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2 years ago

Calculate the aceleration of a vehicle wich start with a zero meter per second, and acelerates to 34 m/s in 21 s.

Physics

1 answer:

Leviafan [203]2 years ago

8 0

Answer:

For the aceleration we have:

Vf = Vo + a * t

Clearing "a":

a = (Vf - Vo) / t

Replacing and resolving:

a = (34 m/s - 0 m/s) / 21 s

a = 34 m/s / 21 s

a = 1,61 m/s^2

The aceleration of the vehicle is<u> 1,61 meters per second squared</u>

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2 years ago

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You ride on an elevator that is moving with constant upward acceleration while standing on a bathroom scale. the reading on the

Blababa [14]

The reading on the scale is greater than your actual weight.

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3 years ago

List several examples of applied force, normal force, and friction that you’ve observed in your life.

grandymaker [24]

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2 years ago

The velocity of an object is the distance it travels per unit time. Suppose the velocity of a gilding bird is measured to be 52.

Elanso [62]

Answer:

$d=7.115s$

Explanation:

What problem says can be written mathematically as:

$v=\frac{d}{t}$

Where:

$v=Velocity\\t=Time\\d=Distance$

The problem itself it's really simple, we only need to replace the data provided in the previous equation, but first, let's convert the units of the velocity from cm/s to m/s because we have to work with the same units and working in meters is the most apropiate action, because is the base unit of length in the International System of Units:

$52\frac{cm}{s} *\frac{1m}{100cm} =0.52\frac{m}{s}$

Now, we can replace the data in the equation and find the time it will take the bird to travel 3.7 m:

$0.52=\frac{3.7}{t}$

Solving for t, multiplying by t both sides, and dividing by 0.52 both sides:

$t=\frac{3.7}{0.52} =7.115384615s\approx7.115s$

5 0

2 years ago

A 0.520-kg object attached to a spring with a force constant of 8.00 N/m vibrates in simple harmonic motion with an amplitude of

Veseljchak [2.6K]

Answer:

Explanation:

Given that,

Mass attached to spring

M = 0.52kg

Force constant K = 8N/m

Amplitude A = 11.6 cm

a. Maximum speed?

Angular velocity is calculated using

w = √k/m

w = √8/0.52

w = √15.385

w = 3.922rad/s

Then, the relation ship between angular velocity and linear velocity is given as

v = - w•A

v = - 3.922 × 11.6

v = - 45.5 cm/s

Then, the maximum velocity is

vmax = |v|= 45.5cm/s

b. Acceleration a?

Acceleration can be determine using the formula

a = -w²• A

a = -3.922² × 11.6

a = -178.46 cm/s²

Magnitude of the acceleration is 178.46cm/s²

c. Speed when the object is at 9.6cm from equilibrium position?

Generally,

The position of the object at equilibrium is

x(t) = A•Cos(wt)

x(t) = 11.6 Cos (3.922t)

Then, when x(t) = 9.6cm

9.6 = 11.6 Cos(3.92t)

Cos(3.922t) = 9.6/11.6

Cos(3.922t) = 0.8276

3.922t = ArcCos(0.8276)

Note: the angle is in radiant

3.922t = 0.596

t = 0.596/3.922

t = 0.152 second

Then, v(t) at that time is

v(t) = x'(t) = -11.6×3.92Sin(3.922t)

v(t) = -45.5Sin(3.922t)

Now, when t =0.152

v(t) = -45.5 Sin(3.922×0.152)

v(t) = -45.5Sin(0.596)

v(t) = -25.5 cm/s

Then, it's magnitude is 25.5cm/s

d. Acceleration at same position

t = 0.152s

a(t) = v'(t) = - 45.5×3.922Cos(3.922t)

a(t) = -178.46Cos(3.92t)

a(t) = -178.46 Cos(3.92×0.152)

a(t) = -178.46 Cos(0.596)

a(t) = -147.68 cm/s²

Magnitude of the acceleration is 147.68 cm/s²

5 0

3 years ago