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Rina8888 [55]
3 years ago
5

Patrick is repairing the roof on his house. He uses a hammer having a mass of 1 kilogram. While working at the apex of the roof,

the hammer slips and falls to the ground at a velocity of 12 meters/second. Find the potential energy of the hammer just before slipping.
Physics
1 answer:
pav-90 [236]3 years ago
5 0

Answer:

72 joules

Explanation:

The potential energy of that hammer is a function of its displacement against gravity. Considering that it fell with a velocity of 12 m/s, it was its displacement against gravity that gave it this velocity. It will continue to move until its displacement to gravity is zero.

since the body is in motion; it has converted its potential energy (mgh, m is mass, g is acceleration due to gravity, and h is the height) to kinetic energy (energy due to motion, 1/2mv^2; m = mass, v = velocity or speed)

therefore the potential energy is equal to kinetic energy

mgh = 1/2mv^2 = 1/2 *1kg* 12*12 = 72 joules.

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AleksandrR [38]

The question is about unclear since no picture provided. But from the question, it could be guessed that the box is moving back and forth on the frictionless plane at the amplitude of A in simple harmonic motion.

Answer:

D. At x=0, it's acceleration is at a maximum

Explanation:

As the box move forward, it reaches point A and than move backward. Theoretically, the box will move backwards, through its origin, to point -A and then going forward.

Point A is the maximum displacement of the box in this case. At this point, the box instantaneously stop to go backward. Therefore the velocity at that moment is zero.

From point -A, the box travel forward and keep building up speed due to the release in potential energy of the spring. And at point x=0, the velocity become maximum. After point x=0, the velocity of the box slows down due to the conversion of kinetic energy to potential energy of the spring. And as it reaches point A, it reaches zero velocity.

The same can be said as the box travels backward from point A to -A

8 0
3 years ago
A family took a trip in a car traveling East from Greensboro to Wilmington, NC. Use the Graph to answer the questions below.
kotykmax [81]

Answer:

1). Average speed = 1.5 m per second

2). Average velocity = 1.5 m per second

Explanation:

1). Since, speed is a scalar quantity

   Therefore, average speed of the trip = \frac{\text{Total distance covered}}{\text{Total time taken}}

    From the graph attached,

   Total distance covered = 10 + 10 + 20 + 0 + 20 + 30

                                           = 90 meters

   Total time taken = 60 seconds

    Average speed = \frac{90}{60}

                               = 1.5 meter per second

2). Velocity is a vector quantity.

    Therefore, average velocity = \frac{\triangle d}{\triangle t}

                                                   = \frac{d_{60}-d_0}{60-0}

                                                   = \frac{90-0}{60-0}

                                                   = 1.5 meter per second                        

7 0
3 years ago
A freshly prepared sample of radioactive isotope has an activity of 10 mCi. After 4 hours, its activity is 8 mCi. Find: (a) the
Maurinko [17]

Answer:

(a). The decay constant is 1.55\times10^{-5}\ s^{-1}

The half life is 11.3 hr.

(b). The value of N₀ is 2.38\times10^{11}\ nuclei

(c). The sample's activity is 1.87 mCi.

Explanation:

Given that,

Activity R_{0}=10\ mCi

Time t_{1}=4\ hours

Activity R= 8 mCi

(a). We need to calculate the decay constant

Using formula of activity

R=R_{0}e^{-\lambda t}

\lambda=\dfrac{1}{t}ln(\dfrac{R_{0}}{R})

Put the value into the formula

\lambda=\dfrac{1}{4\times3600}ln(\dfrac{10}{8})

\lambda=0.0000154\ s^{-1}

\lambda=1.55\times10^{-5}\ s^{-1}

We need to calculate the half life

Using formula of half life

T_{\dfrac{1}{2}}=\dfrac{ln(2)}{\lambda}

Put the value into the formula

T_{\dfrac{1}{2}}=\dfrac{ln(2)}{1.55\times10^{-5}}

T_{\dfrac{1}{2}}=44.719\times10^{3}\ s

T_{\dfrac{1}{2}}=11.3\ hr

(b). We need to calculate the value of N₀

Using formula of N_{0}

N_{0}=\dfrac{3.70\times10^{6}}{\lambda}

Put the value into the formula

N_{0}=\dfrac{3.70\times10^{6}}{1.55\times10^{-5}}

N_{0}=2.38\times10^{11}\ nuclei

(c). We need to calculate the sample's activity

Using formula of activity

R=R_{0}e^{-\lambda\times t}

Put the value intyo the formula

R=10e^{-(1.55\times10^{-5}\times30\times3600)}

R=1.87\ mCi

Hence, (a). The decay constant is 1.55\times10^{-5}\ s^{-1}

The half life is 11.3 hr.

(b). The value of N₀ is 2.38\times10^{11}\ nuclei

(c). The sample's activity is 1.87 mCi.

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8 0
3 years ago
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3 years ago
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