What does the lines<br> between atoms represent?and howThe lines between molecules?

a 0.25 kg arrow is moving at 5 m/s and hits a 0.10 kg apple. The apple sticks to the arrow, and both move to the right together.

MatroZZZ [7]

Answer:

The arrow-apple combo will move with a velocity 3.6 m/s to the right.

Explanation:

Use the law of conservation of momentum to solve this problem. In this case the law can be written as follows:

$m_{arrow}v_{arrow}+m_{apple}\cdot 0= (m_{arrow}+m_{apple})v_{both}$

from which the desired velocity can be isolated:

$\frac{m_{arrow}v_{arrow}}{m_{arrow}+m_{apple}}= v_{both}\\v_{both} = \frac{0.25kg\cdot 5\frac{m}{s}}{0.35kg}=3.6\frac{m}{s}$

The arrow-apple combo will move with a velocity 3.6 m/s to the right.

3 0

4 years ago

How do you reconcile the law of falling bodies (that all objects fall to earth at the same acceleration despite their weight) wi

ASHA 777 [7]

From the gravity acceleration theorem due to a celestial body or planet, we have that the Force is given as

$F = \frac {GMm} {r ^ 2}$

Where,

F = Strength

G = Universal acceleration constant

M = Mass of the planet

m = body mass

r = Distance between centers of gravity

The acceleration by gravity would be given under the relationship

$g = \frac {F} {m}$

$g = \frac {GM} {r ^ 2}$

Here the acceleration is independent of the mass of the body m. This is because the force itself depended on the mass of the object.

On the other hand, the acceleration of Newton's second law states that

$a = \frac {F} {m}$

Where the acceleration is inversely proportional to the mass but the Force does not depend explicitly on the mass of the object (Like the other case) and therefore the term of the mass must not necessarily be canceled but instead, considered.

5 0

3 years ago

Describe how could you use an electromagnete to sort a mixture of iron and copper pieces into two seprate piles of iron and copp

mote1985 [20]

Answer:

Electromagnetic cranes are used to separate copper from iron in a scrap yard. The current is switched on to energies the electromagnet and pick up the iron pieces from the scrap. Then these iron pieces are moved to another position, the electromagnet in switched off and the iron pieces are released.

Explanation:

3 0

3 years ago

How do we know the sun rotates?

Eduardwww [97]

Hi pupil Here's your answer ::::

➡➡➡➡➡➡➡➡➡➡➡➡➡

The sun Doesnt rotates as we know all . But it rotates with the wjole our Milky Way Galaxy. This movement is so slow that we cant even recognise that we are shifted.

This we know because we all had studied about the solar system and space.

By this study we can say that Sun Rotates.

⬅⬅⬅⬅⬅⬅⬅⬅⬅⬅⬅⬅⬅

Hope this helps ......

6 0

3 years ago

Read 2 more answers

There is a parallel plate capacitor. Both plates are 4x2 cm and are 10 cm apart. The top plate has surface charge density of 10C

liberstina [14]

Answer:

1) The total charge of the top plate is 0.008 C

b) The total charge of the bottom plate is -0.008 C

2) The electric field at the point exactly midway between the plates is 0

3) The electric field between plates is approximately 1.1294 × 10¹² N/C

4) The force on an electron in the middle of the two plates is approximately 1.807 × 10⁻⁷ N

Explanation:

The given parameters of the parallel plate capacitor are;

The dimensions of the plates = 4 × 2 cm

The distance between the plates = 10 cm

The surface charge density of the top plate, σ₁ = 10 C/m²

The surface charge density of the bottom plate, σ₂ = -10 C/m²

The surface area, A = 0.04 m × 0.02 m = 0.0008 m²

1) The total charge of the top plate, Q = σ₁ × A = 0.0008 m² × 10 C/m² = 0.008 C

b) The total charge of the bottom plate, Q = σ₂ × A = 0.0008 m² × -10 C/m² = -0.008 C

2) The electrical field at the point exactly midway between the plates is given as follows;

$V_{tot} = V_{q1} + V_{q2}$

$V_q = \dfrac{k \cdot q}{r}$

Therefore, we have;

The distance to the midpoint between the two plates = 10 cm/2 = 5 cm = 0.05 m

$V_{tot} = \dfrac{k \cdot q}{0.05} + \dfrac{k \cdot (-q)}{0.05} = \dfrac{k \cdot q}{0.05} - \dfrac{k \cdot q}{0.05} = 0$

The electric field at the point exactly midway between the plates, $V_{tot}$ = 0

3) The electric field, 'E', between plates is given as follows;

$E =\dfrac{\sigma }{\epsilon_0 } = \dfrac{10 \ C/m^2}{8.854 \times 10^{-12} \ C^2/(N\cdot m^2)} \approx 1.1294 \times 10^{12}\ N/C$

E ≈ 1.1294 × 10¹² N/C

The electric field between plates, E ≈ 1.1294 × 10¹² N/C

4) The force on an electron in the middle of the two plates

The charge on an electron, e = -1.6 × 10⁻¹⁹ C

The force on an electron in the middle of the two plates, $F_e$ = E × e

∴ $F_e$ = 1.1294 × 10¹² N/C × -1.6 × 10⁻¹⁹ C ≈ 1.807 × 10⁻⁷ N

The force on an electron in the middle of the two plates, $F_e$ ≈ 1.807 × 10⁻⁷ N

4 0

3 years ago

What does the lines between atoms represent?and howThe lines between molecules?