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dem82 [27]
3 years ago
15

Using Coulomb's law, rank the particle interactions from most attractive to most repulsive. Question 18 options: a particle with

a 1– charge separated from a particle with a 2+ charge by 150 pm. a particle with a 1– charge separated from a particle with a 3- charge by 150 pm. a particle with a 1– charge separated from a particle with a 2- charge by 150 pm. a particle with a 1– charge separated from a particle with a 3+ charge by 150 pm.

Physics
1 answer:
fgiga [73]3 years ago
4 0

Answer:

If arranged as A, B, C and D, the order of the particle interactions from most attractive to most repulsive will be;

D, A, C, B

Explanation:

Detailed explanation and calculation is shown in the image below

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At a playground, a child slides down a slide that makes a 42° angle with the horizontal direction. The coefficient of kinetic fr
kumpel [21]

Answer:

a = 5.1\ m/s^2

Explanation:

Given,

The angle of the slide=42^\circ

The mass of the child is= m

coefficient of friction = 0.20

when she slides down now apply Newton's law

ma =mg \sin\theta - f

ma = mg\sin \theta -\mu mg \cos \theta

therefore the acceleration

a=g \sin\theta -\mu gcos\theta

a=g[\sin \theta -\mu \cos \theta]

a=9.8\times [\sin 42^\circ -0.2\times \cos 42^\circ]

a = 5.1\ m/s^2

hence, the magnitude of acceleration during her sliding is equal to  a = 5.1\ m/s^2

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3 years ago
Why we cannt cook food on high attitude (like mountain where exist low temperatures )? ​
zloy xaker [14]
Cooking takes longer because water and other liquids evaporate faster and boil at lower temperatures.At sea level water boils at 212degrees Fahrenheit but at an altitude of 7,500 feet, it boils at about 198 degrees. Foods that are prepared by boiling or simmering will cook at a lower temperature and it will take longer to cook them. Hope this helps!! ; )
4 0
3 years ago
1. Plot the following graphs:
VLD [36.1K]

Answer:

(a) The distance-time graph for an object with uniform speed is giving by a straight line sloped graph with a constant positive or negative gradient as shown in the attached diagram

(b) The distance-time graph for an object with non-uniform speed is giving by a curved line sloped graph with varying gradient as shown in the attached diagram

(c) The velocity-time graph for a car with uniform motion is giving by a horizontal line graph at the speed of constant motion with a zero gradient as shown in the attached diagram

(d) The velocity-time graph for a car moving with uniform acceleration is giving by a straight line sloped graph with a constant positive or negative gradient as shown in the attached diagram

(e) The velocity-time graph for a car moving with non-uniform acceleration is giving by a curved line sloped graph with varying gradient as shown in the attached diagram

(f) According to Newton's first law of motion, an object at rest will remain at rest with no motion unless acted by a force, an therefore, will have no motion with time

Explanation:

8 0
3 years ago
In a certain process, the energy change of the system is 250 \rm kJ. The process involves 480 \rm kJ of work done by the system.
alisha [4.7K]

Answer:

\Delta Q=-230kJ

Explanation:

Using the first law of thermodynamics:

\Delta U=\Delta Q-W

Where \Delta U is the change in the internal energy of the system, in this case  \Delta U=250kJ, \Delta Q is the heat tranferred, and W is the work,  W=-480kJ with a negative sign since the work is done by the system.

From the previous equation we solve for heat, because it is the unknown variable in this problem

\Delta Q=\Delta U +W

And replacing the known values:

\Delta Q=250kJ +(-480kJ)

\Delta Q=250kJ -480kJ

\Delta Q=-230kJ

The negative sign shows us that the heat is tranferred from the system into the surroundings.

3 0
3 years ago
A prismatic bar AB of length L and solid circular cross section (diameter d) is loaded by a distributed torque of constant inten
Lyrx [107]

Answer:

a) the maximum shear stress τ_{max} the bar is 16T_{max} /πd³

b) the angle of twist between the ends of the bar is 16tL² / πGd⁴  

Explanation:

Given the data in the question, as illustrated in the image below;

d is the diameter of the prismatic bar of length AB

t is the intensity of distributed torque

(a) Determine the maximum shear stress tmax in the bar

Maximum Applied torque  T_max = tL

we know that;

shear stress τ = 16T/πd³

where d is the diameter

so

τ_{max} = 16T_{max} /πd³

Therefore, the maximum shear stress τ_{max} the bar is 16T_{max} /πd³

(b) Determine the angle of twist between the ends of the bar.

let theta (\theta) be the angle of twist

polar moment of inertia I_p} = πd⁴/32

now from the second image;

lets length dx which is at distance of "x" from "B"

Torque distance x

T(x) = tx

Elemental angle twist = d\theta = T(x)dx / GI_{p}

so

d\theta = tx.dx / G(πd⁴/32)

d\theta = 32tx.dx / πGd⁴

so total angle of twist \theta will be;

\theta =  \int\limits^L_0  \, d\theta

\theta =  \int\limits^L_0  \, 32tx.dx / πGd⁴

\theta = 32t / πGd⁴  \int\limits^L_0  \, xdx

\theta = 32t / πGd⁴ [ L²/2]

\theta = 16tL² / πGd⁴  

Therefore,  the angle of twist between the ends of the bar is 16tL² / πGd⁴  

7 0
3 years ago
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