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aksik [14]
3 years ago
15

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
Physics
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
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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

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I think its D im not sure tho
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