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4 years ago

Which example best represents an object with balanced forces acting upon it?

Physics

1 answer:

Tomtit [17]4 years ago

5 0

Answer:

A stationary (still) book.

Explanation:

For an object to move, unbalanced forces must be applied. Therefore, if an object is completely still, all forces applied to it are balanced.

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Two equally charged, 2.807 g spheres are placed with 3.711 cm between their centers. When released, each begins to accelerate at

statuscvo [17]

$\begin{gathered} m=\text{ 2.807 g} \\ d=\text{ 3.711cm} \\ a=260.125\text{ m/s}^2 \\ m=\text{ mass of both the spheres} \\ d=\text{ distance between the centers of sphere.} \\ a=\text{ acceleration of spheres.} \end{gathered}$ $\begin{gathered} force\text{ due to the sphere having charge q, outside its surface is given by } \\ \vec{F}=\frac{1}{4\pi\epsilon_o}\frac{q_1q_2}{r^2}\hat{r} \\ q_1=charge\text{ on the source object.} \\ q_2=charge\text{ of the object in which we are observing the force.} \\ F=\text{ the force on the charged particle outside the sphere} \\ r=\text{ distance of the charged particle from the center of the sphere} \\ \hat{r}\text{= direction of the force acting on the charged particle} \end{gathered}$ $\begin{gathered} from\text{ Newton's second law} \\ F=ma \\ F=\text{ force acting on the particle.} \\ m=\text{ mass of the object.} \\ a=\text{ acceleration of the object.} \end{gathered}$ $\begin{gathered} from\text{ both the equation } \\ ma=\frac{1}{4\pi\epsilon_o}\frac{q_1q_{\frac{2}{}}}{r^2}\hat{r} \\ here\text{ q}_1\text{ and q}_2\text{ are the same, according to the question.} \end{gathered}$ $\begin{gathered} converting\text{ all the values in s.i. unit} \\ m=2.807*10^{-3}kg \\ d=3.711*10^{-2}m \\ according\text{ to the question q}_1=\text{ q}_2 \\ value\text{ of }\frac{1}{4\pi\epsilon_o}=9*10^9\text{ Nm}^2\text{/C}^2 \\ now\text{ put all the values in the above equation } \\ 2.807*10^{-3}kg*260.125\text{ m/s\textasciicircum2}=9*10^9Nm^2\text{/C}^2*\frac{q^2}{3.711*10^{-2}m} \\ \end{gathered}$ $\begin{gathered} by\text{ trasformation} \\ q=\sqrt{\frac{2.807*10^{-3}kg*260.125m/s^2*3.711*^10^{-2}m}{9*10^9Nm^2\text{/C}^2}} \\ by\text{ solving this we get } \\ q=17.3514*10^{-7}C \\ q=1.73514\text{ micro coulombs.} \end{gathered}$ Hence the correct answer is q= 1.73514 micro coulombs.

5 0

1 year ago

Please help with this question ASAP!!!

Paladinen [302]

The answer is the second choice

7 0

4 years ago

Why is the metric system used globally, but we use the US customary units?

tatuchka [14]

Answer:

The biggest reasons the U.S. hasn't adopted the metric system are simply time and money. When the Industrial Revolution began in the country, expensive manufacturing plants became a main source of American jobs and consumer products.

Explanation:

8 0

3 years ago

A. A horse pulls a cart along a flat road. Consider the following four forces that arise in this situation.

never [62]

Answer:

a) F₁ = F₂, F₃ = F₄, b) the correct answer is 3

Explanation:

a) In this exercise we have several action and reaction forces, which are characterized by having the same magnitude, but different direction and being applied to different bodies

Forces 1 and 2 are action and reaction forces F₁ = F₂

Forces 3 and 4 are action and reaction forces F₃ = F₄

as it indicates that the

b) how the car increases if speed implies that force 1> force3

F₁ > F₃

therefore the correct answer is 3

3 0

3 years ago

You've been called in to investigate a car accident by the Bluffs. A car went over a 7.93 meter high cliff

valina [46]

Explanation:

Use the height of the cliff to determine how long it took the car to land.

Take down to be positive. Given:

Δy = 7.93 m

v₀ = 0 m/s

a = 9.8 m/s²

Find: t

Δy = v₀ t + ½ at²

7.93 m = (0 m/s) t + ½ (9.8 m/s²) t²

t = 1.27 s

Use the time to calculate the horizontal velocity.

Given:

Δx = 26.7 m

a = 0 m/s²

Find: v₀

Δx = v₀ t + ½ at²

26.7 m = v₀ (1.27 s) + ½ (0 m/s²) (1.27 s)²

v₀ = 21.0 m/s

The driver was going 21.0 m/s, faster than the speed limit of 9.72 m/s.

4 0

4 years ago

Read 2 more answers