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topjm [15]
3 years ago
15

An object is released from rest at time t = 0 and falls through the air, which exerts a resistive force such that the accelerati

on a of the object is given by a = g – bV, where V is the object's speed and b is a constant. If limiting cases for large and small values of t are considered, which of the following is a possible expression for the speed of the object as an explicit function of time?
a. v = g(l - e-bt)/b.
b. v = (gebt)/b.
c. v = gt - bt2.
d. v = (g + a)t/b.
e. v = V0 + gt, v0.
Physics
1 answer:
lubasha [3.4K]3 years ago
6 0

Answer:

The correct option is

a. v = g (1-e^{-bt})/b

Explanation:

Time at which the object start fall t = 0

The acceleration a is given by a = g - bV

Where V = Speed of the object

Speed V² = u² + 2·a·h

However with the drag force the object will approach terminal velocity as t becomes progressively larger whereby v will stop increasing

Option a. is the only option that has  limiting value of v which is in the range of g as t increases ∴ option a. is the correct option.

v = g (1-e^{-bt})/b  as t increases (1-e^{-bt}) → 1 s and v→ g/b m/s

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A positive kaon (K+) has a rest mass of 494 MeV/c² , whereas a proton has a rest mass of 938 MeV/c². If a kaon has a total energ
vitfil [10]

Answer:

<em>0.85c </em>

Explanation:

Rest mass of Kaon M_{0K} = 494 MeV/c²

Rest mass of proton M_{0P}  = 938 MeV/c²

The rest energy is gotten by multiplying the rest mass by the square of the speed of light c²

for the kaon, rest energy E_{0K} = 494c² MeV

for the proton, rest energy E_{0P} = 938c² MeV

Recall that the rest energy, and the total energy are related by..

E = γE_{0}

which can be written in this case as

E_{K} = γE_{0K} ...... equ 1

where E = total energy of the kaon, and

E_{0} = rest energy of the kaon

γ = relativistic factor = \frac{1}{\sqrt{1 - \beta ^{2} } }

where \beta = \frac{v}{c}

But, it is stated that the total energy of the kaon is equal to the rest mass of the proton or its equivalent rest energy, therefore...

E_{K} = E_{0P} ......equ 2

where E_{K} is the total energy of the kaon, and

E_{0P} is the rest energy of the proton.

From E_{K} = E_{0P} = 938c²    

equ 1 becomes

938c² = γ494c²

γ = 938c²/494c² = 1.89

γ = \frac{1}{\sqrt{1 - \beta ^{2} } } = 1.89

1.89\sqrt{1 - \beta ^{2} } = 1

squaring both sides, we get

3.57( 1 - \beta^{2}) = 1

3.57 - 3.57\beta^{2} = 1

2.57 = 3.57\beta^{2}

\beta^{2} = 2.57/3.57 = 0.72

\beta = \sqrt{0.72} = 0.85

but, \beta = \frac{v}{c}

v/c = 0.85

v = <em>0.85c </em>

7 0
3 years ago
Acceleration = change of velocity divided by time interval = Δv/Δt.
MariettaO [177]

Answer:

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Explanation:

a=Δv/Δt------eq(1)

Δv=Vf-Vi=120 km/h-0 km/h=120 km/h

or Δv=33.3 m/sec

or time=t=14s

putting values in eq(1)

a=33.3/14

a=2.378 m/s^2

6 0
3 years ago
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Vera_Pavlovna [14]

Answer:

D. gravity

Explanation:

Gravity keeps the atmosphere from escaping into space.

3 0
3 years ago
How many oxygen atoms are in the following compound?<br> 8 C120
Novosadov [1.4K]

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6 0
3 years ago
An airplane traveling at 201 m/s makes a turn. What is the smallest radius of the circular path (in km) that the pilot can make
egoroff_w [7]

Answer:

the smallest radius of the circular path is 8.1 km

Explanation:

The computation of the smallest radius of the circular path is given below:

Given that

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radius = ?

As we know that

a_c = V^2 ÷ r

5 = 201^2 ÷ r

r = 201^2 ÷ 5

= 8,080.2 g

= 8.1 km

Hence, the smallest radius of the circular path is 8.1 km

4 0
2 years ago
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