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konstantin123 [22]
3 years ago
10

!!31 Points and BRAINLIEST

Physics
2 answers:
Eva8 [605]3 years ago
5 0

All of these problems are solved with the same formula:

Wave Speed = (frequency)x(wavelength) .

For each problem, substitute the two numbers that you KNOW into this formula, and solve it for the missing number.

This explanation is way more useful and helpful to you than a list of the answers.

VARVARA [1.3K]3 years ago
3 0
1) 30m/s
2) 5m/s
3) 40Hz
4) 400Hz
5) .77m
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A fly wheel with a diameter of 1.20m is rotating at an angular speed of 200 Rev/min
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Explanation:

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A cyclist and his bicycle have a combined mass of 88 kg and a combined
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Answer:D

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If an object falls and ends with a velocity of 58.8 m/s, how long was the<br><br> object falling?
tensa zangetsu [6.8K]

Answer:

the time taken for the object to fall is 6 s.

Explanation:

Given;

final velocity of the object, v = 58.8 m/s

initial velocity of the object, u = 0

The height of fall of the object is calculated as;

v² = u² + 2gh

v² = 2gh

h = \frac{v^2}{2g} \\\\h = \frac{(58.8)^2}{2(9.8)} \\\\h = 176.4 \ m

The time to fall through the height is calculated as;

h =ut+  \frac{1}{2} gt^2\\\\h = 0 +  \frac{1}{2} gt^2\\\\h =  \frac{1}{2} gt^2\\\\t= \sqrt{\frac{2h}{g} } \\\\t = \sqrt{\frac{2\times 176.4}{9.8} } \\\\t = 6 \ s

Therefore, the time taken for the object to fall is 6 s.

6 0
3 years ago
If the block is subjected to the force of F = 500 N, determine its velocity at s = 0.5 m. When s = 0, the block is at rest and t
STatiana [176]

Answer:

The velocity is  4.6 m/s^2

Explanation:

Given:

Force = 500N

Distance  s= 0

To find :

Its velocity at s = 0.5 m

Solution:

\sum F_{x}=m a

F\left(\frac{4}{5}\right)-F_{S}=13 a

500\left(\frac{4}{5}\right)-\left(k_{s}\right)=13 a

400-(500 s)=13 a

a = \frac{400 -(500s)}{13}

a = (30.77 -38.46s) m/s^2

Using the relation,

a=\frac{d v}{d t}=\frac{d v}{d s} \times \frac{d s}{d t}

a=v \frac{d v}{d s}

v d v=a d s

Now integrating on both sides

\int_{0}^{v} v d v=\int_{0}^{0.5} a d s

\int_{0}^{v} v d v=\int_{0}^{0.5}(30.77-38.46 s) d s

\left[\frac{v^{2}}{2}\right]_{0}^{2}=\left[\left(30.77 s-19.23 s^{2}\right)\right]_{0}^{0.5}

\left[\frac{v^{2}}{2}\right]=\left[\left(30.77(0.5)-19.23(0.5)^{2}\right)\right]

\left[\frac{v^{2}}{2}\right]=[15.385-4.807]

\left[\frac{v^{2}}{2}\right]=10.578

v^{2}=10.578 \times 2

v^{2}=21.15

v = \sqrt{21.15}

v = 4.6 m/s^2

8 0
3 years ago
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