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crimeas [40]
3 years ago
5

Two identical blocks are connected to opposite ends of a compressed spring. the blocks initially slide together on a frictionles

s surface with a velocity v to the right. at some instant, the left block is moving at v/2 to the left and the other block is moving to the right. what is the speed of the center of the mass of they system at that instant?
Physics
1 answer:
lana66690 [7]3 years ago
7 0

Answer:

v to the right.

Explanation:

<u>If there is no external force on the system, then the velocity of the center of mass is </u><u>constant</u><u>. </u>

The initial velocity of center of mass is v to the right. Therefore, the velocity of the center of mass at any point is v to the right.

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Suppose a free-fall ride at an amusement park starts at rest and Is In free fall.
Fittoniya [83]
Ella has both dimes (d) and nickels (n) in her piggy bank. She has a total of 36 coins in all, and has 4 more dimes than nickels. Find how much money she has in her piggy bank using a two-variable, two-equation system.

Ella has both dimes (d) and nickels (n) in her piggy bank. She has a total of 36 coins in all, and has 4 more dimes than nickels. Find how much money she has in her piggy bank using a two-variable, two-equation system.
4 0
2 years ago
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Cesar and Jill went to a field to play soccer. As the ball downward toward Jill, Jill used her foot to kick the ball and keep it
Kobotan [32]

Answer:

D

Explanation:

8 0
3 years ago
A person who weighs 509,45 N empties her lungs as much as
Masja [62]

Answer:

The weight of the girl = 1045.86 kg/m³

Explanation:

Density: This can be defined as the ratio of the mass of a body to the volume of that body. The S.I unit of density is kg/m³.

From Archimedes principle,

R.d = Density of the person/Density of water = Weight of the person in air/Upthrust.

⇒ D₁/D₂ = W/U............................... Equation 1.

Where D₁ = Density of the person, D₂ = Density of water, W = Weight of the person in air, U = Upthrust in water.

Making D₁ the subject of the equation,

D₁ = D₂(W/U)................................... Equation 2

<em>Given: D₂ = 1000 kg/m³ , W = 509.45 N, U = lost in weight = weight in air - weight in water = 509.45 - 22.34 = 487.11 N</em>

<em>Substituting these values into equation 2</em>

D₁ = 1000(509.45/487.11)

D₁ = 1045.86 kg/m³

Thus the weight of the girl = 1045.86 kg/m³

<em></em>

7 0
3 years ago
Two point charges each carrying a charge of + 4.5 E - 6 C are located 4.5 meters away from each other. How strong is the electro
BARSIC [14]

Answer:

0.009 N, repulsive

Explanation:

The electrostatic force between two electric charges is given by:

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

where

k is the Coulomb's constant

q1 and q2 are the two charges

r is the separation between the two charges

In this problem, we have

q_1 =q_2 = +4.5\cdot 10^{-6}C are the two charges

r = 4.5 m is their separation

Substituting into the equation, we find

F=(9\cdot 10^9 Nm^2 C^{-2})\frac{(+4.5\cdot 10^{-6} C)(4.5\cdot 10^{-6} C)}{(4.5 m)^2}=0.009 N

Moreover, the force is repulsive. In fact, the following rules apply:

- When two charges have same sign, they repel each other

- When two charges have opposite signs, they attract each other

7 0
3 years ago
Not in book
umka2103 [35]

Answer:

x=2.4365\ m

and

x=-1.4365\ m

Explanation:

Given:

  • first charge, q_1=5\times 10^{-3}\ C
  • second charge, q_2=3\times 10^{-3}\ C
  • position of first charge, x_1=-2\ m
  • position of second charge, x_2=-1\ m

Now since there are only 2 charges and of the same sign so they repel each other. This repulsion will be zero at some point on the line joining the charges.

<u>Now, according to the condition, electric field will be zero where the effects of field due to both the charges is equal.</u>

E_1=E_2

  • since first charge is greater than the second charge so we may get a point to the right of the second charge and the distance between the two charges is 1 meter.

\frac{1}{4\pi.\epsilon_0} \frac{q_1}{(r+1)^2} =\frac{1}{4\pi.\epsilon_0} \frac{q_2}{(r)^2}

\frac{5\times 10^{-3}}{(r+1)^2} = \frac{3\times 10^{-3}}{(r)^2}

3(r^2+1+2r)=5r^2

2r^2-6r-3=0

r=3.4365 \&\ r=-0.4365

Since we have assumed that the we may get a point to the right of second charge so we calculate with respect to the origin.

x=-1+3.4365=2.4365\ m

and

x=-1-0.4365=-1.4365\ m

6 0
3 years ago
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