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shutvik [7]
3 years ago
5

A 0.43 kg hammer is moving horizontally at 5.2 m/s when it strikes a nail and comes to rest after driving the nail 0.014 m into

a board. What is the duration (in seconds) of the impact?
Physics
2 answers:
vagabundo [1.1K]3 years ago
5 0

Answer:

The duration of the impact is 0.005384 seconds

Explanation:

Given

m = 0.43 kg

v = 5.2 m/s

x = 0.014 m

Knowing the formulas

v^{2}_{f} = v^{2}_{i} + 2ax\\0 = 5.2^{2}  + 2a*0.014\\a = - 965.71 m/s^{2} \\\\vf = vi + at\\0 = 5.2 + (965.71)t\\t = 0.005384 s

Katena32 [7]3 years ago
4 0

Answer:

5.385×10⁻³

Explanation:

First we find the acceleration

Using the equation of motion,

v² = u²+2as........................ Equation 1

Where v = Final velocity, u = initial velocity, a = acceleration, s = distance.

make a the subject of the equation,

a = (v²-u²)/2s................. Equation 2

Given: v = 0 m/s (comes to rest), u = 5.2 m/s, s = 0.014 m

Substitute into equation 2

a = (0²-5.2²)/(2×0.014)

a = -27.04/0.028

a = -965.71 m/s²

Finally Using

a = (v-u)/t

where t = Duration of impact

make t the subject of the equation

t = (v-u)/a.................... Equation 3

Given: v = 0 m/s, u = 5.2 m/s, a = -965.71 m/s²

Substitute into equation 3

t = (0-5.2)/-965.71

t = -5.2/-965.71

t = 5.385×10⁻³ s.

Hence the duration of impact =  5.385×10⁻³

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Answer:

Explanation:

a). Find the graph attached for the motion.

b). If a shopper walk 5.4 m westwards then 7.8 m eastwards,

   Distance traveled by the shopper = Distance traveled in eastwards + Distance traveled westwards

                                                            = 5.4 + 7.8

                                                            = 13.2 m

c). Displacement of the shopper = Distance walked westwards - Distance traveled eastwards

                                                      = 5.4 - 7.8

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Therefore, magnitude of the displacement of the shopper is = 2.4 m

And the direction of the displacement is eastwards.

8 0
2 years ago
A 0.29 kg particle moves in an xy plane according to x(t) = - 19 + 1 t - 3 t3 and y(t) = 20 + 7 t - 9 t2, with x and y in meters
Artist 52 [7]

Answer:

Part a)

F = 7.76 N

Part b)

\theta = -137.7 degree

Part c)

\theta = -127.7 degree

Explanation:

As we know that acceleration is rate of change in velocity of the object

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x = -19 + t - 3t^3

y = 20 + 7t - 9t^2

Part a)

differentiate x and y two times with respect to time to find the acceleration

a_x = \frac{d^2}{dt^2}(-19 + t - 3t^3)

a_x = \frac{d}{dt}(0 +1 - 9t^2)

a_x = -18t

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Now the acceleration of the object is given as

\vec a = (-18t)\hat i + (-18)\hat j

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\vec a = -19.8 \hat i - 18 \hat j

now the net force of the object is given as

\vec F = m\vec a

\vec F = (0.29 kg)(-19.8 \hat i - 18 \hat j)

\vec F = -5.74 \hat i - 5.22 \hat j

now magnitude of the force will be

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Part b)

Direction of the force is given as

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tan\theta = \frac{-5.22}{-5.74}

\theta = -137.7 degree

Part c)

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v_x = (0 +1 - 9t^2)

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