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2 years ago

Find the instantaneous acceleration at t=ls for an object moving along a straight axis with velocity function:

1 answer:

8 0

The answer is A.
Explanation:
We know that the average acceleration a for an interval of time Δt is expressed as:

a = Δv
Δt
where Δv is the change in velocity that occurs during Δt.
e formula for the instantaneous acceleration a is almost the same, except that we need to indicate that we're interested in knowing what the ratio of Δv to Δt approaches as Δt approaches zero.

We can indicate that by using the limit notation.

So, the formula for the instantaneous acceleration is:

a = lim Δv
Δt→0 Δt

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1) A thin ring made of uniformly charged insulating material has total charge Q and radius R. The ring is positioned along the x

allochka39001 [22]

Answer:

(A) considering the charge "q" evenly distributed, applying the technique of charge integration for finite charges, you obtain the expression for the potential along any point in the Z-axis:

$V(z)=\frac{Q}{4\pi (\epsilon_{0}) \sqrt{R^{2} +z^{2}} }$

With $(\epsilon_{0})$ been the vacuum permittivity

(B) The expression for the magnitude of the E(z) electric field along the Z-axis is:

$E(z)=\frac{QZ}{4\pi (\epsilon_{0}) (R^{2} +z^{2})^{\frac{3}{2} } }$

Explanation:

(A) Considering a uniform linear density $λ_{0}$ on the ring, then:

$dQ=\lambda dl$ (1)⇒ $Q=\lambda_{0} 2\pi R$ (2)⇒ $\lambda_{0}=\frac{Q}{2\pi R}$ (3)

Applying the technique of charge integration for finite charges:

$V(z)= 4\pi (ε_{0})\int\limits^a_b {\frac{1}{ r' }} \, dQ$ (4)

Been r' the distance between the charge and the observation point and a, b limits of integration of the charge. In this case a=2π and b=0.

Using cylindrical coordinates, the distance between a point of the Z-axis and a point of a ring with R radius is:

$r'=\sqrt{R^{2} +Z^{2}}$ (5)

Using the expressions (1),(4) and (5) you obtain:

$V(z)= 4\pi (\epsilon_{0})\int\limits^a_b {\frac{\lambda_{0}R}{ \sqrt{R^{2} +Z^{2}} }} \, d\phi$

Integrating results:

$V(z)=\frac{Q}{4\pi (\epsilon_{0}) \sqrt{R^{2} +z^{2}} }$ (S_a)

(B) For the expression of the magnitude of the field E(z), is important to remember:

$|E| =-\nabla V$ (6)

But in this case you only work in the z variable, soo the expression (6) can be rewritten as:

$|E| =-\frac{dV(z)}{dz}$ (7)

Using expression (7) and (S_a), you get the expression of the magnitude of the field E(z):

$E(z)=\frac{QZ}{4\pi (\epsilon_{0}) (R^{2} +z^{2})^{\frac{3}{2} } }$ (S_b)

4 0

3 years ago

Answer it or not your choice

Zigmanuir [339]

Answer:

Explanation:

(a) 0 to 90

(b) 90 to 180

3 0

2 years ago

Which one of the following is a measurement of energy used?

Leokris [45]

The correct answer to the question is D) 6 joules.

EXPLANATION:

As per the question, four measurements are given i.e 24 m/s, 60 mg, 15 L and 6 J.

We are asked to determine the measurement which corresponds to energy.

Energy is measured in various units like joule, erg, electron volt etc.

Joule is the unit of energy in S.I system.

Hence, the correct measurement which corresponds to energy is 6 joules.

All other measurements like 24 m/s, 60 mg and 15 L are not the measurements of energy. These are the measurements of speed, mass and volume respectively.

5 0

4 years ago

Read 2 more answers

Equations E = 1 2πε0 qd z3 and E = 1 2πε0 P z3 are approximations of the magnitude of the electric field of an electric dipole,

tamaranim1 [39]

Answer:

The ratio of $E_{app}$ and $E_{act}$ is 0.9754