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Aneli [31]
2 years ago
8

In Hooke's law, Fspring=kΔx , what does the Fspring stand for?

Physics
2 answers:
Mrrafil [7]2 years ago
6 0

Answer:

The amount of force acting on the spring.

Explanation:

Here, Fspring= kΔx  is a equation denoting the amount of force acting on the spring.

where, k = spring constant.

           Δx= change in the length of the spring.

so, for every Δx change in spring length, kΔx force acts on the spring.

Agata [3.3K]2 years ago
6 0

Answer:

The amount of force acting on the spring.

Explanation:

Here,

Fspring= kΔx  is a equation denoting the amount of force acting on the  spring.

where, k = spring constant.

          Δx= change in the length of the spring.

so, for every Δx change in spring length, kΔx force acts on the spring.

Click to let others know, how helpful is it

- CP

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A rock with density 1900 kg/m3 is suspended from the lower end of a light string. When the rock is in air, the tension in the st
wel

Answer:

the tension T2 when the rock is completely immersed is T2 =  29.05 N

Explanation:

from Newton's second law

F= m*a

where F= force , m= mass , a= acceleration

when the rock is suspended ,a=0 since it is at rest. Then

T1 - m*g = 0 , T1= tension when suspended in air , g= gravity

assuming constant density of the rock

m= ρ rock *V , where  ρ rock = density of the rock , V= volume

thus

T1= m*g = ρ rock *g*V

V=  T1/(ρ rock *g)

when the rock is submerged in oil , it receives an upward force that equals the weight of the volume of displaced oil (V displaced). Since it is completely submerged the volume displaced is the volume of the rock V=Vdisplaced  

When the rock is at rest , then

F= m*a=0

T2 + ρ oil *g*V displaced - ρ rock *g*V  =0

T2 = ρ rock *g*V - ρ oil *g*V = g*V (ρ rock - ρ oil)

T2 = g*V (ρ rock - ρ oil) = T1/(ρ rock *g) *g * (ρ rock - ρ oil)

T2 = T1 * (ρ rock - ρ oil)/ρ rock

replacing values

T2 = 48 N * (1900 kg/m3- 750 kg/m3)/ 1900 kg/m3 = 29.05 N

T2 =  29.05 N

3 0
2 years ago
What is the mass of an object that creates 33,750 joules of energy by traveling at 30 m/sec?
nikklg [1K]
The Energy is Kinetic Energy.

Kinetic Energy = 1/2*mv²,  Where m is mass in kg, v is velocity in m/s

Energy is 33750 Juoles,  v = 30m/s

1/2*mv² = E

1/2*m*30² = 33750

m = (2*33750) / (30²)     Using a calculator

m = 75 kg

Mass of object is 75 kg.
5 0
2 years ago
Read 2 more answers
a lady bug walks 10 cm forward then 5 cm backwards in 20 seconds. what is the average speed of the ladybug ?
igor_vitrenko [27]

A lady bug moves 10 cm forward and 5 cm backwards

so total distance moved by lady bug = 10 + 5 = 15 cm

total time taken by the lady bug

t =  20 s

so the average speed is given as

v = \frac{d}{t}

v = \frac{15}{20}

v = 0.75 cm/s

so its average speed is 0.75 cm/s

5 0
2 years ago
Trained dolphins are capable of a vertical leap of 7.0 m straight up from the surface of the water - an impressive feat. Suppose
dmitriy555 [2]

Answer:14 m

Explanation:

Given

Vertical jump make by the dolphin is given by h=7\ m

Suppose the dolphin jump with an initial velocity of u

so u is given by u^2=2\cdot g\cdot h

If dolphin launches at an angle \theta then maximum horizontal range is given by

assuming the of Dolphin to be Projectile so range is given by

R=\frac{u^2\sin 2\theta }{g}

substitute the value of u^2

R=\frac{2\times 9.8\times 7\sin 2\theta }{9.8}

R=2h\sin 2\theta

Range will be maximum for \theta =45^{\circ}

thus R_{max}=2\times 7\times 1=14\ m

                                     

3 0
2 years ago
A 4.00-kg mass is attached to a very light ideal spring hanging vertically and hangs at rest in the equilibrium position. The sp
Ahat [919]

Answer:

|v| = 8.7 cm/s

Explanation:

given:

mass m = 4 kg

spring constant k = 1 N/cm = 100 N/m

at time t = 0:

amplitude A = 0.02m

unknown: velocity v at position y = 0.01 m

y = A cos(\omega t + \phi)\\v = -\omega A sin(\omega t + \phi)\\ \omega = \sqrt{\frac{k}{m}}

1. Finding Ф from the initial conditions:

-0.02 = 0.02cos(0 + \phi) => \phi = \pi

2. Finding time t at position y = 1 cm:

0.01 =0.02cos(\omega t + \pi)\\ \frac{1}{2}=cos(\omega t + \pi)\\t=(acos(\frac{1}{2})-\pi)\frac{1}{\omega}

3. Find velocity v at time t from equation 2:

v =-0.02\sqrt{\frac{k}{m}}sin(acos(\frac{1}{2}))

5 0
2 years ago
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