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3 years ago

6

Which of the following astronomical bodies or collection of astronomical bodies is likely to be a part of the others listed?

2 answers:

Bond [772]3 years ago

5 0

<span>The correct answer is: Sun

Explanation:
Sun is the only star in our solar system, and planets orbit around it. Our solar system is the part of Milky way (located in one of the outer bands of the Milky way), and the Milky way is the part of the Universe (Universe is made up of billions of galaxies, and Milky way is one of those galaxies). Hence, in the given options, the Sun is the part of all of the other options given.</span>

storchak [24]3 years ago

3 0

Answer:

SUNNNNNNNNNNNNNNNNNNNNNNNNNNNNNNNNN

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A small car with mass of 0.800 kg travels at a constant speed

Alexandra [31]

Answer:

The equation of equilibrium at the top of the vertical circle is:

\Sigma F = - N - m\cdot g = - m \cdot \frac{v^{2}}{R}

The speed experimented by the car is:

\frac{N}{m}+g=\frac{v^{2}}{R}

v = \sqrt{R\cdot (\frac{N}{m}+g) }

v = \sqrt{(5\,m)\cdot (\frac{6\,N}{0.8\,kg} +9.807\,\frac{kg}{m^{2}} )}

v\approx 9.302\,\frac{m}{s}

The equation of equilibrium at the bottom of the vertical circle is:

\Sigma F = N - m\cdot g = m \cdot \frac{v^{2}}{R}

The normal force on the car when it is at the bottom of the track is:

N=m\cdot (\frac{v^{2}}{R}+g )

N = (0.8\,kg)\cdot \left(\frac{(9.302\,\frac{m}{s} )^{2}}{5\,m}+ 9.807\,\frac{m}{s^{2}} \right)

N=21.690\,N

7 0

2 years ago

Which direction will the box move?

Yuki888 [10]

Net force

F1-F2-F3
5-10-20
5-30
-25N

As it's negative the box will move left

3 0

2 years ago

When a cup is placed on a table, which force prevents the cup from falling to the ground?

pshichka [43]

I believe it is normal force.

5 0

3 years ago

Read 2 more answers

A single-turn current loop, carrying a current of 4.03 A, is in the shape of a right triangle with sides 68.1, 151, and 166 cm.

SashulF [63]

Answer:

Part a)

$F = 0$

Part b)

$F = 0.25 N$

Part c)

$F = 0.25 N$

Part d)

Net force on a closed loop in uniform magnetic field is always ZERO

$F_{net} = 0$

Explanation:

As we know that force on a current carrying wire is given as

$\vec F = i(\vec L \times \vec B)$

now we have

Part a)

current in side 166 cm and magnetic field is parallel

so we have

$F = i(\vec L \times \vec B)$

here we know that L and B is parallel to each other so

$F = 0$

Part b)

For 68.1 cm length wire we have

$F = iLB sin\theta$

here we know that

$cos\theta = \frac{68.1}{166}$

$\theta = 65.8$

so we have

$F = (4.03)(0.681)(99.3 \times 10^{-3})sin65.8$

$F = 0.25 N$

Part c)

For 151 cm length wire we have

$F = iLB sin\phi$

here we know that

$cos\phi = \frac{151}{166}$

$\theta = 24.5$

so we have

$F = (4.03)(1.51)(99.3 \times 10^{-3})sin24.5$

$F = 0.25 N$

Part d)

Net force on a closed loop in uniform magnetic field is always ZERO

$F_{net} = 0$

4 0

3 years ago

3. The car's mass is 400 kg. It moves at a velocity of 20 m/s. Calculate the car's momentum. *

Orlov [11]

Answer:

momentum=mass×velocity

momentum =400kg×20m/s=8000kg.m/s

7 0

3 years ago