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4 years ago

An object accelerated from rest with a constant acceleration of 5.2 m/s^2, what will its velocity be after 3 s

Physics

1 answer:

GarryVolchara [31]4 years ago

4 0

Answer: The velocity after 3 s will be $15.6\frac{m}{s}$

Explanation:

Since the initial velocity v0 is zero:

$v(t) = at + v_0 = at\\v(3) = 3a = 3\cdot5.2 \frac{s\cdot m}{s^2}= 15.6\frac{m}{s}\\$

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An elevator accelerates uniformly from rest to a speed of 2.5 m/s in 12s. What is the distance the elevator travels during this

Aleks04 [339]

We use the kinematic equation,

$s= u t +\frac{1}{2}a t^2$

Here, s is the distance traveled by elevator u is the initial velocity and t is time.

As elevator accelerates uniformly from rest to a speed of 2.5 m/s in 12 s, so

$a = \frac{v_{f}- v_{i} }{t} = \frac{2.5 m/s-0}{12 s} = 0.208 \ m/s^2$

Thus, the distance the elevator travels during 12 s time is

$s= 0+\frac{1}{2} 0.208 \ m/s^2(12)^2 = 14.97 m \simeq 15 m$

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4 years ago

A man stands on a scale and holds a heavy object in his hands. What happens to the scale reading if the man quickly lifts the ob

Novosadov [1.4K]

Answer:

Explanation:

When he accelerates the heavy object up , the reading increases because an extra downward normal force acts on it, then scale reading returns to the same reading as when standing stationary, and then decreases as although he is lifting the heavy object , the acceleration is decreasing ,so the extra upward normal force acts.

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4 years ago

Why is the meteor shower is best observed after midnight?

harkovskaia [24]

Answer:

At dawn your location on earth is pointed straight in the direction of the Earth's travel in its orbit. Between midnight and dawn you are moving head-on through the location of the meteors in space, which means that you will, on average, observe more of them.

- public.nrao.edu

Explanation:

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3 years ago

The cable of the 1800kg elevator cab in Fig. 8−56 snaps when the cab is at rest at the first floor, where the cab bottom is a di

drek231 [11]

a ) The speed of the cab just before it hits the spring = 7.4 m / s

b ) The maximum distance x that the spring is compressed = 0.9 m

c ) The distance that the cab will bounce back up the shaft = 2.8 m

a ) The speed of the cab just before it hits the spring,

Ki + Pi = K final + P final + W

1 / 2 m vo² + m g hi = 1 / 2 m v² + m g h + f d

0 + ( 1800 * 9.8 * 3.7 m ) = ( 1 / 2 * 1800 * v² ) + 0 + ( 4400 * 3.7 )

v = 7.4 m / s

b ) The maximum distance x that the spring is compressed,

Ki + Pi = K final + P final + W + Fs

1 / 2 m vo² + m g x = 1 / 2 m v² + m g h + f d + 1 /2 k x²

( 1/2 * 1800 * 7.4² ) + ( 1800 * 9.8 * x ) =0 + 0+ ( 4400 * x ) + ( 1/2 * 1800 * x² )

75000 x² + 13420 x - 50625 = 0

x = 0.9 m

c ) The distance that the cab will bounce back up the shaft,

Ki + Pi + Fs = K final + P final + W

1 / 2 m vo² + m g x + 1 /2 k x² = 1 / 2 m v² + m g h + f d

0 + 0 + ( 1 / 2 * 0.15 * 0.9² ) = 0 + ( 1800 * 9.8 * h ) + ( 4400 * h )

h = 2.8 m

Therefore,

a ) The speed of the cab just before it hits the spring = 7.4 m / s

b ) The maximum distance x that the spring is compressed = 0.9 m

c ) The distance that the cab will bounce back up the shaft = 2.8 m

To know more about law of conservation of energy

brainly.com/question/12050604

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3 0

1 year ago

Suppose a wheel with a tire mounted on it is rotating at the constant rate of 3.37 times a second. A tack is stuck in the tire a

zhannawk [14.2K]

Answer:

The tangential speed of the tack is 8.19 m/s.

Explanation:

The wheel rotates 3.37 times a second that means wheel complete 3.37 revolutions in a second. Therefore, the angular speed ω of the wheel is given as follows:

$\omega =3.37rev/s \times(\frac{2\pi rad}{1s} )\\\\=21.174rad/s$

Use the relation of angular speed with tangential speed to find the tangential speed of the tack.

The tangential speed v of the tack is given by following expression

v = ω r

Here, r is the distance to the tack from axis of rotation.

Substitute 21.174 rad/s for ω, and 0.387 m for r in the above equation to solve for v.

v = 21.174 × 0.387

v = 8.19m/s

Thus, The tangential speed of the tack is 8.19 m/s.

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3 years ago

Read 2 more answers