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babymother [125]
3 years ago
9

block of mass m sits at rest on a rough inclined ramp that makes an angle with the horizontal. What must be true about normal fo

rce F on the block due to the ramp
Physics
1 answer:
oee [108]3 years ago
6 0

Answer:

Explanation:

For a body on a ramp with mass m, the forces acting on the body along the vertical component are the weight and the normal reaction.

The weight of the body acts in the negative y direction while the normal reaction acts in the positive y direction

Taking the sum of forces along the y component

Sum Fy = -W+R = ma

Since acceleration is zero

-W+R = m(0)

-W+R = 0

-W = -R

W = R

Hence the Normal reaction force acting on the on the body is equal to normal force

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In example 20.3 in the text, the net force on a 1.0 nC charge located between two 10 nC charges is calculated. How would the ans
VashaNatasha [74]

Answer:

The forces experienced by the middle particle are attractive, and the net force will remain the same (0) if and only if the distances of the sides particles to the middle particle are the same.

Explanation:

In example 20.3 the forces experienced by the middle particle are repulsive because all the particles are positive, for the case in which the particles on the sides are replaced for negative charge particles the forces experienced by the middle particle are attractive. Regarding the net force, because we don't know the distances we can not give a definitive answer, what we can say is that if the distances from the middle particle to the sides particles are the same the net force is zero for both cases (remain unchanged).  

7 0
4 years ago
In this reaction diagram which part represents the doffrence in energy between the reactants and the products?
Annette [7]

Answer:

The correct answer is - option C. G.

Explanation:

In this reaction diagram, there is a representation of the reaction profile. The reaction profile shows the change that takes place during a reaction in the energy of reactants or substrate and products. In this profile, activation energy looks like a hump in the line, and the minimum energy required to initiate the reaction.

The overall energy of the reaction, including or excluding activation energy depends on the nature of the reaction if it is exothermic or endothermic. and products are represented by the G which shows the difference between the energy of the reactants and products.

3 0
3 years ago
Read 2 more answers
1. How do we tell how much energy a ray of light has?
andrew-mc [135]

Ce clasă ești sa văd daca te pot ajuta

6 0
3 years ago
A bicycle rider travels 50.0 km in 2.5 hours. What is his average speed?
ycow [4]
50km / 2.5 = 20 km per hour
4 0
4 years ago
2. Three charged particles are placed at the corners of an equilateral triangle of side 1.20 m. The charges are +7.0μC, -8.0 μC
Scorpion4ik [409]

Answer:

0.53 N, 25.6°

Explanation:

side of triangle, a = 1.2 m

q = 7 μC

q1 = - 8 μC

q2 = - 6 μC

Let F1 be the force between q and q1

By using the coulomb's law

F_{1}=\frac{Kq_{1}q}{a^{2}}

F_{1}=\frac{9\times 10^{9}\times 7\times 10^{-6}\times 8\times 10^{-6}}{1.2^{2}}

F1 = 0.35 N

Let F2 be the force between q and q2

By using the coulomb's law

F_{2}=\frac{Kq_{2}q}{a^{2}}

F_{2}=\frac{9\times 10^{9}\times 7\times 10^{-6}\times 6\times 10^{-6}}{1.2^{2}}

F2 = 0.26 N

Write the forces in the vector form

\overrightarrow{F_{1}}=0.35\widehat{i}

\overrightarrow{F_{2}}=0.26\left ( Cos60 \widehat{i}+Sin60\widehat{j}\right )

\overrightarrow{F_{2}}=0.13 \widehat{i}+0.23\widehat{j}

Net force

\overrightarrow{F}=\overrightarrow{F_{1}}+\overrightarrow{F_{2}}

\overrightarrow{F}=0.48 \widehat{i}+0.23\widehat{j}

Magnitude of the force

F=\sqrt{0.48^{2}+0.23^{2}}

F = 0.53 N

Direction of force with x axis

tan\theta =\frac{0.23}{0.48}

θ = 25.6°

7 0
3 years ago
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