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3 years ago

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The Earth belongs to the group of rocky planets that orbits between the sun and the asteroid belt. Place these planets in order,

starting closest to the sun. A) Mercury-Earth-Venus-Mars B) Mercury-Venus-Earth-Mars C) Mercury-Venus-Earth-Saturn D) Venus-Mars-Earth-Jupiter

2 answers:

galina1969 [7]3 years ago

6 0

Answer:

B. Mercury-Venus-Earth-Mars

Explanation:

kari74 [83]3 years ago

3 0

B. Mercury-Venus-Earth-Mars

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Arrange the balls in order from greatest amount of gravitational potential energy to least

Aloiza [94]

F e d b c a
Hope this helps !!

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3 years ago

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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

Two protons in an atomic nucleus are typically separated by a distance of 2 ✕ 10-15 m. The electric repulsion force between the

castortr0y [4]

Answer:

The magnitude of the electric force between the to protons will be 57.536 N.

Explanation:

We can use Coulomb's law to find out the force, in scalar form, will be:

$F \ = \ \frac{1}{4 \pi \epsilon_0 } \frac{q_1 q_2}{d^2}$ .

Now, making the substitutions

$d \ = \ 2.00 * 10 ^{-15} \ m$ ,

$q_1 = q_2 = 1.60 * 10 ^ {-19} \ C$ ,

$\frac{1}{4\pi\epsilon_0}=8.99 * 10^9 \frac{Nm^2}{C^2}$ ,

we can find:

$F \ = \ 8.99 * 10^9 \frac{Nm^2}{C^2} \frac{(1.60 * 10 ^ {-19} \ C)^2}{(2.00 * 10 ^{-15} \ m)^2}$ .

$F \ = 57.536 N$ .

Not so big for everyday life, but enormous for subatomic particles.

4 0

3 years ago

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The overall length of a piccolo is 32.0 cm. The resonating air column vibrates as in a pipe that is open at both ends. (a) Find

lana66690 [7]

Answer:

lowest frequency = 535.93 Hz

distance between adjacent anti nodes is 4.25 cm

Explanation:

given data

length L = 32 cm = 0.32 m

to find out

frequency and distance between adjacent anti nodes

solution

we consider here speed of sound through air at room temperature 20 degree is approximately v = 343 m/s

so

lowest frequency will be = $\frac{v}{2L}$ ..............1

put here value in equation 1

lowest frequency will be = $\frac{343}{2(0.32)}$

lowest frequency = 535.93 Hz

and

we have given highest frequency f = 4000Hz

so

wavelength = $\frac{v}{f}$ ..............2

put here value

wavelength = $\frac{343}{4000}$

wavelength = 0.08575 m

so distance = $\frac{wavelength}{2}$ ..............3

distance = $\frac{0.08575}{2}$

distance = 0.0425 m

so distance between adjacent anti nodes is 4.25 cm

3 0

3 years ago

At which part of its orbit is the planet traveling at the slowest speed?<br> A, B, C, or D

Natali [406]

I think its D im not sure tho

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3 years ago

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