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notsponge [240]
3 years ago
14

How does the vertical component of a projectile’s motion compare with the motion of vertical free fall when air resistance is ne

gligible?
Physics
1 answer:
likoan [24]3 years ago
5 0

Answer:

Explanation:

The vertical component of velocity remains same as the free fall. The vertical motion of the projectile is same as the free fall motion.

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A piece of iron of mass 200g and tempreture 300°C is dropped into 1.00 kg of water of tempreture 20°C. Predict the final equilib
Alekssandra [29.7K]

Answer:

The final equilibrium T_{f} = 25.7[°C]

Explanation:

In order to solve this problem we must have a clear concept of heat transfer. Heat transfer is defined as the transmission of heat from one body that is at a higher temperature to another at a lower temperature.

That is to say for this case the heat is transferred from the iron to the water, the temperature of the water will increase, while the temperature of the iron will decrease. At the end of the process a thermal balance is found, i.e. the temperature of iron and water will be equal.

The temperature of thermal equilibrium will be T_f.

The heat absorbed by water will be equal to the heat rejected by Iron.

Q_{iron} = Q_{water}

Heat transfer can be found by means of the following equation.

Q_{iron}=m*C_{piron}*(T_{i}-T_{f})

where:

Qiron = Iron heat transfer [kJ]

m = iron mass = 200 [g] = 0.2 [kg]

T_i = Initial temperature of the iron = 300 [°C]

T_f = final temperature [°C]

Q_{water}=m*C_{pwater}*(T_{f}-T_{iwater})

Cp_iron = 437 [J/kg*°C]

Cp_water = 4200 [J/kg*°C]

0.2*437*(300-T_{f})=1*4200*(T_{f}-20)\\26220-87.4*T_{f}=4200*T_{f}-84000\\26220+84000=4200*T_{f}+87.4*T_{f}\\110220 = 4287.4*T_{f}\\T_{f}=25.7[C]

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3 years ago
A proton is released from rest at the positive plate of a parallel-plate capacitor. It crosses the capacitor and reaches the neg
Natali5045456 [20]

Answer: Vf = 2,400,000 m/s

Explanation:

1) The only relevant force is the electrostatic force

2) The formula for the electrostatic force is F = E×q

Where E is the electric field and q is the magnitude of the charge.

3) Since the electric field is the same in both cases, and the charge of the protons and electrons have the same magnitude, you can state that the magntitude of the electric forces acting in both proton and electron are the same

4) Fe = Fp (Fe stands for force on the electron and Fp stands for force on the proton).

5) Using second law of Newton, Force = mass × acceleration

Fe = Me × Ae (Me is mass of the electron, and Ae is acceleration of the electron)

Fp = Mp × Ap (Mp is mass of the proton, and Ap is acceleration of the proton)

⇒Me × Ae = Mp × Ap

⇒ Ae = Mp × Ap / Me

6) Now, state the equations for the velocity in uniformly accelerated motion:

i) Vf² = Vo² + 2ad

Vo² = 0 for both cases, and d is the same distance.

⇒ Vf² = 2ad

ii) For the proton Vf² = 2(Ap)(d) ⇒ Ap = Vf² / (2d)

⇒ Ap = (55,000 m/s)² / (2d)

iii) For the electron Vf² = 2(Ae)² (2d)

iv) Using Ae = Mp × Ap / Me (found prevously):

Vf² = Mp × (55,000 m/s)² / (2d) × (2d) / Me

⇒ Vf² = Mp × (55,000 m/s)² / Me

Taking square root in both sides:

⇒ Vf = 55,000 m/s × √ [Mp / Me]

7) These are the values for the masses of a proton and an electron:

Mp = 1.67 × 10⁻²⁷ kg

Me = 9.11×10⁻³¹ kg

8) Replace and compute:

Vf = 55,000 m/s × √ [ 1.67 × 10⁻²⁷ kg / 9.11×10⁻³¹ kg] = 2,354,841.8 m/s

Round to two significan digits: Vf = 2,400,000 m/s

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