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3 years ago

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If the change in position Dx is related to velocity v (with units of m/s) in the equation Dx=Av, the constant A has which dimens

ion ?

1 answer:

-Dominant- [34]3 years ago

4 0

Answer: seconds

Explanation:

We have the following equation:

$Dx=A.V$

Where $Dx$ has unit of meters ( $m$ ) and $V$ has units of meters per second ( $m/s$ ).

If we input these units in the equation, we will have:

$m=A(m/s)$

Clearing $A$ :

$A=\frac{m.s}{m}=s$

Hence $A$ is related to time, and its unit in this case is second.

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Rock composed of many thin layers. This image appears in an Earth science magazine with this caption: "A non-foliated rock found

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Answer:

its B

Explanation:

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3 years ago

Read 2 more answers

Use the values from PRACTICE IT to help you work this exercise. Suppose the same two vehicles are both traveling eastward, the c

Mariulka [41]

Answer:

A. $v_{3}=12.17m/s$

B. $v_{car}=6.3m/s\\v_{truck}=-6.3m/s$

C. ΔK $=-4.13x10^3J$

Explanation:

From the exercise we know that the car and the truck are traveling eastward. I'm going to name the car 1 and the truck 2

$v_{1}=5.79m/s\\m_{1}=102kg\\v_{2}=18.5m/s\\m_{2}=103kg$

A. Since the two vehicles become entangled the final mass is:

$m_{3}=102kg+103kg=205kg$

From linear momentum we got that:

$p_{1}=p_{2}$

$m_{1}v_{1}+m_{2}v_{2}=m_{3}v_{3}$

$v_{3}=\frac{m_{1}v_{1}+m_{2}v_{2}}{m_{3} }=\frac{(102kg)(5.79m/s)+(103kg)(18.5m/s)}{(205kg)}$

$v_{3}=12.17m/s$

B. The change in velocity of both vehicles are:

For the car

$v_{car}=v_{f}-v_{o}=12.17m/s-5.79m/s=6.38m/s$

For the truck

$v_{truck}=12.17m/s-18.5m/s=-6.3m/s$

C. The change in kinetic energy is:

ΔK= $K_{2}-K_{1} =\frac{1}{2}m_{3}v_{3}^{2}-(\frac{1}{2}m_{1}v_{1}^{2}+\frac{1}{2}m_{2}v_{2}^{2})$

ΔK= $\frac{1}{2}(205)(12.17)^{2}-(\frac{1}{2}(102)(5.79)^{2}+\frac{1}{2}(103)(18.5)^{2})=-4.13x10^{3}J$

ΔK $=-4.13x10^{3}J$

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3 years ago

What additional information do you need to prove ∆ABC ≅ ∆DEF by the SAS Postulate?

miv72 [106K]

Answer:

Option A

You need a Angle C congruent to angle F

Explanation:

EX) Side angle Side = sas

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3 years ago

Two charges +q and -q are situated at certain distance, at the point exactly midway betwee them what happens on electric field a

Luden [163]

Answer: Electric field is not zero but potential is zero

Explanation:

In science, charge, also recognized as electric charge, electrical charge, or electrostatic charge and expressed q, is a component of a unit of body that reveals the extent to which it has more or fewer electrons than protons.

Considering the direction of the electric field is of positive to negative charge. Therefore, it determination be toward due −q also due to both the charges.

But for potential, it is a scalar quantity, so the sum of potential due to all the charges will be zero.

Henceforth, the option (c) Electric field is not zero but potential is zero is the correct answer.

a incorect

From the above statements, we could conclude that option (c) Electric field is not zero, but potential is zero is the correct answer. So, the option (a) Electric field and potential both are zero is an incorrect solution.

b incorect

Among all the options, the option (c) Electric field is not zero but potential is zero is the correct answer.

Therefore, the option (b) Electric field is zero but potential is not zero is an incorrect solution

d) is incorrect

It has been explained that the option (c) Electric field is not zero but potential is zero is the correct answer.

Hence, the option (d) Electric field is not zero but potential is zero is an incorrect solution.

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3 years ago

A point charge is placed 100 m from a 6 uC charge generating an electric field. What is the

kvv77 [185]

The strength of the electric field is 5 N/C

Explanation:

The magnitude of the electric field produced by a single-point charge is given by:

$E=\frac{kQ}{r^2}$

where

$k=8.99\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

Q is the magnitude of the charge

r is the distance from the charge

In this problem, we have:

$Q=6 \mu C = 6\cdot 10^{-6}C$ is the charge producing the field

r = 100 m is the distance from the charge at which we want to calculate the field

Substituting into the equation, we find the s trength of the electric field:

$E=\frac{(8.99\cdot 10^9)(6\cdot 10^{-6})}{(100)^2}=5.4 N/C \sim 5 N/C$

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3 years ago