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3 years ago

a bus starts to move with the speed of 20 m/s.calculate the time taken by the bus to cover a distance of 72km

Physics

1 answer:

kkurt [141]3 years ago

8 0

speed(s)=distance(d)/timetaken(t)

20m/s=72000m/t

t=72000/20m/s

ANS:

time taken (t)=3600s(60mins or 1 hr)

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(fraction equal to ' 1 ') ^

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nataly862011 [7]

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2 years ago

A uniform log of length L is inclined 30° from the horizontal when supported by a frictionless rock located 0.6L from its left e

mafiozo [28]

Answer:

x = 0.974 L

Explanation:

given,

length of inclination of log = 30°

mass of log = 200 Kg

rock is located at = 0.6 L

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mass of engineer = 53.5 Kg

let x be the distance from left at which log is horizontal.

For log to be horizontal system should be in equilibrium

∑ M = 0

mass of the log will be concentrated at the center

distance of rock from CM of log = 0.1 L

now,

∑ M = 0

$m_{log} g \times 0.1 L = m_{engineer} g \times (x - 0.6 L)$

$200 \times 0.1 L = 53.5 \times (x - 0.6 L)$

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3 years ago

an audio CD has a diameter of 120 mm and spins at up to 540 rpm. When a CD is spinning at its maximum rate, how much time is req

Andru [333]

Answer:

t = 0.1111 s

Explanation:

Let's reduce the magnitudes to the SI system

d = 120 mm (1m / 1000 mm)

d= 0.120 m

w = 540 rpm (2pi rad / 1 rev) (1 min / 60s)

w= 56.55 rad / s

When at maximum speed we can use angular kinematic relationships to find the time for a sperm revolution with zero angular acceleration

W = θ / t

t = θ / w

t = 2π / 56.55

t = 0.1111 s

6 0

3 years ago

A 0.950 kg block is attached to a spring with spring constant 16.0 N/m . While the block is sitting at rest, a student hits it w

Bumek [7]

Answer:

Explanation:

Given that,

Mass attached m = 0.95kg

Spring constant k = 16N/m

Instantaneous speed v = 36cm/s = 0.36m/s

Amplitude A=?

When x = 0.7A

Using conservation of energy

∆K.E + ∆P.E = 0

K.E(final) — K.E(initial) + P.E(final) — P.E(initial) = 0

At the beginning immediately the hammer hits the mass, the potential energy is 0J, Therefore, P.E(initial) = 0J, so the speed is maximum.

Also, at the end, at maximum displacement, the speed is zero, therefore, K.E(final) = 0

So, the equation becomes

— K.E(initial) + P.E(final) = 0

K.E(initial) = P.E(final)

½mv² = ½kA²

mv² = kA²

0.95 × 0.36² = 16×A²

0.12312 = 16•A²

A² = 0.12312/16

A² = 0.007695

A = √0.007695

A = 0.088 m

A = 8.8cm

B. Speed at x = 0.7A

Using the same principle above

K.E(initial) = P.E(final)

½mv² = ½kA²

Where A = 0.7A = 0.7 × 0.088 = 0.0614m

Then,

½× 0.95 × v² = ½ × 16 × 0.0614²

0.475v² = 0.0310644

v² = 0.0310644/0.475

v² = 0.0635

v = √0.0635

v = 0.252 m/s

v = 25.2 cm/s

8 0

3 years ago

a bus starts to move with the speed of 20 m/s.calculate the time taken by the bus to cover a distance of 72km​

a bus starts to move with the speed of 20 m/s.calculate the time taken by the bus to cover a distance of 72km