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andrew-mc [135]
3 years ago
6

A wheelbarrow is experiencing two forces. One is pushing 5 N to the east and the other 5 N in the exact opposite direction. What

must be true of the wheelbarrow's motion?
Physics
2 answers:
Tamiku [17]3 years ago
7 0

Answer:

its acceleration must be 0.

Explanation:

According the second Newton's law, the net force applied in a object is equal to the mass of the object multiplied by its acceleration:

F_{net} =m.a

In this case, the net force is

F_{net}=5N-5N=0

What makes sense since both forces have the same value and opposite directions.

Since the mass of the object can't be zero, the acceleration is the one that must be 0 in order fulfill the equation of the second newton's law.

Another option is that, since both forces are opposite, the wheelbarrow could be at rest. Though this is correct, this is not always true.

In conclusion, the only  statement that always is true about the wheelbarrow's motion is that its acceleration must be 0.

mezya [45]3 years ago
4 0
Assuming no other forces are acting on the wheelbarrow, it must be stationary.
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Answer:

F_x = -\frac{6 C_6}{2^7}

Attractive

Explanation:

Data provided in the question

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Based on the given information, the force that one atom exerts on the other is

Potential energy μ = \frac{C_6}{X_6}

Force exerted by one atom upon another

F_x = \frac{\partial U}{\partial X} = \frac{\partial}{\partial X}  (-\frac{C_6}{X^6})

or

F_x = \frac{\partial}{\partial X}  (\frac{C_6}{X^6})

or

F_x = -\frac{6 C_6}{2^7}

As we can see that the C_6 comes in positive and constant which represents that the force is negative that means the force is attractive in nature

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3 years ago
Three systems of two charged particles are shown below. All the particles have the same mass,
Len [333]

Answer:D

Explanation:

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3 years ago
A car starts from rest and undergoes an acceleration of 4.0 m/s/s for a time of 5.0 s. What is the final velocity of the car?
Kruka [31]

Answer:

20m/s

Explanation:

acceleration=final velocity-initial velocity/time

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20m/s=v

7 0
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Two metal plates 15mm apart have a potential difference of 750v between them. The force on a small charged sphere placed between
Romashka [77]

Answer:

50,000 V/m

Explanation:

The electric field between two charged metal plates is uniform.

The relationship between potential difference and electric field strength for a uniform field is given by the equation

\Delta V=Ed

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E is the magnitude of the electric field

d is the  distance between the plates

In this problem, we have:

\Delta V=750 V is the potential difference between the plates

d = 15 mm = 0.015 m is the distance between the plates

Therefore, rearranging the equation we find the strength of the electric field:

E=\frac{\Delta V}{d}=\frac{750}{0.015}=50,000 V/m

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3 years ago
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Gennadij [26K]

Answer:

46.67 N/m

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4 0
3 years ago
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