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4 years ago

What is the weight of a 8-kg substance in N, kN, kg·m/s2, kgf, lbm·ft/s2, and lbf?

Physics

1 answer:

kvv77 [185]4 years ago

5 0

Answer:

$W = 78.48N\\W =0.0784kN\\W = 8kgf\\$

$W= 3.6lbf\\W= 115.2 lbm*ft/s2$

Explanation:

$m=8kg=3.6lb\\$

and g=9.81 m/s2=32.16 ft/s2

and

W=m*g

we can just replace de mass and gravity and we have

$W = 8kg * 9.81 \frac{m}{s^{2} } =78.48N\\W = \frac{78.48N}{1000} =0.0784kN\\W = 8kgf\\$

$W= 3.6lbf\\W= 3.6lbm *32.16 ft/s2 =115.2 lbm*ft/s2$

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sergij07 [2.7K]

Answer:

Approximately $(-1.6 \times 10^{5}\; \rm J)$ (assuming that the car was on level ground.)

Explanation:

When an object of mass $m$ is moving at a speed of $v$ , the kinetic energy of that object would be $(1/2) \, m \cdot v^{2}$ .

Initial kinetic energy of the car:

$\begin{aligned}&\frac{1}{2}\, m \cdot v^{2} \\ &= \frac{1}{2}\times 1120\; \rm kg \times (17\; \rm m \cdot s^{-1})^{2} \\ &\approx 1.6\times 10^{5}\; \rm J \end{aligned}$ .

After the car comes to a stop, the kinetic energy of this car would be $0\; \rm J$ because the car would not be moving.

Change to the kinetic energy of the car: $\text{Final KE} - \text{Initial KE} = -1.6 \times 10^{5}\; \rm J$ .

If the car is traveling on level ground, friction would be the only force that contributed to this energy change. Hence:

$\text{Work of friction} = \text{Energy change} = -1.6\times 10^{5}\; \rm J$ .

The value of the work that friction did is negative. The reason is that at any instant before the car comes to a stop, friction would be exactly opposite to the direction of the movement of the car.

The work of a force on an object is the dot product of that force and the displacement of that object. The dot product of two vectors of opposite directions is negative. Hence, in this question, the work that friction did on the car would be negative because the friction vector would be opposite to the movement of the car.

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3 years ago

Under which conditions do both convex and concave mirrors form virtual images? State whether or not each can form a real image.I

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Concave mirrors can make a real image.when the object is away from the mirror it is called real image and when the object is closer to the mirror it is called virtual mirror.

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3 years ago

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Oksi-84 [34.3K]

It would have to be 36,719 Km high in order to be to be in geosynchronous orbit.

To find the answer, we need to know about the third law of Kepler.

<h3>What's the Kepler's third law?</h3>

It states that the square of the time period of orbiting planet or satellite is directly proportional to the cube of the radius of the orbit.
Mathematically, T²∝a³

<h3>What's the radius of geosynchronous orbit, if the time period and altitude of ISS are 90 minutes and 409 km respectively?</h3>

The time period of geosynchronous orbit is 24 hours or 1440 minutes.
As the Earth's radius is 6371 Km, so radius of the ISS orbit= 6371km + 409 km = 6780km.
If T1 and T2 are time period of geosynchronous orbit and ISS orbit respectively, a1 and a2 are radius of geosynchronous orbit and ISS orbit, as per third law of Kepler, (T1/T2)² = (a1/a2)³
a1= (T1/T2)⅔×a2

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= 43,090 km

Altitude of geosynchronous orbit = 43,090 - 6371= 36,719 km

Thus, we can conclude that the altitude of geosynchronous orbit is 36,719km.

Learn more about the Kepler's third law here:

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2 years ago

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Explanation:

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Their general electronic configuration is given by:

$ns^2np^6$

n = principle quantum number

c) In an ionic bonding, electrons are transferred to the electronegative atom from the electropositive atom.

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Electropositive : An atom which looses electrons.Positive charge carrying ion is formed.

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Answer:

Explanation:

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