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MA_775_DIABLO [31]
3 years ago
9

Under what circumstances will the distance traveled by an object be the same as the magnitude of the displacement of an object?​

Physics
1 answer:
KiRa [710]3 years ago
5 0
Hi Randa, want me to help you with physics? The answer is when the object moves in a straight line.
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A train travels 100 km east and then 75 km west. What is the magnitude of<br> its displacement?
tamaranim1 [39]

Answer:

displacement = 100 - 75 = 25 km east

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3 years ago
Se o Universo tem uma idade definida, de aproximadamente 13,7 bilhões de anos, o que existiu antes do big-bang?
UkoKoshka [18]
Ninguém sabe exatamente
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3 years ago
If a charge at 60c flow in a conductor for 30 second then the current that flow in a conductor is​
saw5 [17]

Explanation:

<h3>Given</h3>

- Charge = 60c

time = 30 sec

<h3>To find -</h3>

current

<h3>Solution </h3>

Current = Charge/time

I = V/T

I = 60/30

I = 2 ampere

More to know -

I = Current

V = Charge

T = Time

3 0
3 years ago
To analyze the motion of a body that is traveling along a curved path, to determine the body's acceleration, velocity, and posit
DiKsa [7]

To solve this problem we will apply the kinematic equations of linear motion and centripetal motion. For this purpose we will be guided by the definitions of centripetal acceleration to relate it to the tangential velocity. With these equations we will also relate the linear velocity for which we will find the points determined by the statement. Our values are given as

R = 350ft

a_t = 1.1ft/s^2

PART A )

a_c = \frac{V^2}{R}

a_c = \frac{V^2}{350}

Calculate the velocity of the motorcycle when the net acceleration of the motorcycle is 5.25ft/s^2

a = \sqrt{a_t^2+a_r^2}

5.25 = \sqrt{(1.1)^2+(\frac{v^2}{350})^2}

27.5625 = 1.21 + \frac{v^4}{122500}

v=42.3877ft/s

Now calculate the angular velocity of the motorcycle

v = r\omega

42.3877 = 350\omega

\omega = 0.1211rad/s

Calculate the angular acceleration of the motorcycle

a_t = r\alpha

1.1 = 350\alpha

\alpha = 3.1428*10^{-3}rad/s^2

Calculate the time needed by the motorcycle to reach an acceleration of

5.25ft/s^2

\omega = \alpha t

0.1211 = 3.1428*10^{-3}t

t = 38.53s

PART B) Calculate the velocity of the motorcycle when the net acceleration of the motorcycle is 6.75ft/s^2

a = \sqrt{a_t^2+a_r^2}

6.75 = \sqrt{(1.1)^2+(\frac{v^2}{350})^2}

45.5625 = 1.21 + \frac{v^4}{122500}

v=48.2796ft/s

PART C)

Calculate the radial acceleration of the motorcycle when the velocity of the motorcycle is 21.5ft/s

a_r = \frac{v^2}{R}

a_r = \frac{21.5^2}{350}

a_r =1.3207ft/s^2

Calculate the net acceleration of the motorcycle when the velocity of the motorcycle is 21.5ft/s

a = \sqrt{a_t^2+a_r^2}

a = \sqrt{(1.1)^2+(1.3207)^2}

a = 1.7187ft/s^2

PART D) Calculate the maximum constant speed of the motorcycle when the maximum acceleration of the motorcycle is 6.75ft/s^2

a = \sqrt{a_t^2+a_r^2}

6.75 = \sqrt{(1.1)^2+(\frac{v^2}{350})^2}

45.5625 = 1.21 + \frac{v^4}{122500}

v=48.2796ft/s

3 0
3 years ago
One half of a balanced chemical equation is shown.
Zarrin [17]

Answer:

A: 3 Mg, 2 P, 14 O, 12 H

Explanation:

I took the test on Edgenuity

6 0
3 years ago
Read 2 more answers
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