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Marysya12 [62]
3 years ago
11

In the parallel spring system, the springs are positioned so that the 37 N weight stretches each spring equally. The spring cons

tant for the left-hand spring is 2.7 N/cm and the spring constant for the righ-hand spring is 4.3 N/cm. How far down will the 37 N weight stretch the springs?
Physics
1 answer:
Reptile [31]3 years ago
3 0

Answer:

x = 5.29 m

Explanation:

given,

weight of stretch = 37 N

left-hand spring constant (k₁)= 2.7 N/cm

right hand spring constant(k₂)= 4.3 N/ cm

spring are connected in parallel

F = F₁ + F₂

F = k₁x + k₂x

F = (k₁+ k₂)x

37= (4.3+ 2.7)x

7 x = 37

x = 5.29 m

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valentina_108 [34]

Answer:

v=1.6\times10^{5}m/s

Explanation:

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Newton's 2nd Law states F=ma, which for our case will mean:

a=\frac{F}{m}=\frac{qE}{m}

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6 0
3 years ago
Work out the current through a television with a power of 0.8 kW if it uses the 230 V mains supply. Give your answer to 1 decima
Y_Kistochka [10]

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Vector A⃗ has magnitude 8.00 m and is in the xy-plane at an angle of 127∘ counterclockwise from the +x–axis (37∘ past the +y-axi
WINSTONCH [101]

Answer:

293.7 degrees

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5 0
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