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Elena L [17]
2 years ago
14

Help pls ... will give brainlist

Physics
1 answer:
Ghella [55]2 years ago
6 0
I, i’m pretty sure sorry if it’s wrong
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Emmett is lifting a box vertically. Which forces are necessary for calculating the total force?
GrogVix [38]

When Emmett is lifting a box vertically, the forces that must be added to calculate the total force are: the gravitational force, tension force(the force exerted by Emmett to the box and the force exerted by the box to Emmett), and air resistance force.

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2 years ago
When you walk across the ground and push on it with your feet...
marissa [1.9K]

Answer:

The ground pushes back on your feet with equal force.

Explanation:

Newton's Laws of Motion

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3 years ago
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While driving on a rural road, your right wheels run off the pavement. You should hold the steering wheel firmly and
Vsevolod [243]

Answer:

The answer is C. Steer in a straight line while gently slowing down

Explanation:

The following are advised when your cars go off the pavement while driving;

firstly, Do not panic.

ensure you hold on to your steering wheel tightly.

keep Steering straight ahead.

ensure you Stay on the shoulder.

Ease up on the accelerator and brake gently.

When you  know you can safely do so, turn back on the road at a much lower speed.

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3 years ago
Three charges lie along the x axis. The positive charge q1 = 15 μC is at x= 2.0 m and the positive charge q2 = 6.0μC is at the o
Delicious77 [7]

Answer:

x = 0.775m

Explanation:

Conceptual analysis

In the attached figure we see the locations of the charges. We place the charge q₃ at a distance x from the origin. The forces F₂₃ and F₁₃ are attractive forces because the charges have an opposite sign, and these forces must be equal so that the net force on the charge q₃ is zero.

We apply Coulomb's law to calculate the electrical forces on q₃:

F_{23} = \frac{k*q_{2}*q_{3}}{x^2} (Electric force of q₂ over q₃)

F_{13} = \frac{k*q_{1}*q_{3}}{(2-x)^2} (Electric force of q₁ over q₃)

Known data

q₁ = 15 μC = 15*10⁻⁶ C

q₂ = 6 μC = 6*10⁻⁶ C

Problem development

F₂₃ = F₁₃

\frac{k*q_{2}*q_{3}}{x^2} = \frac{k*q_{1}*q_{3}}{(2-x)^2} (We cancel k and q₃)

\frac{q_2}{x^2}=\frac{q_1}{(2-x)^2}

q₂(2-x)² = q₁x²

6×10⁻⁶(2-x)² = 15×10⁻⁶(x)² (We cancel 10⁻⁶)

6(2-x)² = 15(x)²

6(4-4x+x²) = 15x²

24 - 24x + 6x² = 15x²

9x² + 24x - 24 = 0

The solution of the quadratic equation is:

x₁ = 0.775m

x₂ = -3.44m

x₁ meets the conditions for the forces to cancel in q₃

x₂ does not meet the conditions because the forces would remain in the same direction and would not cancel

The negative charge q₃ must be placed on x = 0.775 so that the net force is equal to zero.

6 0
3 years ago
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Hey guys i need help (potential energy)
Vlad1618 [11]

Explanation:

Michael should put the vase at the bottom of the shelf to reduce the potential energy because the height of the vase to the floor is nearly zero.

8 0
3 years ago
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