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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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4 0
2 years ago
Read 2 more answers
PLEASEEE HELP, thank you :)
telo118 [61]

Answer:

The answer is B.

Explanation:

Given that the <em>current </em>(Ampere) in a series circuit is same so we can ignore it. We can assume that the total voltage is 60V and all the 3 resistance are different, 20Ω, 40Ω and 60Ω. So first, we have to find the total resistance by adding :

Total resistance = 20Ω + 40Ω + 60Ω

= 120Ω

Next, we have to find out that 1Ω is equal to how many voltage by dividing :

120Ω = 60V

1Ω = 60V ÷ 120

1Ω = 0.5V

Lastly, we have to calculate the voltage at R1 so we have to multiply by 20 (R1) :

1Ω = 0.5V

20Ω = 0.5V × 20

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8 0
3 years ago
A van de Graaff generator accelerates electrons so that they have energies equivalent to that attained by falling through a pote
Marat540 [252]

Answer:

Hello your question is incomplete hence I will give you a general answer on how A van de Graaff generator works

answer :

If the electrons falls through a PD of 150mV the electron will gain energy of   150MeV

Explanation:

when a Van de Graff generator is used to accelerate an electron through a PD ( potential difference ) of any value the particle ( electron )  the electron will gain energy  ( eV )  which is is equivalent in value of the PD it accelerated through

hence if the electrons falls through a PD of 150mV the electron will gain energy of   150MeV

4 0
3 years ago
A. What is the RMS speed of Helium atoms when the temperature of the Helium gas is 343.0 K? (Possibly useful constants: the atom
kkurt [141]

Answer:

(a) 1462.38 m/s

(b) 2068.13 m/s

Explanation:

(a)

The Kinetic energy of the atom can be given as:

K.E = (3/2)KT

where,

K = Boltzman's Constant = 1.38 x 10⁻²³ J/k

K.E = Kinetic Energy of atoms = 343 K

T = absolute temperature of atoms

The K.E is also given as:

K.E = (1/2)mv²

Comparing both equations:

(1/2)mv² = (3/2)KT

v² = 3KT/m

v = √[3KT/m]

where,

m = mass of Helium = (4 A.M.U)(1.66 X 10⁻²⁷ kg/ A.M.U) = 6.64 x 10⁻²⁷ kg

v = RMS Speed of Helium Atoms = ?

Therefore,

v = √[(3)(1.38 x 10⁻²³ J/K)(343 K)/(6.64 x 10⁻²⁷ kg)]

<u>v = 1462.38 m/s</u>

(b)

For double temperature:

T = 2 x 343 K = 686 K

all other data remains same:

v = √[(3)(1.38 x 10⁻²³ J/K)(686 K)/(6.64 x 10⁻²⁷ kg)]

<u>v = 2068.13 m/s</u>

8 0
3 years ago
Doe anyone get this ​
8_murik_8 [283]

Answer:

we know a = F/ M

Explanation:

  • 2 m/s²
  • 0.19 m/s²
  • 9.25 m/ s²
  • 0.04 m/s²
  • 100.39 m/s²

4 0
2 years ago
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