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3 years ago

Using hooke's law find the elastic constant of a spring that stretches 2 cm when 4newton force is applied to it

Physics

1 answer:

erica [24]3 years ago

5 0

<u>Answer:</u>

2N/cm

<u>Step-by-step explanation:</u>

According to the Hooke's Law, the force required to extend or compress a spring is directly proportional distance you can stretch it, which is represented as:

$F=kx$

where, $F$ is the force which is stretching or compressing the spring,

$k$ is the spring constant; and

$x$ is the distance the spring is stretched.

Substituting the given values to find the elastic constant $k$ to get:

$F=kx$

$4=k(2)$

$k=\frac{4}{2}$

$k=2$

Therefore, the elastic constant is 2 Newton/cm.

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A student produces a wave in a long spring by

Sliva [168]

Answer: 2

Explanation:

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2 years ago

There is a clever kitchen gadget for drying lettuce leaves after you wash them. It consists of a cylindrical container mounted s

lukranit [14]

Answer:

Option B

$a=19 m/s^{2}$

Explanation:

Given information

Radius of container, r=12cm=12/100=0.12m

Angular velocity= 2 rev/s, converted to rad/s we multiply by 2π

Angular velocity, $\omega=2*2\pi =12.56637061$

We know that speed, $v=r\omega$

Centripetal acceleration, $a=\frac {v^{2}}{r}$ and substituting $v=r\omega$ we obtain that

$a=r\omega^{2}$

Substituting \omega for 12.56637061 and r for 0.12

$a=0.12*(12.56637061)^{2}=18.94964045 m/s^{2}$

Rounded off, $a=19 m/s^{2}$

7 0

3 years ago

An object of mass 4kg is moving along a horizontal plane. If the coefficient of kinetic friction is 0.2 find the friction force

lana66690 [7]

Answer:

The friction force acting on the object is 7.84 N

Explanation:

Given;

mass of object, m = 4 kg

coefficient of kinetic friction, μk = 0.2

The friction force acting on the object is calculated as;

F = μkN

F = μkmg

where;

F is the frictional force

m is the mass of the object

g is the acceleration due to gravity

F = 0.2 x 4 x 9.8

F = 7.84 N

Therefore, the friction force acting on the object is 7.84 N

5 0

3 years ago

If the radius of an atom is 60 pm and the radius of the Earth is 6000 km, by how many orders of

kherson [118]

Answer:

1 x 10¹⁷

Explanation:

Given data:

Radius of the earth = 6000km

Radius of an atom = 60pm

Now, how many orders is the radius of the earth larger than an atom

Solution:

To solve this problem, let us express both quantity as the same unit;

1000m = 1km

6000km = 6000 x 10³m = 6 x 10⁶m

60pm;

1 x 10⁻¹²m = 1pm

60pm = 60 x 1 x 10⁻¹²m = 6 x 10⁻¹¹m

Now;

The order: $\frac{6 x 10^{6} }{6 x 10^{-11} }$ = 1 x 10¹⁷

6 0

2 years ago

The atlas stone’s is a strong man competition where athletes have to load 5 stones of masses 100kg, 120kg, 140kg, 160kg and 180k

lyudmila [28]

Answer: 6250 joules

Explanation:

The work needed to lift an object of mass M by a height H is equal to:

w = M*g*H

where h = 10m/s^2

then the total work that he did is equal to the sum of the work for every stone:

W = (100kg*g*H) + (120kg*g*H) + (140kg*g*H) + (160kg*g*H) + (180kg*g*H)

= (100kg + 120kg + 140kg + 160kg + 180kg)*g*H

= (500kg)*g*H

and now we can repalce g by 10m/s^2 and H by 125cm

But you can notice that we have two different units of distance, so knowing that 100cm = 1m

we can write H = 125cm = (125/100) m = 1.25 m

Then we have:

H = 500kg*10m/s^2*1.25m = 6250 J

3 0

2 years ago