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madreJ [45]
3 years ago
9

An athlete stretches a spring an extra 40.0 cm beyond its initial length. How much energy has he transferred to the spring, if t

he spring constant is 52.9 N/cm?
a) 4230 kJ
b) 4230 J
c) 423 kJ
d) 423 J
Physics
1 answer:
notsponge [240]3 years ago
4 0

Answer:

<h2>b) 4230 J </h2>

Explanation:

Step one:

given data

extension= 40cm

Spring constant K= 52.9N/cm

Step two:

Required

the Kinetic Energy KE

the expression to find the kinetic energy is

KE= 1/2ke^2

substituting our data we have

KE= 1/2*52.9*40^2

KE=0.5*52.9*1600

KE= 42320Joules

<u>The answer is b) 4230 J </u>

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<span>Back in the day, one measured a printer's speed in CPM, which stands for characters per minute. Most of the modern printers that exist today, including the inkjet printer measure their speed in PPM, which is also known as pages per minute.</span>
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Given the height of rod,length of shadow of tree and length of shadow of the rod, estimate the height of the tree? Given:height
diamong [38]

Answer:

1000 cm.

Explanation:

To obtain the estimated tree height :

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Substituting values into the formula :

(150cm / 120 cm) = (height of tree / 800 cm)

Using cross multiplication :

Height of tree * 120 = 150 * 800

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2 years ago
If a 160W light bulb consumes 1000J of electrical energy in a given time, then in the same time interval, how much energy will a
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7 0
2 years ago
Read 2 more answers
I’d really appreciate it if someone could help
Vanyuwa [196]

Answer:

0.25

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6 0
3 years ago
Suppose you stand with one foot on ceramic flooring and one on a wool carpet, making contact over a 77.2 cm2 area with each foot
Finger [1]

Answer:

P_{c= 5.74W

P_{w}=0.27 W

Explanation:

d= 2.6cm =>0.026m (for both the ceramic and the carpet )

Thermal conductivity of wool 'k_{w}'= k_{carpet} = k_{wool}=  0.04J/(sm °C)

Thermal conducticity of carpet 'k_c}' = 0.84 J/(sm °C)

Area 'A'= A_{carpet}= A_{ceramic}= 77.2cm²=> 77.2 x 10^{-4}m²

T_h}=33.0°C

T_c}=10.0°C

Average Power P_{avg} is determined by dividing amount of energy'Q' by time taken for the transfer't':

P_{avg} = Q/Δt

Due to conductivity, heat of flow rate will be P= dQ/dt

P=\frac{dQ}{dt} = \frac{kA[T_{h}-T_{c} ]}{d}

For CERAMIC:

P_{c=\frac{k_{c} A[T_{h}-T_{c} ]}{d} => [0.84 x 77.2 x 10^{-4}(33-10) ]/0.026

P_{c= 5.74W

For WOOL CARPET:

P_{w}= \frac{k_{w} A[T_{h}-T_{c} ]}{d}=> [0.04 x 77.2 x 10^{-4}(33-10) ]/0.026

P_{w}=0.27 W

8 0
2 years ago
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