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LekaFEV [45]
3 years ago
11

If it requires 7.0 JJ of work to stretch a particular spring by 1.7 cmcm from its equilibrium length, how much more work will be

required to stretch it an additional 3.6 cm ?
Physics
1 answer:
Lynna [10]3 years ago
3 0

Answer:

Explanation:

First of all, well calculate the spring constant k

K = 2Ei/x^2

Where Ei = initial work required

x = initial stretch length

k = 2×7/0.017^2 = 48443J/m^2

Now work done in stretching it to 5.3cm (1.7 + 3.6) or 0.053m

EF = kx^2/2

48443 × 0.053^2/2 = 68J

Work done in stretching additional 3.6cm is equal to

68J-7J = 61J

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Answer:

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Which composition of water moves to begin a deepwater current?
Paraphin [41]

<u>Answer:</u>

<h2>Cold, Salty water.</h2>

<u>Explanation:</u>

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6 0
3 years ago
At a certain temperature, the rate constant for this reaction is 5.64×10−4 s−1 . Calculate the half-life of cyclopropane at this
oee [108]

Answer:

t_{1/2}= 20.47\ minute

Explanation:

given,

constant rate for the reaction(k) =  5.64  ×  10⁻⁴ s⁻¹

to calculate the half life  = ?

t_{1/2} = \dfrac{0.693}{k}

t_{1/2} = \dfrac{0.693}{5.64\times 10^{-4}}

on solving the above equation

t_{1/2} = 1228.72 s

t_{1/2} = \dfrac{1228.72}{60}

t_{1/2}= 20.47\ minute

hence, the half-life of cyclopropane at this temperature =t_{1/2}= 20.47\ minute

5 0
3 years ago
A long, straight wire carries a current of 8.60
Kay [80]

First of all, we need to calculate the magnetic field generated by the wire at a distance r=3.90 cm=0.039 m, which is given by:

B=\frac{\mu_0 I}{2 \pi r}

where I=8.60 A is the current. Substituting numbers in the equation, we find

B=\frac{(4 \pi \cdot 10^{-7})(8.6 A)}{2 \pi (0.039 m)}=4.4 \cdot 10^{-5} T

And now we can calculate the force exerted on the electron, which is given by:

F=qvB

where q=1.6 \cdot 10^{-19} C is the charge of the electron and v=5.0 \cdot 10^4 m/s is its speed. Substituting data in the formula, we find

F=(1.6 \cdot 10^{-19}C)(5.00 \cdot 10^4 m/s)(4.4 \cdot 10^{-5} T)=3.5 \cdot 10^{-19} N

6 0
3 years ago
What is the resistance of al lamp that allows a current of 10 amps with 120 volts
BartSMP [9]
V=ir
I=10
v=120
r=?
r=v/i
r=120/10
r=12 ohm
5 0
3 years ago
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