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Vinvika [58]
3 years ago
12

. When using straight ladders, the base of the ladder should be _____ away from the wall or other vertical surface for every ___

__ of height to the point of support.
Physics
2 answers:
nalin [4]3 years ago
8 0
<h3><u>Answer;</u></h3>

one foot; four feet

<h3><u>Explanation</u>;</h3>
  • When using straight ladders, the base of the ladder should be <u>one foot</u> away from the wall or other vertical surface for every <u>four feet</u> of height to the point of support.
  • The Four-to-one rule ladder safety requires that for every four feet of height climbed, move the base one foot away from the wall.
  • <em><u>It is always important to remember the 4-to-1 rule for ladder safety so as to avoid home improvement injuries.</u></em>

guapka [62]3 years ago
3 0

one foot; four feet is the answer

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Read 2 more answers
A force of 960 newtons stretches a spring 4 meters. A mass of 60 kilograms is attached to the end of the spring and is initially
Drupady [299]

Answer:

x(t) = - 6 cos 2t

Explanation:

Force of spring = - kx

k= spring constant

x= distance traveled by compressing

But force = mass × acceleration

==> Force = m × d²x/dt²

===> md²x/dt² = -kx

==> md²x/dt² + kx=0   ------------------------(1)

Now Again, by Hook's law

Force = -kx

==> 960=-k × 400

==> -k =960 /4 =240 N/m

ignoring -ve sign k= 240 N/m

Put given data in eq (1)

We get

60d²x/dt² + 240x=0

==> d²x/dt² + 4x=0

General solution for this differential eq is;

x(t) = A cos 2t + B sin 2t   ------------------------(2)

Now initially

position of mass spring

at time = 0 sec

x (0) = 0 m

initial velocity v= = dx/dt=  6m/s

from (2) we have;

dx/dt= -2Asin 2t +2B cost 2t = v(t) --- (3)

put t =0 and dx/dt = v(0) = -6 we get;

-2A sin 2(0)+2Bcos(0) =-6

==> 2B = -6

B= -3

Putting B = 3 in eq (2) and ignoring first term (because it is not possible to find value of A with given initial conditions) - we get

x(t) = - 6 cos 2t

==>  

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