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3 years ago

A baseball travels 200 metes in 6 seconds, what is the baseball’s velocity?

Physics

1 answer:

goldenfox [79]3 years ago

8 0

Answer:

33.33 m/sec

Explanation:

A baseball travels 200 metes in 6 seconds,

what is the baseball’s velocity?

use the formula: velocity = distance over time

where (d) distance = 200 m

and (t) time = 6 sec.

plugin values into the formula:

v = d / t

= 200 m / 6 sec

= 33.33 m/sec.

therefore, the baseball's velocity is 33.33 m/sec

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2. A 1.54 kΩ resistor is connected to an AC voltage source with an rms voltage of 240 V.

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(a) The maximum potential difference across the resistor is 339.41 V.

(b) The maximum current through the resistor is 0.23 A.

(d) The average power dissipated by the resistor is 38.4 W.

<h3>Maximum potential difference</h3>

Vrms = 0.7071V₀

where;

V₀ is peak voltage

V₀ = Vrms/0.7071

V₀ = 240/0.7071

V₀ = 339.41 V

<h3> rms current through the resistor </h3>

I(rms) = V(rms)/R

I(rms) = (240)/(1,540)

I(rms) = 0.16 A

<h3>maximum current through the resistor </h3>

I₀ = I(rms)/0.7071

I₀ = (0.16)/0.7071

I₀ = 0.23 A

<h3> Average power dissipated by the resistor</h3>

P = I(rms) x V(rms)

P = 0.16 x 240

P = 38.4 W

Learn more about maximum current here: brainly.com/question/14562756

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2 years ago

How long dose long term alcohol abuse affect the liver?

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</span><span>How can long term alcohol abuse affect the liver?

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3 years ago

Read 2 more answers

Cars A and B are racing each other along the same straight road in the following manner: Car A has a head start and is a distanc

4vir4ik [10]

The question is incomplete. Here is the complete question.

Cars A nad B are racing each other along the same straight road in the following manner: Car A has a head start and is a distance $D_{A}$ beyond the starting line at t = 0. The starting line is at x = 0. Car A travels at a constant speed $v_{A}$ . Car B starts at the starting line but has a better engine than Car A and thus Car B travels at a constant speed $v_{B}$ , which is greater than $v_{A}$ .

Part A: How long after Car B started the race will Car B catch up with Car A? Express the time in terms of given quantities.

Part B: How far from Car B's starting line will the cars be when Car B passes Car A? Express your answer in terms of known quantities.

Answer: Part A: $t=\frac{D_{A}}{v_{B}-v_{A}}$

Part B: $x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}$

Explanation: First, let's write an equation of motion for each car.

Both cars travels with constant speed. So, they are an uniform rectilinear motion and its position equation is of the form:

$x=x_{0}+vt$

where

$x_{0}$ is initial position

v is velocity

t is time

Car A started the race at a distance. So at t = 0, initial position is $D_{A}$ .

The equation will be:

$x_{A}=D_{A}+v_{A}t$

Car B started at the starting line. So, its equation is

$x_{B}=v_{B}t$

Part A: When they meet, both car are at "the same position":

$D_{A}+v_{A}t=v_{B}t$

$v_{B}t-v_{A}t=D_{A}$

$t(v_{B}-v_{A})=D_{A}$

$t=\frac{D_{A}}{v_{B}-v_{A}}$

Car B meet with Car A after $t=\frac{D_{A}}{v_{B}-v_{A}}$ units of time.

Part B: With the meeting time, we can determine the position they will be:

$x_{B}=v_{B}(\frac{D_{A}}{v_{B}-v_{A}} )$

$x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}$

Since Car B started at the starting line, the distance Car B will be when it passes Car A is $x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}$ units of distance.

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The nucleus of most atoms is composed of which of the following sub-atomic particles?

Blababa [14]

Answer:

protons (+ charge) & neutrons (neutral charge)

however protons has a positive charge so it determined what atom it is.

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The roller coaster is designed according to safety regulations that prohibit the speed of the car from exceeding 20 m/s. find th

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maximum allowed value of the speed in roller coaster is given as