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xenn [34]
3 years ago
14

A bus makes a 10-kilometer journey through town. It's top speed is 50 km/h and it's mean speed is 16kn/h. Explain why these spee

ds are different.
Somebody pls help me with this question!
Physics
1 answer:
Maru [420]3 years ago
6 0

Answer:

speed are different at different places

Explanation:

because it's top speed is 50km/h,so it's initial speed may be less. when it covers some distance it's speed changes again. then we have given a mean speed .mean speed means sum of all speed divided by sum of total time.so the mean speed and final speed differs from each other

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A football has a mass of 2 kg and it accelerates at 20 m/s2. What is its force? (You can use a calculator)
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3 years ago
Read 2 more answers
Consider a step-down transformer with 15 turns in the primary and 6 turns in the secondary windings. Calculate the input impedan
AURORKA [14]

Answer:

Input impedance of this transformer is 50 ohms.  

Explanation:

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Number of turns in the primary coil, N_p=15

Number of turns in the secondary coil, N_s=6

Output impedance of the transformer, V_o=8\ \Omega

The number of turns and the impedance ratio in the step down transformer is given by :

\dfrac{15}{6}=\sqrt{\dfrac{Z_p}{Z_s}}\\\\\dfrac{15}{6}=\sqrt{\dfrac{Z_p}{8}}\\\\Z_p=50\ \Omega

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8 0
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Two Questions Worth 14 points(Gradpoint)
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The frictionless system shown is released from rest. After the right-hand mass has risen 75 cm, the object of mass 0.50m falls l
Andreas93 [3]

Let a be the acceleration of the masses. By Newton's second law, we have

• for the masses on the left,

1.3mg - T = 1.3ma

where T is the magnitude of tension in the pulley cord, and

• for the mass on the right,

T - mg = ma

Eliminate T to get

(1.3mg - T) + (T - mg) = 1.3ma + ma

0.3mg = 2.3ma

\implies a = \dfrac{0.3}{2.3}g \approx 0.13g \approx 1.3 \dfrac{\rm m}{\mathrm s^2}


Starting from rest and accelerating uniformly, the right-hand mass moves up 75 cm = 0.75 m and attains an upward velocity v such that

v^2 = 2a(0.75\,\mathrm m) \\\\ \implies v \approx \sqrt{2\left(1.3\frac{\rm m}{\mathrm s^2}\right)(0.75\,\mathrm m)} \approx 1.4\dfrac{\rm m}{\rm s}

When the 0.5m mass is released, the new net force equations change to

• for the mass on the right,

mg - T' = ma'

where T' and a' are still tension and acceleration, but not having the same magnitude as before the mass was removed; and

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T' - 0.8mg = 0.8ma'

Eliminate T'.

(mg - T') + (T' - 0.8mg) = ma' + 0.8ma'

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3 0
2 years ago
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