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Naddika [18.5K]
3 years ago
5

In Hooke’s law, what does the x represent?

Physics
2 answers:
Arte-miy333 [17]3 years ago
7 0
X Represents the distance the spring is stretched or compressed away from its equilibrium or rest position.
masha68 [24]3 years ago
6 0

Answer:

x is the displacement of spring

Explanation:

The Hooke's law gives the force acting on the spring when it is compressed or stretched. The mathematical expression for force is given by :

F=-kx

Where

k is the spring constant of and it shows the stiffness of spring.

x shows the displacement of spring when it is stretched or compressed from equilibrium position and it acts in opposite direction.

You might be interested in
2. Can you place three forces of 5g, 6g, and 12g so they are in equilibrium. Justify your answer.
Bond [772]

Answer:

We cannot place three forces of 5g, 6g, and 12g in equilibrium.

Explanation:

Equilibrium means their sum must be zero.

Here the forces are 5g, 6g, and 12g.

For number of forces to be in equilibrium the magnitude of largest vector should be less than sum of the magnitude of other vectors.

Here

        Magnitude of largest force = 12 g

        Sum of magnitudes of other forces = 5g + 6g = 11g

       Magnitude of largest force >   Sum of magnitudes of other forces

So this forces cannot form equilibrium.

We cannot place three forces of 5g, 6g, and 12g in equilibrium.

4 0
2 years ago
The helicopter was deformed and destroyed in the __?___ collision.
MariettaO [177]

Answer:

The helicopter was deformed and destroyed in the inelastic collision.

Explanation:

  • When two object collide there exist two way of colliding: elastic collision and inelastic collision.
  • Two terms are considered during the collision: kinetic energy and momentum.
  • If both of these terms are conserved in any collision then there is no significant loss of property, this is called as elastic collision.
  • If only momentum is conserved but kinetic energy is converted into other forms then it is inelastic collision. In inelastic collision, the energy is lost in the form of vibration, sound etc. causing the damage to colliding object.
  • Hence the deformation of helicopter was due to inelastic collision.
5 0
3 years ago
A bomb is dropped from a bomber traveling at the speed of 120 km / h, destroying a military objective located at a distance of 2
schepotkina [342]

Answer:

18 km

Explanation:

Convert km/h to m/s:

120 km/h × (1000 m/km) × (1 h / 3600 s) = 33.3 m/s

The time it takes the bomb to travel the 2000 meters is:

2000 m / (33.3 m/s) = 60 s

So it takes 60 seconds for the bomb to fall.  The distance it fell is therefore:

Δy = v₀ᵧ t + ½ aᵧ t²

Δy = (0 m/s) (60 s) + ½ (10 m/s²) (60 s)²

Δy = 18,000 m

Δy = 18 km

7 0
2 years ago
An object is dropped from a bridge. A second object is thrown downward 1.48 s later. They both reach the water 48.1 m below at t
pashok25 [27]

To solve this problem we will apply the linear motion kinematic equations. With the data provided we will calculate the time of the first object to fall. Later we will get the time difference between the two. This difference will allow us to find the free fall distance. Through the distance we will find the initial velocity, that is,

x = v_0 t +\frac{1}{2}at^2

48.1 = 0*t + \frac{1}{2} (9.8)t^2

t = 3.13s

The second object is thrown downward at one second later and it meets the first object at the water is

t' = 3.13 -1.48

t' = 1.65s

The distance of the object will travel due to free fall acceleration is

x = v_0 t+\frac{1}{2} at^2

x = 0*(1.65) +\frac{1}{2}(9.8)(1.65)^2

x = 13.34m

The distance of the object will travel due to its initial velocity is

v_0 = \frac{d_0}{t}

d_0 = v_0 t

48.1-13.34 = v_0 (1.65)

v_0 = 21.06m/s

Therefore the initial speed of the second object is 21.06m/s

8 0
3 years ago
1) Subatomic particles called muons can be created in the upper atmosphere by collisions of cosmic rays (energetic particles com
Vsevolod [243]

Answer and explanation:

A.

Muon travelled straight down towards the earth. Therefore the tree moves up in the rest frame of muon (option a)

B.

In muon rest frame it travels Zero meters

C.

Distance, d = Velocity, v * Time, s

where, v = 0.9c = 0.9 \times 8 \times 10^8 , s = 2.2 \mu s

d = 0.9 \times 3 \times 10^8 \times 2.2 \times 10^{-6}\\\\d = 594m

D.

Distance from the top of the mountain to the tree is the same as the distance travelled by the tree in the muons rest frame

that is same as in part C which is 594m

E.

Using lorentz contraction

In the rest frame of someone standing on the mountain

the distance is given by

d' = \frac{d}{\gamma} = d\sqrt{1 - \frac{v^2}{c^2}}, where, \frac{1}{\gamma}= \sqrt{1 - \frac{v^2}{c^2}}

d' = 594\sqrt{1 - \frac{(0.9c)^2}{c^2}}

d' = 594\sqrt{1 - 0.81}

d' = 594 \times 0.4359

d' = 258.92m

F.

in the rest frame of someone standing on the mountain,

muon moves straight down

3 0
3 years ago
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