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Ronch [10]
3 years ago
7

Suppose that two point charges, each with a charge of +3.00 Coulomb are separated by a distance of 3.00 meters. Determine the ma

gnitude of the electrical force of repulsion between them
Physics
1 answer:
Sveta_85 [38]3 years ago
3 0

Answer:

9\cdot 10^9 N

Explanation:

The magnitude of the electrical force between the two point charges is

F=k\frac{q_1 q_2}{r^2}

where

k is the Coulomb's constant

q_1 =q_2 = +3.0 C is the magnitude of each charge

r = 3.00 m is the separation between the two charges

Substituting the numbers into the formula, we find

F=(9\cdot 10^9 Nm^2C^{-2})\frac{(+3 C)^2}{(3.00 m)^2}=9\cdot 10^9 N

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Explanation:

the block will move to the right side with small velocity because the force from the left side greater than force from right side. Velocity will be less because of friction and gravitational attraction.

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Find the ratio of the gravitational force between two planets if the masses of both planets are quadrupled but the distance betw
Snezhnost [94]

Answer:

The ratio of the new force over the original force is 16

Explanation:

Recall the formula for the gravitational force between two masses M1 and M2 separated a distance D:

F_G=G\,\frac{M_1\,\,M_2}{D^2}

So now, if the masses M1 and M2 are quadrupled and the distance stays the same, the new force becomes:

F'_G=G\,\frac{4M_1\,\,4M_2}{D^2}=G\,\frac{16\,\,M_1\,\,M_2}{D^2}=16\,\,G\,\frac{M_1\,\,M_2}{D^2}= 16\,\,F_G

which is 16 times the original force.

So the ratio of the new force over the original force is 16

5 0
3 years ago
Solve 3x+1<br><img src="https://tex.z-dn.net/?f=%20%5Cleqslant%201" id="TexFormula1" title=" \leqslant 1" alt=" \leqslant 1" ali
Rzqust [24]

3x + 1 ≤ 1

Subtract 1 from each side:  3x ≤ 0

Divide each side by  3 :     <em>x ≤ 0</em>

8 0
3 years ago
The cabinet is mounted on coasters and has a mass of 45 kg. The casters are locked to prevent the tires from rotating. The coeff
stira [4]

Answer:

the force P required for impending motion is 132.3 N

the largest value of "h" allowed if the cabinet is not to tip over is 0.8 m

Explanation:

Given that:

mass of the cabinet  m = 45 kg

coefficient of static friction μ =  0.30

A free flow body diagram illustrating what the question represents is attached in the file below;

The given condition from the question let us realize that ; the casters are locked to prevent the tires from rotating.

Thus; considering the forces along the vertical axis ; we have :

\sum f_y =0

The upward force and the downward force is :

N_A+N_B = mg

where;

\mathbf { N_A  \ and  \ N_B} are the normal contact force at center point A and B respectively .

N_A+N_B = 45*9.8

N_A+N_B = 441    ------- equation (1)

Considering the forces on the horizontal axis:

\sum f_x = 0

F_A +F_B  = P

where ;

\mathbf{ F_A \ and \ F_B } are the static friction at center point A and B respectively.

which can be written also as:

\mu_s N_A + \mu_s N_B  = P

\mu_s( N_A +  N_B)  = P

replacing our value from equation (1)

P = 0.30 ( 441)    

P = 132.3 N

Thus; the force P required for impending motion is 132.3 N

b) Since the horizontal distance between the casters A and B is 480 mm; Then half the distance = 480 mm/2 = 240 mm = 0.24 cm

the largest value of "h" allowed for  the cabinet is not to tip over is calculated by determining the limiting condition  of the unbalanced torque whose effect is canceled by the normal reaction at N_A and it is shifted to N_B:  

Then:

\sum M _B = 0

P*h = mg*0.24

h =\frac{45*9.8*0.24}{132.3}

h = 0.8 m

Thus; the largest value of "h" allowed if the cabinet is not to tip over is 0.8 m

6 0
3 years ago
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