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Sidana [21]
3 years ago
8

 what is the difference between repelling and attracting

Physics
2 answers:
VikaD [51]3 years ago
7 0

Answer: Repelling is when something will not connect with another object. The force will cause a repel between the two objects. Attracting is when something is attracted or being pulled to another object.

Explanation: Hope this helps!

enyata [817]3 years ago
3 0

Answer:

Attracting means pulling toward you and repelling means pushing away

Explanation:

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Find the kinetic energy of a 0.1 kg toy truck moving at a speed of 1.1 m/s
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0.0605J is your answer. Use the formula KE=1/2mv^2
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4 years ago
How can groundwater flow, elevation can transport contaminates
Pavel [41]

Answer: it's under explanation

Explanation:Groundwater contamination occurs when man-made products such as gasoline, oil, road salts and chemicals get into the groundwater and cause it to become unsafe and unfit for human use. Materials from the land's surface can move through the soil and end up in the groundwater.

3 0
3 years ago
What do these letters stand for<br> P=mv
victus00 [196]

Answer:

The equation for momentum of a piece of matter.

In either case, the momentum would be less than a linebacker hitting you at full speed. The equation for momentum is written: p = mv where p stands for momentum. That is, mass times velocity equals momentum.

Explanation:

Hope This Helps

Have A Great Day

4 0
2 years ago
Ice skaters often end their performances with spin turns, where they spin very fast about their center of mass with their arms f
vampirchik [111]

Answer:

\large \boxed{\text{30 rev/s}}

Explanation:

This question is based on the Law of Conservation of Angular Momentum.

Angular momentum (L) equals the moment of inertia (I) times the angular speed (ω).

L = Iω

If momentum is conserved,

I₁ω₁ = I₂ω₂

Data:

 I₁ = 3.5    kg·m²s⁻¹

ω₁ = 6.0    rev·s⁻¹

 I₂ = 0.70 kg·m²s⁻¹

Calculation:

\begin{array}{rcl}I_{1}\omega_{1} &= &I_{2}\omega_{2}\\\text{3.5 kg$\cdot$m$^{2}$}\times \text{6.0 rev/s} &= &\text{0.70 kg$\cdot$m$^{2}$}\times\omega_{2}\\\text{21 rev/s} &= &0.70\omega_{2}\\\omega_{2} & = & \dfrac{\text{21 rev/s}}{0.70}\\\\&=&\textbf{30 rev/s}\\\end{array}\\\text{The skater's final rotational speed is $\large \boxed{\textbf{30 rev/s}}$}

8 0
3 years ago
The engine in an imaginary sports car can provide constant power to the wheels over a range of speeds from 0 to 70 miles per hou
Inessa05 [86]

Answer:

a) 4.40 s

b) 2.20 s

Explanation:

Given parameters are:

At constant power  ,

initial speed of the car, v_0=0

final speed of the car, v=32 mph

At full power,

initial speed of the car, v_0=0

final speed of the car, v=64 mph

a)

At constant power, KE = \frac{1}{2} mv^2

At full power, KE = \frac{1}{2} m(2v)^2

So KE_f = 4KE_i

So, time to reach 64 mph speed is 4 times more than the initial time

t = 4*1.10 =4.40 s

b)

v=v_0+at\\a=\frac{v-v_0}{t}=\frac{32-0}{1.1/3600}=104727.27 miles/hours^2

For final 64 mph speed,

v=v_0+at\\t=\frac{v-v_0}{a}=\frac{64-0}{104727.27} = 6.111*10^{-4} hours = 6.111*10^{-4}*3600=2.20 s

7 0
3 years ago
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