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3 years ago

What is it called when an atom can pass electrons?

Physics

1 answer:

fiasKO [112]3 years ago

7 0

Answer:

Ionic bonding.

Explanation:

According to our crude, conceptual definition, chemical bonds can form either by electron transfer between atoms or by sharing of electrons. ... When atoms lose or gain electrons, they become what are called ions. Loss of electrons leaves an atom with a net positive charge, and the atom is called a cation.

Hope this helped!

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Definition: A command economy is where a central government makes all economic decisions. The government or a collective owns th

denis-greek [22]

A command economy is a system where the government, rather than the free market, determines what goods should be produced, how much should be produced and the price at which the goods are offered for sale.

7 0

3 years ago

Why is the mass of carbon atom greater than the total mass of its protons and electrons

VLD [36.1K]

One or more neutrons in the nucleus add mass to the atom.

8 0

4 years ago

A 1400.0 kg car crests a 3200.0 m pass in the mountains and briefly comes to rest. The car descends 1000 m before climbing and c

rjkz [21]

Answer:

a) v = 88.54 m/s

b) vf = 26.4 m/s

Explanation:

Given that;

m = 1400.0 kg

by using the energy conservation

loss in potential energy is equal to gain in kinetic energy

mg × ( 3200-2800) = 1/2 ×m×v²

1400 × 9.8 × 400 = 0.5 × 1400 × v²

5488000 = 700v²

v² = 5488000 / 700

v² = 7840

v = √7840

v = 88.54 m/s

Work done by all forces is equal to change in KE

W_gravity + W_non - conservative = 1/2×m×(vf² - vi²)

we substitute

1400 × 9.8 × ( 3200-2800) - (5 × 10⁶) = 1/2 × 1400 × (vf² -0 )

488000 = 700 vf²

vf² = 488000 / 700

vf² = 697.1428

vf = √697.1428

vf = 26.4 m/s

4 0

4 years ago

Two rockets are fired at each other with initial velocities of 150m/s150m/s and are 6000m6000m apart. The first rocket is accele

Nataliya [291]

Answer:

3469.788 m

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

First rocket

$s=ut+\frac{1}{2}at^2\\\Rightarrow s=150\times t+\frac{1}{2}\times 5\times t^2\\\Rightarrow s=150t+2.5t^2\ m$

Second rocket

$s=ut+\frac{1}{2}at^2\\\Rightarrow s=150\times t+\frac{1}{2}\times 15\times t^2\\\Rightarrow s=150t+7.5t^2\ m$

When this will collide the total distance they would have covered would be 6000 m.

$6000=150t+2.5t^2+150t+7.5t^2\\\Rightarrow 6000=300t+10t^2\\\Rightarrow 10t^2+300t-6000=0$

$t=5\left(\sqrt{33}-3\right),\:t=-5\left(3+\sqrt{33}\right)\\\Rightarrow t=13.72, -43.72$

Hence at 13.72 seconds they will collide assuming they are launched at the same time.

$s=150t+7.5t^2\\\Rightarrow s=150\times 13.72+7.5\times 13.72^2\\\Rightarrow s=3469.788\ m$

The second rocket would have gone 3469.788 m when they collide

7 0

4 years ago

[9]

ad-work [718]

Answer:

9.66 m/s 15° with +y

2.59 m/s 75° with +y

Explanation:

Momentum is conserved in the y direction.

mu₁ + mu₂ = mv₁ + mv₂

u₁ + u₂ = v₁ + v₂

10 m/s + 0 m/s = v₁ cos 15° + v₂ cos 75°

10 = v₁ cos 15° + v₂ cos 75°

Momentum is conserved in the x direction.

mu₁ + mu₂ = mv₁ + mv₂

u₁ + u₂ = v₁ + v₂

0 m/s + 0 m/s = v₁ sin 15° − v₂ sin 75°

0 = v₁ sin 15° − v₂ sin 75°

v₁ sin 15° = v₂ sin 75°

v₂ = v₁ sin 15° / sin 75°

Substitute.

10 = v₁ cos 15° + (v₁ sin 15° / sin 75°) cos 75°

10 = v₁ cos 15° + v₁ sin 15° / tan 75°

10 = v₁ (cos 15° + sin 15° / tan 75°)

v₁ ≈ 9.66 m/s

v₂ ≈ 2.59 m/s

7 0

3 years ago