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Angelina_Jolie [31]
2 years ago
14

A spring is 14cm long. Three masses are hung from it and then it is measured again. Now it is 19.5cm long. What force did the th

ree masses provide? The spring constant for the spring is 30N/m.
Physics
1 answer:
lukranit [14]2 years ago
3 0

Answer:

Gravity

Explanation:

The answer is gravity because when the 3 masses were hung from the spring, gravity pulled the spring towards the ground.

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A large ebony wood log, totally submerged, is rapidly floating down a flooded river. If the mass of the log is 165 kg, what is t
Sedbober [7]

Answer:

F = 1618.65[N]

Explanation:

To solve this problem we use the following equation that relates the mass, density and volume of the body to the floating force.

We know that the density of wood is equal to 750 [kg/m^3]

density = m / V

where:

m = mass = 165[kg]

V = volume [m^3]

V = m / density

V = 165 / 750

V = 0.22 [m^3]

The floating force is equal to:

F = density * g * V

F = 750*9.81*0.22

F = 1618.65[N]

3 0
2 years ago
A bike first accelerates from 0.0 m/s to 5.0 m/s in 4.5s, then continues at this constant speed for another 4.5 s. What is the t
Veseljchak [2.6K]
Assuming constant acceleration, the distance travelled in the first 4.5s is:

0.5*5.0*4.5 = 11.25m

The distance travelled in the next 4.5s is:

5.0*4.5 = 22.5m

The total distance travelled is:

11.25 + 22.5 = 33.75m
3 0
2 years ago
Read 2 more answers
A solenoid with 3,000.0 turns is 70.0 cm long. If its self-inductance is 25.0 mH, what is its radius? (The value of μ0 is 4π x 1
nevsk [136]

Answer:

A. 2.2*10^-2m

Explanation:

Using

Area = length x L/ uo xN²

So A = 0.7m * 25 x 10^-3H /( 4π x10^-7*

3000²)

A = 17.5*10^-3/ 1.13*10^-5

= 15.5*10^-2m²

Area= π r ²

15.5E-2/3.142 = r²

2.2*10^2m

Explanation:

5 0
2 years ago
Under the Big Top elephant, Ella (2500 kg), is attracted to Phant, the 3,000 kg
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Under the Big Top elephant, Ella (2500 kg), is attracted to Phant, the 3,000 kg elephant. They are separated by 8

4 0
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A 2100 kg car starts from rest and accelerates at a rate of 2.6 m/s2 for 4.0 s. Assume that the force acting to accelerate the c
Reika [66]
When the system is experiencing a uniformly accelerated motion, there are a set of equations to work from. In this case, work is energy which consist solely of kinetic energy. That is, 1/2*m*v2. First, let's find the final velocity.

a = (vf - v0)/t
2.6 = (vf - 0)/4
vf = 10.4 m/s

Then W = 1/2*(2100 kg)*(10.4 m/s)2
W = 113568 J = 113.57 kJ
8 0
2 years ago
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