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2 years ago

6

A kayak took 5 hours to finish its trip on a river. If it traveled at an

1 answer:

snow_tiger [21]2 years ago

5 0

It would mean that it was a 90 mile trip. All you would have to do is multiply 18 by 5

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Concept Simulation 20.4 provides background for this problem and gives you the opportunity to verify your answer graphically. Ho

77julia77 [94]

Answer:

The time constant is 1.049.

Explanation:

Given that,

Charge $q{t}= 0.65 q_{0}$

We need to calculate the time constant

Using expression for charging in a RC circuit

$q(t)=q_{0}[1-e^{-(\dfrac{t}{RC})}]$

Where, $\dfrac{t}{RC}$ = time constant

Put the value into the formula

$0.65q_{0}=q_{0}[1-e^{-(\dfrac{t}{RC})}]$

$1-e^{-(\dfrac{t}{RC})}=0.65$

$e^{-(\dfrac{t}{RC})}=0.35$

$-\dfrac{t}{RC}=ln (0.35)$

$-\dfrac{t}{RC}=-1.049$

$\dfrac{t}{RC}=1.049$

Hence, The time constant is 1.049.

6 0

2 years ago

a car moving at 5 m/s accelerates at a rate of 10 m/s2 for 25 seconds. How far does it move during this time?

Alexus [3.1K]

Answer:

t is time in s For example, a car accelerates in 5 s from 25 m/s to 3 5m/s. Its velocity changes by 35 - 25 = 10 m/s. Therefore its acceleration is 10 ÷ 5 = 2 m/s2

Explanation:

3 0

2 years ago

What type of energy comes from the food that we eat? A) chemical B) electrical C) kinetic D) mechanical

ella [17]

Chemical energy comes from the food that we eat

7 0

2 years ago

Read 2 more answers

How does the plant lichen quicken the weathering process?

konstantin123 [22]

C I Believe , Have Ah Good Day

4 0

2 years ago

Water flows through a cylindrical pipe of varying cross-section. The velocity is 3.0 m/s at a point where the pipe diameter is 1

Setler79 [48]

Answer:

The flow rate is $R =2.357 *10^{-4} \ m^3/s$

Explanation:

From the question we are told that

The velocity is $v = 3.0 \ m/s$

The diameter of the pipe is $d = 1.0 \ cm = 0.01 \ m$

The radius of the pipe is mathematically represented as

$r = \frac{d}{2}$

substituting values

$r = \frac{0.01}{2}$

$r = 0.005 \ m$

The flow rate is mathematically represented as

$R = v * A$

Where is the cross-sectional area of the pipe which is mathematically evaluated as

$A = \pi r^2$

substituting values

$A = 3.142 * (0.005)^2$

$A = 7.855 * 10^{-5} \ m^2$

So

$R = 3.0 * 7.855 *10^{-5}$

$R = 2.357*10^{-4} \ m^3 /s$

8 0

2 years ago