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3 years ago

If the period of a spring is 5 seconds what is the frequency

Physics

1 answer:

Scilla [17]3 years ago

7 0

Answer:

C. 0.2 Hertz

Explanation:

The frequency of a spring is equal to the reciprocal of the period:

$f=\frac{1}{T}$

where

f is the frequency

T is the period

For the spring in this problem,

T = 5 s

therefore, the frequency is

$f=\frac{1}{5 s}=0.2 Hz$

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77julia77 [94]

Answer:

1) 51 m

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Explanation:

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3 years ago

( 8c5p79) A certain force gives mass m1 an acceleration of 13.5 m/s2 and mass m2 an acceleration of 3.5 m/s2. What acceleration

Talja [164]

Answer:

$4.725 m/s^{2}$

Explanation:

We know that from Newton's second law of motion, F=ma hence making acceleration the subject then $a=\frac {F}{m}$ where a is acceleration, F is force and m is mass

Also making mass the subject of the formula $m=\frac {F}{a}$

For $m1= \frac {F}{13.5}$ and $m2=\frac {F}{3.5}$ hence $F=(m2-m1)a= (\frac {F}{3.5}-\frac {F}{13.5})a=0.2116402116\\\frac {1}{a}=0.2116402116\\a=4.725 m/s^{2}$

4 0

3 years ago

A ceiling fan with 90-cm-diameter blades is turning at 64 rpm . Suppose the fan coasts to a stop 28 s after being turned off. Wh

mixas84 [53]

Answer:

the speed of the tip of a blade 10 s after the fan is turned off is 16.889 m/s.

Explanation:

Given;

diameter of the ceiling fan, d = 90 cm = 0.9 m

angular speed of the fan, ω = 64 rpm

time taken for the fan to stop, t = 28 s

The distance traveled by the ceiling fan when it comes to a stop is calculated as;

$d = vt = \omega r\times t= ( \frac{64 \ rev}{\min} \times \frac{2 \pi \ rad}{rev} \times \frac{1 \min}{60 \ s} \times 0.9 \ m) \times 28 \ s\\\\d = 168.89 \ m$

The speed of the tip of a blade 10 s after the fan is turned off is calculated as;

$v = \frac{d}{t} \\\\v = \frac{168.89}{10} \\\\v = 16.889 \ m/s$

Therefore, the speed of the tip of a blade 10 s after the fan is turned off is 16.889 m/s.

4 0

3 years ago

Please help (will mark brainliest)

serg [7]

Answer:

if im not mistaken i think its d let me know if correct plz

7 0

2 years ago

Read 2 more answers

A student connects an object with mass m to a rope with a length r and then rotates the rope around her head parallel to the gro

Alexandra [31]

The object takes 0.5 seconds to complete one rotation, so its rotational speed is 1/0.5 rot/s = 2 rot/s.

Convert this to linear speed; for each rotation, the object travels a distance equal to the circumference of its path, or 2π (1.2 m) = 2.4π m ≈ 7.5 m, so that

2 rot/s = (2 rot/s) • (2.4π m/rot) = 4.8π m/s ≈ 15 m/s

thus giving it a centripetal acceleration of

a = (4.8π m/s)² / (1.2 m) ≈ 190 m/s².

Then the tension in the rope is

T = (50 kg) a ≈ 9500 N.

7 0

3 years ago