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3 years ago

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A construction worker drops a 2.20 kg hammer from a roof. At a distance of 4.50 m above the ground, the hammer's total mechanica

l energy is 188 J. Calculate the kinetic energy of the hammer at this height.

1 answer:

Rina8888 [55]3 years ago

3 0

Answer:

Kinetic energy = 90.98 J

Explanation:

Given that,

The mass of a hammer, m = 2.2 kg

At a distance of 4.50 m above the ground, the hammer's total mechanical energy is 188 J.

We need to find the kinetic energy of the hammer at this height.

We know that,

Mechanical energy = kinetic energy + potential energy

$188=K_E+mgh\\\\K_E=188-mgh\\\\K_E=188-2.2\times 9.8\times 4.5\\\\K_E=90.98\ J$

So, the kinetic energy of the hammer at this height is equal to 90.98 J.

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The three stages of a train route took 1 hour ,2 hours ,and 4 hours . The first two stages were 80km and 200km of the train aver

luda_lava [24]

Answer:

the third stage was 480 km long

Explanation:

Stage 1:

Time = 1 hours

Speed = 80km

Stage 2:

Time = 2 hours

Speed = 200km

Stage 3:

Time = 4 hours

Let the Distance at the stage 3 be x

Average speed of the train route = 100 km/h

So

$\frac{ \text{speed at stage 1} + \text{speed at stage 2} + \text{speed at stage 3}}{3} = 0$

$\frac{ \text{speed at stage 1} + \text{speed at stage 2} + \text{speed at stage 3}}{3} = 100$

Lets find the speed at stage 1

Speed = $\frac{Distance }{Time}$

Speed = $\frac{80}{1}$

Speed 1= 80 km/hr

The speed at stage 2

Speed = $\frac{Distance }{Time}$

Speed = $\frac{200}{2}$

Speed 2 = 100 km/hr

The speed at stage 3

Speed = $\frac{Distance }{Time}$

Speed = $\frac{x}{4}$

Speed 3 = $\frac{x}{4}$

we kow that average is ,

$\frac{ \text{speed 1} + \text{speed 2} + \text{speed 3}}{3} = 100$

$\frac{ 80 + 100+ \frac{x}{4} }{3} = 100$

$\frac{ 180 + \frac{x}{4} }{3} = 100$

$\frac{ \frac{720 +x}{4} }{3} = 100$

$\frac{720 +x}{4} \times \frac{1}{3} = 100$

$\frac{720 +x}{12} = 100$

$720 +x = 100 \times 12$

$720 +x = 1200$

$x = 1200- 720$

x = 480

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Answer:

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Two cars A and B are moving with velocities 20 m/s and 15 m/s in the direction east and west respectively. If

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Answer:

<u>Distance</u><u> </u><u>between</u><u> </u><u>them</u><u> </u><u>is</u><u> </u><u>4</u><u>,</u><u>2</u><u>0</u><u>0</u><u> </u><u>meters</u><u>.</u>

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substitute:

$distance = 20 \times (2 \times 60) \\ = 2400 \: m$

Consider car B:

$distance = 15 \times (2 \times 60) \\ = 1800 \: m$

since these cars move in opposite directions, distance between them is their summation:

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In the 1980s, the term picowave was used to describe food irradiation in order to overcome public resistance by playing on the w

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Answer:

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Explanation:

The photon travels at the speed of light, 3.0 × $10^{8}$ m/s, and given that its frequency = 1 picometer = 1.0 × $10^{-12}$ m.

Its energy can be determined by;

E = hf

= (hc) ÷ λ

where E is the energy, h is the Planck's constant, 6.626 × $10^{-34}$ Js, c is the speed of the light and f is its frequency.

Therefore,

E = (6.626 × $10^{-34}$ × 3.0 × $10^{8}$ ) ÷ 1.0 × $10^{-12}$

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E = 1.9878 × $10^{-13}$ J

But, 1 eV = 1.6 × $10^{-19}$ J. So that;

E = $\frac{1.9878*10^{-13} }{1.6*10^{-19} }$

= 1242375 eV

∴ E = 1.24MeV

The energy of the photon is 1.24MeV.

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