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andreyandreev [35.5K]
3 years ago
6

What equation gives the position at a specific time for an object with constant acceleration?

Physics
1 answer:
melisa1 [442]3 years ago
6 0

Answer:

x = x_{0}+v_{0}t+\frac{1}{2}at^{2} equation gives the position at a specific time for an object with constant acceleration

Explanation:

x = x_{0}+v_{0}t+\frac{1}{2}at^{2} equation gives the position at specific time for an object having constant acceleration. Constant acceleration is referred as the change of velocity with respect to the time is known as the acceleration, but when the velocity changes occurs at constant rate this rate is termed as the constant acceleration. The constant acceleration can never be zero. The velocity changes but that change occurs in the consistently. The acceleration is affected by the mass.

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Two points charge of 4\mu C and 2\mu C are placed at theopposite corners of a rectangle. What is the potential difference Va- Vb
bulgar [2K]

Answer:

Va-Vb=168KV

Explanation:

From the question we are told that

Two points charge of 4\mu C and 2\mu C

Generally we find the  Va and Vb individually to find there difference

Given a rectangle with two equal sides each,Assume lengths for bot sides

Length L=0.3

Breath B=0.4

Diagonal D=\sqrt{0.3^2+0.4^2} =0.5

at  opposite sides

Mathematically Va can represented as

Va =k(\frac{4*10^_-_6}{0.3} +\frac{-2*10^_-_6}{0.5} )

Va =9*10^9(\frac{4*10^_-_6}{0.3} +\frac{-2*10^_-_6}{0.5} )

Va =9*10^9(0.00001333333-0.000004} )

Va =84000V

Va =84KV

Mathematically Vb is  represented as

Va =k(\frac{-4*10^_-_6}{0.3} +\frac{2*10^_-_6}{0.5} )

Va =9*10^9(\frac{-4*10^_-_6}{0.3} +\frac{+2*10^_-_6}{0.5} )

Va =9*10^9(-0.00001333333+0.000004} )

Va =-84000V

Va =-84KV

Therefore

Va-Vb=84-(-84)\\Va-Vb=84+84\\Va-Vb=168KV

7 0
3 years ago
What is the average speed between the times t = 4s and t = 12 s?
ziro4ka [17]
You need distance and time to find average speed.
3 0
3 years ago
True or False?
WINSTONCH [101]

Answer:

true is the answer of the question

5 0
3 years ago
Read 2 more answers
What is the electrical force between q2 and q3? recall that k = 8.99 × 109 n•meters squared over coulombs squared.. 1.0 × 1011 n
max2010maxim [7]

The magnitude of the electrical force between q2 and q3 is given as a ratio between the product of their charges and the square of the distance of separation.

<h3>What is the magnitude of electrical forces between two charges?</h3>

The magnitude of the electrical force between two charges refers to the attractive or repulsive forces that exists between two charges separated by a given distance in an electric field.

The magnitude of the electrical force, F between the two charges q2 and q3 is given be my the formula below

F = \frac{K \times q_2 \times q_3}{d^{2}}

Therefore, the magnitude of the electrical force between q2 and q3 is given as a ratio between the product of their charges and the square of the distance of separation.

Learn more about electrical force at: brainly.com/question/17692887

#SPJ4

6 0
2 years ago
A light beam in glass (n = 1.5) reaches an air-glass interface, at an angle of 60 degrees from the surface. What is the angle of
tester [92]

Answer:

θ₂ = 35.26°

Explanation:

given,

refractive index of air, n₁ = 1

refractive index of glass, n₂ = 1.5

angle of incidence, θ₁ = 60°

angle of refracted light, θ₂ = ?

using Snell's Law

n₁ sin θ₁ = n₂ sin θ₂

1 x sin 60° = 1.5 sin θ₂

sin θ₂ = 0.577

θ₂ = sin⁻¹(0.577)

θ₂ = 35.26°

Hence, the refracted light is equal to  θ₂ = 35.26°

7 0
3 years ago
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