ * 0.034875 - 297 * 1.0256)

≈ - 135.5285 kJ

At final state :

U = 84.44 kJ / kg , V = 0.0008261 m³/kg, S = 0.31958 kJ/kg.k

exergy ( ( Ф ) at final state = M ( U + P₂V - T₀S )

= -51.96 kJ

B) Determine the exergy destroyed

exergy destroyed = To * M ( S2 - S1 )

= 297 * 5 ( 0.31958 - 1.0256 )

= - 1048.4397 KJ

Hence we can conclude that A) The exergy of the refrigerant at the initial and final states are : Initial state = - 135.5285 kJ, Final state = -51.96 kJ and The exergy destroyed during this process is : - 1048.4397 kJ

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Final answer:

Exergy of refrigerant-134a at initial and final states is obtained from property tables and by multiplying the mass of the refrigerant with its specific exergy at each state. The difference in exergy between the two states represents the exergy destroyed.

Explanation:

To solve the given question, we need the property values of

refrigerant-134a

at the initial and the final states.

At an initial state of 0.7 MPa and 60°C, the specific exergy for refrigerant-134a can be obtained from property tables which are standard in thermodynamics textbooks. Same for the final state at 0.7 MPa and 24°C, the specific exergy can be obtained from the same property tables.

The exergy of the refrigerant at the initial and the final states can be calculated by multiplying the mass of the refrigerant with its specific exergy at each state.

Exergy destruction during this process can be calculated using the relation between exergy change and exergy destruction. The exergy change of a system between initial and final states is equal to the difference of the exergy of the system at final and initial states.

Based on the second law of thermodynamics, the difference in exergy should be equal to the exergy destroyed during the process.

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A 750-kg automobile is moving at 26.2 m/s at a height of 5.00 m above the bottom of a hill when it runs out of gasoline. The car coasts down the hill and then continues coasting up the other side until it comes to rest. Ignoring frictional forces and air resistance, what is the value of h, the highest position the car reaches above the bottom of the hill?

Answers

To solve this problem it is necessary to apply to the concepts related to energy conservation. For this purpose we will consider potential energy and kinetic energy as the energies linked to the body. The final kinetic energy is null since everything is converted into potential energy, therefore

Potential Energy can be defined as,

$PE = mgh$

Kinetic Energy can be defined as,

$K= (1)/(2) mv^2$

Now for Conservation of Energy,

$KE_i+PE_i = PE_f$

$(1)/(2)mv_i^2+mgh_1 = mgh_2$

$(1)/(2) (750kg) (26.2m/s)^2 + (750)(9.8)(5) = (750)(9.8)h_2$

$h_2 = 40.0224m$

Therefore the highets position the car reaches above the bottom of the hill is 40.02m

Air contains 78.08% nitrogen, 20.095% oxygen, and 0.93% argon. calculate the partial pressure of oxygen if the total pressure of the air sample is 1.7 atm.a.