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4 years ago

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A 3.5 kilogram object is swung in a circular path on the end of a 0.4 meter long string. the object makes one trip around the ci

rcle every 0.2 seconds. determine the speed of the object.

1 answer:

jeyben [28]4 years ago

6 0

Path length is 2*pi*0.4=2.512
Speed=distance/time
Speed =2.512/0.2=12.56m/s

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A 15 kg bucket of water is suspended by a very light ropewrapped around a solid uniform cylinder 0.300 m in diamter withmass 12.

matrenka [14]

Answer:

Part a)

$T = 42 N$

Part b)

$v_f = 11.8 m/s$

Part c)

$t = 1.7 s$

Part d)

$F = 159.7 N$

Explanation:

Part a)

While bucket is falling downwards we have force equation of the bucket given as

$mg - T = ma$

for uniform cylinder we will have

$TR = I\alpha$

so we have

$T = \frac{1}{2}MR^2(\frac{a}{R^2})$

$T = \frac{1}{2}Ma$

now we have

$mg = (\frac{M}{2} + m)a$

$a = \frac{mg}{(\frac{M}{2} + m)}$

$a = \frac{15 \times 9.81}{(6 + 15)}$

$a = 7 m/s^2$

now we have

$T = \frac{12 \times 7}{2}$

$T = 42 N$

Part b)

speed of the bucket can be found using kinematics

so we have

$v_f^2 - v_i^2 = 2 a d$

$v_f^2 - 0 = 2(7)(10)$

$v_f = 11.8 m/s$

Part c)

now in order to find the time of fall we can use another equation

$v_f - v_i = at$

$11.8 - 0 = 7 t$

$t = 1.7 s$

Part d)

as we know that cylinder is at rest and not moving downwards

so here we can use force balance

$F = T + Mg$

$F = 42 + (12 \times 9.81)$

$F = 159.7 N$

5 0

3 years ago

Electrical energy can be transformed into other types of energy. We often experience this transformation of energy in our everyd

vitfil [10]

Well I’m not sure because you don’t have anything listed

3 0

4 years ago

Determine the value of the resultant and its location from O.<br>see attach image.

xxTIMURxx [149]

Answer:

Explanation:

In the x direction the force will be

½(-w₀)L/2 = -¼w₀L

acting ⅔(L/2) = L/3 below the x axis.

In the y direction the force will be

½(-w₀)L + ½w₀L/2 = -¼w₀L

the magnitude of the resultant will be

F = w₀L √((-¼)² + (-¼)²) = w₀L√⅛

in the direction

θ = arctan(-¼w₀L / -¼w₀L) = 225°

to find the distance, we balance moments

(w₀L√⅛)[d] = ½(w₀)L[⅔L] + ¼w₀L[⅔L/2] - ¼w₀L[L - ⅓L/2]

(√⅛)[d] = ½ [⅔L] + ¼ [⅔L/2] - ¼ [L - ⅓L/2]

(√⅛)[d] = ½[⅔L] + ¼[⅔L/2] - ¼[L - ⅓L/2]

(√⅛)[d] = ⅓L + ⅟₁₂L - ¼L + ⅟₂₄L

(√⅛)[d] = 5L/24

d = 5L/24 / (√⅛)

d = 5√⅛L/3

8 0

3 years ago

2. A 56N force is exerted on an 8kg object. What is the object's acceleration?

White raven [17]

acceleration = 7 m/s²

Explanation:

To determine the object's acceleration we use the following formula:

force (N) = mass (kg) × acceleration (m/s²)

acceleration (m/s²) = force (N) / mass (kg)

acceleration = 56 N / 8 kg

acceleration = 7 m/s²

Learn more about:

acceleration

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7 0

3 years ago

A train whistle has a sound intensity level of 70. dB, and a library has a sound intensity level of about 40. dB. How many times

kodGreya [7K]

Answer:

The sound intensity of train is 1000 times greater than that of the library.

Explanation:

We have expression for sound intensity level,

$L=10log_{10}\left ( \frac{I}{I_0}\right )$

A train whistle has a sound intensity level of 70 dB

We have

$70=10log_{10}\left ( \frac{I_1}{I_0}\right )$

A library has a sound intensity level of about 40 dB

We also have

$40=10log_{10}\left ( \frac{I_2}{I_0}\right )$

Dividing both equations

$\frac{70}{40}=\frac{10log_{10}\left ( \frac{I_1}{I_0}\right )}{10log_{10}\left ( \frac{I_2}{I_0}\right )}\\\\\frac{7}{4}=\frac{log_{10}\left ( \frac{I_1}{I_0}\right )}{log_{10}\left ( \frac{I_2}{I_0}\right )}\\\\10^7\frac{I_2}{I_0}=10^4\frac{I_1}{I_0}\\\\\frac{I_1}{I_2}=10^3=1000$

The sound intensity of train is 1000 times greater than that of the library.

3 0

3 years ago