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FrozenT [24]
3 years ago
8

A spring is used as part of a lift system and follows Hooke's law. If the spring is

Physics
1 answer:
salantis [7]3 years ago
6 0

Answer:

1.05m or 105cm

Explanation:

Using the hooke's law equation as follows;

F = –k.x

Where;

F = force (N)

x = extension length (m)

k = constant of proportionality (N/m)

According to the information given in this question;

Displacement (x) = 85cm = 85/100 = 0.85m

Force = 12500N

Using F = kx, we find the proportionality constant

k = F/x

K = 12500/0.85

K = 14705.8N/m.

Also, since K = 14705.8N/m, the displacement (x), when the force increases to 15500N is;

F = kx

x = F/k

x = 15500/14705.8

x = 1.05m or 105cm

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A wire lying along a y axis form y=0 to y=0.25 m carries a current of 2.0 mA in the negative direction of the axis. The wire ful
balu736 [363]

Answer:

FB = 0.187 N

Explanation:

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the angle between B and L is given by:

\vec{L}\cdot\vec{B}=LBcos\theta\\\\\theta=cos^{-1}(\frac{\vec{L}\cdot\vec{B}}{LB})=cos^{-1}(\frac{0*0.3y+0.25*0.4y}{0.25*0.5y})=36.86\°

Due to B depends on "y" you take into account the contribution of each element dy of the wire to the magnitude of the magnetic force. Thus, you have to integrate the following expression:

|\vec{F_B}|=Isin\theta\int_0^{0.25}B(y)dy=Isin\theta\int_0^{0.25}(0.5y)dy\\\\|\vec{F_B}|=(2.0*10^{-3}A)(sin36.86\°)(0.5T)[\frac{0.25^2}{2}m]=0.187\ N

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3 years ago
Through what potential difference would you need to accelerate an alpha particle, starting from rest, so that it will just reach
Svetlanka [38]

Answer:

\Delta V    = 1.8 \times 10^7 V

Explanation:

GIVEN

diameter = 15 fm  =15 \times 10^{-15}m

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just reach the {238}_U nucleus after being accelerated through a potential difference  ∆V

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