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horsena [70]
3 years ago
8

A sports car has an average acceleration of 5.81 m/s2. How long does it take for the car to reach 60.0 mi/h, if it starts from r

est?
Physics
2 answers:
stepladder [879]3 years ago
8 0

Answer:

4.62 s

Explanation:

We have 3 known variables and 1 unknown variable that we want to solve for.

  • a = 5.81 m/s²
  • v₀ = 0 m/s
  • v = 60.0 mi/h
  • t = ?

Convert 60 mi/h to m/s.

  • 60 mi/h -> 26.8224 m/s

Use the kinematic equation that contains all four of these variables.

  • v = v₀ + at
  • 26.8224 = 0 + (5.81)t
  • 26.8224 = 5.81t
  • t = 4.61659208262

It takes the car 4.62 seconds to reach a velocity of 60 mi/h (26.82 m/s) if it starts from rest.

Aleks04 [339]3 years ago
5 0

Answer:

  4.617 s

Explanation:

The speed of 60 mi/h can be converted to m/s:

  (60 mi/h) × (1609.344 m/mi) × (1 h)/(3600 s) = 26.8244 m/s

The relationship between speed and acceleration is ...

  v = at

  t = v/a = (26.8244 m/s)/(5.81 m/s²) ≈ 4.617 s

It will take the car 4.617 seconds to reach 60 mi/h starting from rest.

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6 0
4 years ago
A point charge Q is held at a distance r from the center of a dipole that consists of two charges ±q separated by a distance s.
marishachu [46]

Answer:

a) the magnitude of the force is

F= Q(\frac{kqs}{r^3}) and where k = 1/4πε₀

F = Qqs/4πε₀r³

b)  the magnitude of the torque on the dipole

τ = Qqs/4πε₀r²

Explanation:

from coulomb's law

E = \frac{kq}{r^{2} }

where k = 1/4πε₀

the expression of the electric field due to dipole at a distance r is

E(r) = \frac{kp}{r^{3} } , where p = q × s

E(r) = \frac{kqs}{r^{3} } where r>>s

a) find the magnitude of force due to the dipole

F=QE

F= Q(\frac{kqs}{r^3})

where k = 1/4πε₀

F = Qqs/4πε₀r³

b) b) magnitude of the torque(τ) on the dipole is dependent on the perpendicular forces

τ = F sinθ × s

θ = 90°

note: sin90° = 1

τ = F × r

recall  F = Qqs/4πε₀r³

∴ τ = (Qqs/4πε₀r³) × r

τ = Qqs/4πε₀r²

8 0
3 years ago
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