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CaHeK987 [17]
4 years ago
6

If the swimmer could cross a 14 km channel maintaining the same average velocity as for the first 50 m in the pool, how long wou

ld it take?
Physics
1 answer:
alukav5142 [94]4 years ago
7 0

Complete question:

If the swimmer could cross a 14 km channel maintaining the same average velocity as for the first 50 m in the pool, how long would it take?

For the first 50m in the pool, the average velocity was 2.08 m/s

Answer:

It would take for the swimmer approximately 1.87 hours.

Explanation:

If the swimmer maintains the average velocity on the channel, we should find and approximate value of the time it takes to cross the channel with the Galileo’s kinematic equation:

x=vt

With x the displacement, v the average velocity and t the time, solving for t:

t=\frac{x}{v}=\frac{14000m}{2.08\frac{m}{s}}

t=6730.7 s= 1.87 h

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3 years ago
Why do we need to convert mass to moles in stoichiometry problems.
laiz [17]
Since a mole represents a large number of molecules, it is not possible to count the number of molecules directly, the only method is through weight. The mass is converted to moles so that the number of molecules in the reaction are kept track of.
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3 years ago
A wave pulse travels along a stretched string at a speed of 100 cm/s. What will be the speed in cm/s if the string's tension is
makvit [3.9K]

Answer:

The new velocity of the string is 100 centimeters per second (1 meter per second).

Explanation:

The speed of a wave through a string (v), in meters per second, is defined by the following formula:

v = \sqrt{\frac{T\cdot L}{m} } (1)

Where:

T - Tension, in newtons.

L - Length of the string, in meters.

m - Mass of the string, in kilograms.

The expression for initial and final speeds of the wave are:

Initial speed

v_{o} = \sqrt{\frac{T_{o}\cdot L_{o}}{m_{o}} } (2)

Final speed

v = \sqrt{\frac{(4\cdot T_{o})\cdot (0.5\cdot L_{o})}{2\cdot m_{o}} }

v = \sqrt{\frac{T_{o}\cdot L_{o}}{m_{o}} } (3)

By (2), we conclude that:

v =v_{o}

If we know that v_{o} = 1\,\frac{m}{s}, then the new speed of the wave in the string is v = 1\,\frac{m}{s}.

5 0
3 years ago
What is the prefix notation of 0.0000738?​
MrMuchimi

Answer:

7.38 × 10-5

Explanation:

All numbers in scientific notation or standard form are written in the form m × 10n, where m is a number between 1 and 10 ( 1 ≤ |m| < 10 ) and the exponent n is a positive or negative integer.

To convert 0.0000738 into scientific notation, follow these steps:

Move the decimal 5 times to the right in the number so that the resulting number, m = 7.38, is greater than or equal to 1 but less than 10

Since we moved the decimal to the right the exponent n is negative

n = -5

Write in the scientific notation form, m × 10n

= 7.38 × 10-5

Therefore, the decimal number 0.0000738 written in scientific notation is 7.38 × 10-5 and it has 3 significant figures.

Answer = 7.38 × 10-5

4 0
3 years ago
Read 2 more answers
Ocean waves pass through two small openings, 20.0 m apart, in a breakwater. You're in a boat 70.0 m from the breakwater and init
Klio2033 [76]

Answer:

λ = 5.65m

Explanation:

The Path Difference Condition is given as:

δ=(m+\frac{1}{2})\frac{lamda}{n}  ;

where lamda is represent by the symbol (λ) and is the wavelength we are meant to calculate.

m = no of openings which is 2

∴δ= \frac{3*lamda}{2}

n is the index of refraction of the medium in which the wave is traveling

To find δ we have;

δ= \sqrt{70^2+(33+\frac{20}{2})^2 }-\sqrt{70^2+(33-\frac{20}{2})^2 }

δ= \sqrt{4900+(\frac{66+20}{2})^2}-\sqrt{4900+(\frac{66-20}{2})^2}

δ= \sqrt{4900+(\frac{86}{2})^2 }-\sqrt{4900+(\frac{46}{2})^2 }

δ= \sqrt{4900+43^2}-\sqrt{4900+23^2}

δ= \sqrt{4900+1849}-\sqrt{4900+529}

δ= \sqrt{6749}-\sqrt{5429}

δ=  82.15 -73.68

δ= 8.47

Again remember; to calculate the wavelength of the ocean waves; we have:

δ= \frac{3*lamda}{2}

δ= 8.47

8.47 = \frac{3*lamda}{2}

λ = \frac{8.47*2}{3}

λ = 5.65m

3 0
4 years ago
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