Answer:
F = +17.48 N
Explanation:
We have,
Electric charge is +0.638 C and the electric field is 27.4 N/C to the right.
It is required to find the electric force felt by the electric charge.
When a charged particle is placed in electric field, it will posses electric force. The sign of force depends on the charged particle.
The electric force is given by :
![F=qE\\\\F=+0.638\times 27.4\\\\F=17.48\ N](https://tex.z-dn.net/?f=F%3DqE%5C%5C%5C%5CF%3D%2B0.638%5Ctimes%2027.4%5C%5C%5C%5CF%3D17.48%5C%20N)
So, the electric force is 17.48 N and the force is positive.
It will be positive since it will lose 3 electrons. Its in group 13 and the elements there lose 3 electrons to become stable
The layers that make up the upper mantle are the following.
- The lithosphere and the asthenosphere.
- The lithosphere and the asthenosphere are the layers that make up the upper mantle.
- The rocky outer part of the Earth is called the lithosphere.
- It is very rigid, solid, and it is not as hot as the other parts.
- According to scientists, its thickness is about 90 to 100 kilometers from the surface of the earth.
- The layer of the Earth that moves the lithospheric plates is called the asthenosphere.
- Lithospheric plates are 60 miles long on average and are composed of continental crust.
We conclude that the lithosphere and the asthenosphere are the layers that make up the upper mantle. The new lithosphere is being created at oceanic ridges due to plate tectonics. Yes, planet Earth is getting larger constantly.
Learn more about this topic here:
brainly.com/question/11292361?referrer=searchResults
The coefficient of friction between the sled and the snow is 0.119.
To find the answer, we need to know about the friction.
<h3>How to find the coefficient of friction between the sled and the snow?</h3>
- Whenever a body moves over the surface of another body, a force come into play, which acts parallel to the surface of contact and oppose the relative motion. This opposing force is called friction.
- To solve the problem, we have to draw the free body diagram of the given system.
- We have given with the following values,
![a=0\\\alpha =35^0\\T=75N\\m=57kg](https://tex.z-dn.net/?f=a%3D0%5C%5C%5Calpha%20%3D35%5E0%5C%5CT%3D75N%5C%5Cm%3D57kg)
Here, acceleration will be equal to zero, because the velocity is given as constant.
- Thus, from the diagram, we can write the balancing equations as follows,
![ma=Tcos\alpha -f\\\where\\f=kN\\\N+Tsin\alpha=mg\\Thus,\\N=mg-Tsin\alpha](https://tex.z-dn.net/?f=ma%3DTcos%5Calpha%20-f%5C%5C%5Cwhere%5C%5Cf%3DkN%5C%5C%5CN%2BTsin%5Calpha%3Dmg%5C%5CThus%2C%5C%5CN%3Dmg-Tsin%5Calpha)
- Substituting N in f and f in the equation of ma, then we get,
![ma= Tcos\alpha -k(mg-Tsin\alpha )](https://tex.z-dn.net/?f=ma%3D%20Tcos%5Calpha%20-k%28mg-Tsin%5Calpha%20%29)
- Substituting values, we get the coefficient of friction as,
![0=(75*cos35)-k((57*9.8)-(75sin35))\\\\k((57*9.8)-(75sin35))=(75*cos35)\\\\515.6k=61.44\\\\k=\frac{61.44}{515.6}=0.119](https://tex.z-dn.net/?f=0%3D%2875%2Acos35%29-k%28%2857%2A9.8%29-%2875sin35%29%29%5C%5C%5C%5Ck%28%2857%2A9.8%29-%2875sin35%29%29%3D%2875%2Acos35%29%5C%5C%5C%5C515.6k%3D61.44%5C%5C%5C%5Ck%3D%5Cfrac%7B61.44%7D%7B515.6%7D%3D0.119)
Thus, we can conclude that, the coefficient of friction between the sled and the snow is 0.119.
Learn more about the friction here:
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Answer:
not sure. I'll try answering this later
Explanation:
I'm not sure. I'll try answering this later