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Jlenok [28]
3 years ago
13

How many newton-meters are equal to 1600 joules

Physics
1 answer:
Salsk061 [2.6K]3 years ago
7 0
1 Joule IS 1 newton-meter.
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Consider the following mass distribution where the x- and y-coordinates are given in meters: 5.0 kg at (0.0, 0.0) m, 2.9 kg at (
Sauron [17]

Answer:

x = -1.20 m

y = -1.12 m

Explanation:

as we know that four masses and their position is given as

5.0 kg (0, 0)

2.9 kg (0, 3.2)

4 kg (2.5, 0)

8.3 kg (x, y)

As we know that the formula of center of gravity is given as

x_{cm} = \frac{m_1 x_1 + m_2x_2 + m_3x_3 + m_4x_4}{m_1 + m_2 + m_3 + m_4}

0 = \frac{5(0) + 2.9(0) + 4(2.5) + 8.3 x}{5 + 2.9 + 4 + 8.3}

10 + 8.3 x = 0

x = -1.20 m

Similarly for y direction we have

y_{cm} = \frac{m_1 y_1 + m_2y_2 + m_3y_3 + m_4y_4}{m_1 + m_2 + m_3 + m_4}

0 = \frac{5(0) + 2.9(3.2) + 4(0) + 8.3 y}{5 + 2.9 + 4 + 8.3}

9.28 + 8.3 x = 0

x = -1.12 m

5 0
3 years ago
A mass oscillates up and down on a vertical spring with an amplitude of 3 cm and a period of 2 s. What total distance does the m
harina [27]

Explanation:

It is given that,

A mass oscillates up and down on a vertical spring with an amplitude of 3 cm and a period of 2 s. It is a case of simple harmonic motion. If the amplitude of a wave is T seconds, then the distance cover by that object is 4 times the amplitude.

In 2 seconds, distance covered by the mass is 12 cm.

In 1 seconds, distance covered by the mass is 6 cm

So, in 16 seconds, distance covered by the mass is 96 cm

So, the distance covered by the mass in 16 seconds is 96 cm. Hence, this is the required solution.

6 0
3 years ago
The angular momentum of a flywheel having a rotational inertia of 0.140 kg ·m2 about its central axis decreases from 3.00 to 0.8
Rasek [7]

Answer

given,

I = 0.140 kg ·m²

decrease from 3.00 to 0.800 kg ·m²/s in 1.50 s.

a) \tau = \dfrac{\Delta L}{\Delta t}

   \tau = \dfrac{0.8-3}{1.5}

        τ = -1.467 N m

b) angle at which fly wheel will turn

   \theta= \omega t +\dfrac{1}{2}\alpha t^2

   \theta= \dfrac{L}{I} t +\dfrac{1}{2}\dfrac{\tau}{I}t^2

   \theta= \dfrac{3}{0.14}\times 1.5+\dfrac{1}{2}\dfrac{-1.467}{0.14}\times 1.5^2

        θ = 20.35 rad

c) work done on the wheel

     W = τ x θ

     W = -1.467 x 20.35 rad

    W = -29.86 J

d) average power of wheel

    P_{av} =-\dfrac{W}{t}

    P_{av} =-\dfrac{(-29.86)}{1.5}

     P_{av} =19.91\ W          

7 0
3 years ago
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An Earth satellite is orbiting at a distance from the Earth's surface equal to one Earth radius (4 000 miles). At this location,
Taya2010 [7]

Answer:0.25 times

Explanation:

Given

Distance of satellite from earth surface=Radius of earth

Force on the satellite is F=mg'

where g'=acceleration due to gravity at that point

Distance from center of Earth=R+R=2R

Gravitational Force is given by

F=\frac{GM_1M_2}{r^2}

Force F=mg'=\frac{GMm}{4R^2}-----1

Force on earth surface F=mg=\frac{GMm}{R^2}------2

Divide 1 and 2 we get

\frac{g'}{g}=\frac{R^2}{4R^2}

g'=\frac{g}{4}  

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3 years ago
WILL NAME BEST ANSWER THE BRAINLIEST (if you answer well)
mylen [45]
The best and most correct answer among the choices provided by the question is third choice. What happened during the set-up was the e<span>xcess protons move from the positive rod to the negative rod.</span> I hope my answer has come to your help. God bless and have a nice day ahead!
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