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bazaltina [42]
4 years ago
10

What are the possible units for a spring constant

Physics
2 answers:
Serggg [28]4 years ago
6 0
[any unit of force] divided by [any unit of length or distance]

can be a unit of spring constant.

Useful spring units can include [newtons per meter] and [pounds per foot]. 
AleksAgata [21]4 years ago
5 0
Hooke's Law says that F=-kx where k is the spring constant measured in N/m (newtons per meter)
You might be interested in
Blood is more viscous than water. This is because it
JulijaS [17]

Answer:

Blood is thicker than water due primarily to the presence of red blood cells

Answer: <u>This is because it becomes more dense when cells are added</u>

Hope this helped :3

6 0
4 years ago
Conduction of a nerve impulse would be the fastest in _________
Vinil7 [7]

The options of the given are:

A. A large diameter myelinated fiber

B. A small diameter myelinated fiber

C. A large unmyelinated fiber

D. A small unmyelinated fiber

E. A small fiber with multiple Schwann cells

Answer: Option A, A large diameter myelinated fiber.

Explanation:

The conduction of the nerve impulse would be greatest in the myelinated fiber because the main function of the myelin sheath is to increase the speed of the impulse at which the electrical signals propagate.

In case of the unmyelinated sheath the nerve impulse travels slowly as the conduction waves but in case of the large diameter myelinated sheath the signals travel via saltatory conduction( hop)

In this type of propagation the signals are transferred from the node of Ranvier in one neuron to next node which increases the overall velocity of the action potentials.

8 0
3 years ago
Cart A of inertia m has attached to its front end a device that explodes when it hits anything, releasing a quantity of energy E
Leviafan [203]
We need to write down momentum and energy conservation laws, this will give us a system of equation that we can solve to get our final answer. On the right-hand side, I will write term after the collision and on the left-hand side, I will write terms before the collision.
Let's start with energy conservation law:
\frac{mv^2}{2}+\frac{2mv^2}{2}+0.75E=\frac{mv_{A}^2}{2}+\frac{2mv_{B}^2}{2}
\frac{3mv^2}{2}+0.75E=\frac{mv_{A}^2}{2}+mv_{B}^2
This equation tells us that kinetic energy of two carts before the collision and 3 quarters of explosion energy is beign transfered to kinetic energy of the cart after the collision.
Let's write down momentum conservation law:
mv+2mv=mv_A+2mv_B\\ 3mv=mv_A+2mv_B\\
Because both carts have the same mass we can cancel those out:
3v=v_A+2v_B
Now we have our system of equation that we have to solve:
\frac{3mv^2}{2}+0.75E=\frac{mv_{A}^2}{2}+mv_{B}^2\\ 3v=v_A+2v_B
Part A
We need to solve our system for v_a. We will solve second equation for v_b and then plug that in the first equation.
3v=v_A+2v_B\\ 3v-v_A=2v_B\\ v_B=\frac{3v-v_A}{2}
Now we have to plug this in the first equation:
\frac{3mv^2}{2}+0.75E=\frac{mv_{A}^2}{2}+mv_{B}^2\\v_B=\frac{3v-v_A}{2}\\
We will multiply the first equation with 2 and divide by m:
3v^2+\frac{3E}{2m}=v_{A}^2}+2v_{B}^2\\v_B=\frac{3v-v_A}{2}\\
Now we plug in the second equation into first one:
3v^2+\frac{3E}{2m}=v_{A}^2}+2v_{B}^2\\ 3v^2+\frac{3E}{2m}=v_{A}^2}+2\frac{(3v-v_A)^2}{4}\\ 3v^2+\frac{3E}{2m}=v_{A}^2}+\frac{9v^2-6v\cdot v_A+v_{A}^2}{2} /\cdot 2\\ 6v^2+\frac{3E}{m}=2v_{A}^2+9v^2-6v\cdot v_A+v_{A}^2}\\ 3v_A^2-6v\cdot v_a+3(v^2-\frac{E}{m})=0/\cdot\frac{1}{3}\\ v_A^2-3v\cdot v_A+ (v^2-\frac{E}{m})=0
We end up with quadratic equation that we have to solve, I won't solve it by hand. 
Coefficients are:
a=1\\&#10;b=-6v\\&#10;c=v^2-\frac{E}{m}
Solutions are:
v_A=\frac{3v+\sqrt{5v^2+\frac{4E}{m}}}{2},\:v_A=\frac{3v-\sqrt{5v^2+\frac{4E}{m}}}{2}
Part B
We do the same thing here, but we must express v_a from momentum equation:
3v=v_A+2v_B\\&#10;v_A=3v-2v_B
Now we plug this into our energy conservation equation:
3v^2+\frac{3E}{2m}=v_{A}^2}+2v_{B}^2\\v_A={3v-v_B}\\&#10;3v^2+\frac{3E}{2m}=(3v-v_B)^2+2v_B^2\\&#10;3v^2+\frac{3E}{2m}=9v^2-6v\cdot v_B+v_B^2+2v_B^2\\&#10;3v^2+\frac{3E}{2m}=3v_B^2-6v\cdot v_B+9v^2\\&#10;3v_B^2-6v\cdot v_B+9v^2-3v^2-\frac{3E}{2m}=0\\&#10;3v_B^2-6v\cdot v_B+(6v^2-\frac{3E}{2m})=0&#10;
Again we end up with quadratic equation. Coefficients are:
a=3\\&#10;b=-6v\\&#10;c=6v^2-\frac{3E}{2m}
Solutions are:
v_B=\frac{6v+\sqrt{-36v^2+\frac{18E}{m}}}{6},\:v_B=\frac{6v-\sqrt{-36v^2+\frac{18E}{m}}}{6}



8 0
3 years ago
An electric current heats a 221 g (0.221 kg) copper wire from 20.0 °C to 38.0 °C. How much heat was generated by the current?
lorasvet [3.4K]

Answer:

Heat generated by the current = 1547.89 J

Explanation:

We have equation for heat energy H = mCΔT

Mass of copper = 0.221 kg

Specific heat of copper = 0.093 kcal/kgC° = 389.112 J/kgC°

ΔT = 38 - 20 = 18°C

Substituting in H = mCΔT

           H = 0.221 x 389.112 x 18 = 1547.89 J

Heat generated by the current = 1547.89 J

     

3 0
4 years ago
A mass is bobbing up and down on a spring. If you increase the amplitude of the motion, how does this affect the time for one os
NikAS [45]

Answer:

It will not change

Explanation:

The period of oscillation of a mass-spring system is given by

T=2\pi \sqrt{\frac{m}{k}}

where

T is the period

m is the mass hanging on the spring

k is the spring constant

As we see from the formula, the period of oscillation does not depend on the amplitude of the motion: therefore, if we change the amplitude, the time for one oscillation will not change.

6 0
3 years ago
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