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2 years ago

A 2 kg mass is attached to a vertical spring and the spring stretches 15 cm from its original relaxed position.

Physics

1 answer:

Naya [18.7K]2 years ago

5 0

Answer:

A. 130.7 N/m

Explanation:

By Hook's law

Force(F) = Spring constant(k) × Extention (d)

F = k× d ⇒ k = F/d

To find Force(F) you know it is the weight of the object. Considering g = 9.8 ms⁻²

k = 2×9.8/0.15 = 130.7 Nm⁻¹

Hooke’s law,

for small deformations of an object, the displacement is directly proportional to the deforming force or load.

Under these conditions the object returns to its original shape and size upon removal of the load.

The deforming force may be applied to a solid by stretching, compressing,etc. so a spring shows an elastic behavior according to Hooke’s law

Mathematically,

applied force( F) = constant (k)* displacement in length (x)

F = kx.

The value of k depends

on the kind of elastic material
on its dimensions and shape.

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Which of the following is associated with fission but not fusion?

kodGreya [7K]

Answer:

D. Uranium

Explanation:

I just got the answer right on my quiz.

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2 years ago

A 5kg bag falls a verticle height of 10m before hitting the ground.

g100num [7]

Answer:

$u = 7m {s}^{ - 1}$

Explanation:

We know that when we don't have air friction on a free fall the mechanical energy (I will symbololize it with ME) is equal everywhere. So we have:

$me(1) = me(2)$

where me(1) is mechanical energy while on h=10m

and me(2) is mechanical energy while on the ground

Ek(1) + DynamicE(1) = Ek(2) + DynamicE(2)

Ek(1) is equal to zero since an object that has reached its max height has a speed equal to zero.

DynamicE(2) is equal to zero since it's touching the ground

Using that info we have

$m \times g \times h = \frac{1}{2} \times m \times u {}^{2} \\$

we divide both sides of the equation with mass to make the math easier.

$9.8 \times 10 = \frac{1}{2} \times u {}^{2} \\ \frac{98}{2} = u {}^{2} \\ u { }^{2} = 49 \\ u = 7$

7 0

3 years ago

IF you are in Space and push a bowling what happens to you and the bowling ball?

Anastaziya [24]

Answer:

The bowling ball did not change size or shape- the only thing that changed was the amount of gravity that pulls on it. But the mass of the bowling ball would never change. A bowling ball with a mass of 12 pounds on earth will have the mass of 12 pounds on the moon! Mass is the amount of atoms that a space fills.

Explanation:

I hope this helps! :D

4 0

2 years ago

Concrete colums are constructed with reinforcing steel in them to make them stronger and more ductile. The reinforcing bars are

Sergio039 [100]

Answer:

$21678.47223\ lbf-in^2$

$383.1109\ lbf-in^2$

Explanation:

d = Diameter of column = 0.5 inch

$A_c$ = Area of concrete = $119.4\ in^2$

The strain in the system is conserved

$\dfrac{F_sL}{A_sE_s}=\dfrac{F_cL}{A_cE_c}\\\Rightarrow F_c=\dfrac{F_sA_cE_c}{A_sE_s}\\\Rightarrow F_c=\dfrac{F_s \times 119.4\times 4.1\times 10^6}{8\times \dfrac{\pi \dfrac{1}{2^2}}{4}\times 29\times 10^6}\\\Rightarrow F_c=10.74658F_s$

Now

$F_c+F_s=50000\\\Rightarrow 10.74658F_s+F_s=50000\\\Rightarrow F_s=\dfrac{50000}{11.74658}\\\Rightarrow F_s=4256.55807\ lbf$

$F_c=10.74658F_s\\\Rightarrow F_c=10.74658\times 4256.55807\\\Rightarrow F_c=45743.44182\ lbf$

Stress is given by

$\sigma_s=\dfrac{4256.55807}{\pi \dfrac{1}{2^2}}{4}\\\Rightarrow \sigma_s=21678.47223\ lbf-in^2$

The stress in the steel is $21678.47223\ lbf-in^2$

$\sigma_c=\dfrac{45743.44182}{119.4}\\\Rightarrow \sigma_s=383.1109\ lbf-in^2$

The stress in the steel is $383.1109\ lbf-in^2$

4 0

3 years ago

A simple pendulum 2 m long swings through a maximum angle of 30 ? with the vertical.

Alexus [3.1K]

The complete question was calculate the period T assuming the smallest amplitude.
Using the equation;
T = 2 π√(L/g)
Where T is the period in seconds, L is the length of the rod or wire in meters and g is the acceleration due to gravity.
Hence; T = 2×3.14 × √(2/9.81)
= 6.28 × 0.4515
= 2.836 seconds

6 0

3 years ago