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2 years ago

15

two forces act concurrently on an object on a horizontal frictionless surface. their resultatn force has the largest magnitude w

hen the angle between the forces is

1 answer:

Gennadij [26K]2 years ago

5 0

Answer:

54

Explanation:4

4

4

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What would be the length of a day(that is, the time required forone

Alex17521 [72]

Answer:

T = 5,06 10³ s

Explanation:

For this exercise, let's use Newton's second law, where force is the force of gravitational attraction

F = m a

The acceleration is centripetal

a = v² / r

Let's replace

G m $M_{e}$ / $R_{e}$ ² = m v² / $R_{e}$

G $M_{e}$ / $R_{e}$ = v²

The velocity module is constant, so we can use the equation of uniform motion

v = d / t

Where the distance is the length of the circle and in this case the time is called the period

d = 2π $R_{e}$

We replace

(2π $R_{e}$ / T)² = G $M_{e}$ / $R_{e}$

T² = 4π² $R_{e}$ ³ / G $M_{e}$

T = √ 4π² $R_{e}$ ³ / G $M_{e}$

Let's calculate

T = √ (4π² (6.37 10⁶)³ / 6.67 10⁻¹¹ 5.98 10²⁴)

T = √ (255.83 10⁵) = √ (25.583 10⁶)

T = 5,05796 10³ s

4 0

3 years ago

the density of ice is 917.what fraction of the volume of a piece of ice will be above the liquid when floating in fresh water

yulyashka [42]

Answer:

$8.3\,\%$ of that piece of ice would be above the freshwater. Assumptions:

the density of the ice is $\rho(\text{ice}) = 917\; \rm kg \cdot m^{-3}$ , and
the density of freshwater is $\rho(\text{water}) = 1.00 \times 10^3\; \rm kg \cdot m^{-3}$ .

Explanation:

The volume of that chunk of ice can be split into two halves: volume above water $V(\text{above})$ , and volume under water $V(\text{under})$ . The mass of the whole chunk of ice would be:

$m(\text{ice}) = \rho(\text{ice}) \cdot (V(\text{above}) + V(\text{under}))$ .

Let $g$ be the acceleration due to gravity. The gravity on the entire chunk of ice would be

$\begin{aligned}&W(\text{ice}) \\ &= m({\text{ice}}) \cdot g \\ &= \rho(\text{ice}) \cdot (V(\text{above}) + V(\text{under})) \cdot g\end{aligned}$ .

On the other hand, the size of buoyant force on an object is equal to the weight of the liquid that it displaces. That is: $F(\text{bouyancy}) = W(\text{water displaced})$ .

Recall that $V(\text{above})$ is the volume of the ice above the water, and $V(\text{under})$ is the volume of the ice under the water.

The mass of water displaced would be equal to:

$\begin{aligned}& m(\text{water displaced}) \\ &= \rho(\text{water}) \cdot V(\text{water displaced}) \\ &= \rho(\text{water}) \cdot V(\text{under})\end{aligned}$ .

The weight of that much water would be

$\begin{aligned} &W(\text{water displaced}) \\ &= m(\text{water displaced}) \cdot g \\ &= \rho(\text{water}) \cdot V(\text{under}) \cdot g \end{aligned}$ .

Apply the equation $F(\text{bouyancy}) = W(\text{water displaced})$ . The bouyant force on this chunk of ice would be equal to $\begin{aligned} &W(\text{water displaced}) = \rho(\text{water}) \cdot V(\text{under}) \cdot g \end{aligned}$ .

Since the ice is floating, the forces on it need to be balanced. In other words, $\begin{aligned}W(\text{ice}) &= F(\text{bouyancy}) \\ &= \rho(\text{water}) \cdot V(\text{under}) \cdot g\end{aligned}$ .

On the other hand, recall that

$\begin{aligned}&W(\text{ice}) = \rho(\text{ice}) \cdot (V(\text{above}) + V(\text{under})) \cdot g\end{aligned}$ .

Combine the two halves to obtain:

$\begin{aligned}& \rho(\text{ice}) \cdot (V(\text{above}) + V(\text{under})) \cdot g \\ &= W(\text{ice}) = \rho(\text{water}) \cdot V(\text{under}) \cdot g\end{aligned}$ .

$\begin{aligned}& \rho(\text{ice}) \cdot (V(\text{above}) + V(\text{under})) \cdot g = \rho(\text{water}) \cdot V(\text{under}) \cdot g\end{aligned}$ .

Divide both sides by $g$ (assume that $g \ne 0$ ) to obtain:

$\begin{aligned}& \rho(\text{ice}) \cdot (V(\text{above}) + V(\text{under})) = \rho(\text{water}) \cdot V(\text{under})\end{aligned}$ .

Rearrange to obtain:

$\begin{aligned}& \frac{V(\text{under})}{V(\text{above}) + V(\text{under})} = \frac{\rho(\text{water})}{\rho(\text{ice})}\end{aligned}$ .

However, the question is asking for $\displaystyle \frac{V(\text{above})}{V(\text{above}) + V(\text{under})}$ , the fraction of the volume above water. Note that

$\begin{aligned}& \frac{V(\text{under})}{V(\text{above}) + V(\text{under})} + \frac{V(\text{above})}{V(\text{above}) + V(\text{under})} = 1\end{aligned}$ .

Therefore,

$\begin{aligned} &\frac{V(\text{above})}{V(\text{above}) + V(\text{under})} \\ &= 1 - \frac{V(\text{under})}{V(\text{above}) + V(\text{under})} \\ &= 1 - \frac{\rho(\text{water})}{\rho(\text{ice})} = 1 - \frac{917}{10^3} = 0.083\end{aligned}$ .

That's equivalent to $8.3\,\%$ .

5 0

3 years ago

What can you infer from the statement, velocity is an object is zero?

stiks02 [169]

Explanation:

a.Object is in linear motion with constant velocity.

b.Object is moving at a constant speed.

c.Object is either at rest or it returns to the initial point.

d.Object is moving in a straight line without changing its direction.

hope it helps you

6 0

3 years ago

Can an object be in mechanical equilibrium when only a single force acts on it? explain.

elena-s [515]

No because if there is a force, there is acceleration which means the object is getting faster.

3 0

3 years ago

When a potassium atom forms an ion, it loses one electron. What is the electrical charge of the potassium ion? *

Ganezh [65]

+1 An electron has a negative charge so losing a charge of -1 from an uncharged, or neutral, atom will leave an ion with a positive charge.

5 0

4 years ago

Read 2 more answers