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3 years ago

Displacement vectors of 10 m west and 14 m west make a resultant vector that is _____.

Physics

2 answers:

Anastaziya [24]3 years ago

6 0

For this case we can write each one of the vectors in the following way:

$v1 = 10 i\\v2 = 14 i$

This means that the x-axis is defined as positive towards the west.

Then, the resulting vector is:

$v1 + v2 = 10i + 14i\\v1 + v2 = 24 i$

Answer:

Displacement vectors of 10 m west and 14 m west make a resultant vector that is 24 m west

Mars2501 [29]3 years ago

4 0

Since they are both west, they simply add. The resultant is 24m, pointing west

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Based on the graph, which data point is most likely to have experimental

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Answer:

B. 59 kg

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2 years ago

A spring that has a spring constant of 1400 N/m is stretched to a length of 2.5 m. If the normal length of the spring is 1.0 m,

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3 years ago

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Answer:

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3 years ago

Two taut strings of identical mass and length are stretched with their ends fixed, but the tension in one string is 1.10 times g

ollegr [7]

Answer:

The beat frequency when each string is vibrating at its fundamental frequency is 12.6 Hz

Explanation:

Given;

velocity of wave on the string with lower tension, v₁ = 35.2 m/s

the fundamental frequency of the string, F₁ = 258 Hz

velocity of wave on the string with greater tension;

$v_1 = \sqrt{\frac{T_1}{\mu }$

where;

v₁ is the velocity of wave on the string with lower tension

T₁ is tension on the string

μ is mass per unit length

$v_1 = \sqrt{\frac{T_1}{\mu} } \\\\v_1^2 = \frac{T_1}{\mu} \\\\\mu = \frac{T_1}{v_1^2} \\\\ \frac{T_1}{v_1^2} = \frac{T_2}{v_2^2}\\\\v_2^2 = \frac{T_2v_1^2}{T_1}$

Where;

T₁ lower tension

T₂ greater tension

v₁ velocity of wave in string with lower tension

v₂ velocity of wave in string with greater tension

From the given question;

T₂ = 1.1 T₁

$v_2^2 = \frac{T_2v_1^2}{T_1} \\\\v_2 = \sqrt{\frac{T_2v_1^2}{T_1}} \\\\v_2 = \sqrt{\frac{1.1T_1*(35.2)^2}{T_1}}\\\\v_2 = \sqrt{1.1(35.2)^2} = 36.92 \ m/s$

Fundamental frequency of wave on the string with greater tension;

$f = \frac{v}{2l} \\\\2l = \frac{v}{f} \\\\thus, \frac{v_1}{f_1} =\frac{v_2}{f_2} \\\\f_2 = \frac{f_1v_2}{v_1} \\\\f_2 =\frac{258*36.92}{35.2} \\\\f_2 = 270.6 \ Hz$

Beat frequency = F₂ - F₁

= 270.6 - 258

= 12.6 Hz

Therefore, the beat frequency when each string is vibrating at its fundamental frequency is 12.6 Hz

6 0

3 years ago

When catching a baseball, the ball applies a force of 39.6 N to a catcher's glove. If the work done on the catchers glove is 4

cricket20 [7]

Answer:

d = 1.19 m

Explanation:

Given that,

The force applied by the ball, F = 39.6 N

The work done on the catchers glove is 47.5 J

We need to find the distance traveled by the ball. We know that,

Work done, W = Fd

Where

d is the distance traveled

$d=\dfrac{W}{F}\\\\d=\dfrac{47.5 }{39.6 }\\\\d=1.19\ m$

So, it will cover 1.19 m.

4 0

3 years ago