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Inessa [10]
4 years ago
15

A mass resting on a horizontal, frictionless surface is attached to one end of a spring; the other end is fixed to a wall. It ta

kes 3.6 J of work to compress the spring by 0.13 m. If the spring is compressed, and the mass is released from rest, it experiences a maximum acceleration of Find the value of (a) the spring constant and (b) the mass.
Physics
1 answer:
Marina CMI [18]4 years ago
7 0

Answer:

(a) The spring constant is 426N/m

(b) The mass is 5.65kg

Explanation:

(a) Energy stored in the spring (E) = 3.6J, extension (e) = 0.13m

E = 1/2Ke^2

3.6 = 1/2 × K × 0.13^2

K (spring constant) = 3.6×2/0.0169 = 426N/m

(b) F = Ke = 426 × 0.13 = 55.38N

Assuming the maximum acceleration (a) is 9.8m/s^2

Mass (m) = F/a = 55.38/9.8 = 5.65kg

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An Olympic-class sprinter starts a race with an acceleration of 5.10 m/s2. What is her speed 2.40 s later?
ivolga24 [154]

Answer:

12.24 m/s

Explanation:

Speed: This can be defined as the rate of change of distance with time. The S.I unit of speed is m/s.

Using the formula,

a = v/t................ Equation 1

Where a = acceleration of the sprinter, v = speed of the sprinter, t = time.

making v the subject of the equation,

v = at ................. Equation 2

Given: a = 5.1 m/s², t = 2.4 s.

Substitute into equation 2

v = 5.1(2.4)

v = 12.24 m/s.

Hence, the speed of the sprinter = 12.24 m/s

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The eye of a hurricane passes over grand bahama island in a direction 60.0° north of west with a speed of 41.0 km/h. three hour
Ksivusya [100]

initial velocity is given as 41 km/h at 60 degree North of West

v_i = -41 cos60\hat i + 41 sin60 \hat j

v_i = -20.5 \hat i + 35.5 \hat j

After some time the velocity is given as

v_f = 25 \hat j

now we can find the acceleration

a = \frac{v_f - v_i}{t}

a = \frac{25\hat j - (-20.5 \hat i + 35.5 \hat j)}{3}

a = 6.83 \hat i - 3.5 \hat j

now the distance is given by

d = v*t + \frac{1}{2} at^2

d = (-20.5 \hat i + 35.5 \hat j)*4.5 + 0.5*(6.83 \hat i - 1.5 \hat j)* 4.5^2

d = -92.25\hat i + 159.75\hat j + 69.15\hat i - 15.19\hat j

d = -23.1 \hat i + 144.56\hat j

so the magnitude of distance is

d = \sqrt{23.1^2 + 144.56^2} = 146.4 km

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3 years ago
A pendulum of mass 5.0 kg hangs in equilibrium. A frustrated student walks up to it and kicks the bob with a horizontal force of
Phoenix [80]

Answer:

The length is L = 6.206m and the angle is \theta = 37.752^o.

Explanation:

The period T of the pendulum is related to its length L by

T = 2\pi \sqrt{\dfrac{L}{g} },

where g =9.8m/s^2 is the acceleration due to gravity.

Solving for L we get

L = \dfrac{T^2g}{4\pi^2}

putting in T =5.0s and g =9.8m/s^2 we get:

L = \dfrac{(5.0s)^2*9.8m/s^2}{4\pi^2}

\boxed{L = 6.206m.}

There are two forces acting on the pendulum: The gravitational force mg and the F = 30N student's force. Therefore, the angular displacement \theta that these forces give is

sin(\theta ) = \dfrac{F}{mg}

\theta = sin^{-1}( \dfrac{F}{mg})

putting in F =30N, m =5.0kg, and g =9.8m/s^2 we get

\theta = sin^{-1}( \dfrac{30N}{5.0kg*9.8m/s^2})

\boxed{\theta = 37.752^o.}

4 0
4 years ago
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