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Romashka-Z-Leto [24]
3 years ago
13

What percentage of the mass of the solar nebula consisted of elements other than hydrogen and helium?

Physics
2 answers:
Korvikt [17]3 years ago
6 0

Answer:

2%

Explanation:

75% of the mass contains hydrogen 25% helium and all other elements comprising only of 2%

neonofarm [45]3 years ago
5 0

Answer:

C. 2 percent.

Explanation:

75% of the mass in the form of hydrogen

25% as helium

All other elements comprising only 2% of the total

Assume mass of the solar nebula = 100

Remaining % = 2%

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What happens to the position of an object as an unbalanced force acts on?
jarptica [38.1K]

Answer:

Unbalanced forces change the motion of an object. If an object is at rest and an unbalanced force pushes or pulls the object, it will move. Unbalanced forces can also change the speed or direction of an object that is already in motion.

5 0
3 years ago
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Which of the following changes would increase the focal length of a lens? a) decrease rodius of curvature of surfoce c) increase
anastassius [24]

Answer:

Increase the radius of curvature of surface.

Explanation:

Radius of curvature is the measurement that is two times the focal length, for a given lens. It lies on either side of the lens.

Focal length is the distance which is half of the radius of curvature. Radius of curvature is a measure of the radius of the circle . Focal length is the distance between the center of curvature of the lens and the point where all the rays are brought to a focus for a distant object.

8 0
3 years ago
the flame on a gas stove heats the bottom of a metal pot. (A) conduction (B)convection (C)radiation. the cold water in a metal p
Sedbober [7]
The flame on a gas stove heats the bottom of a metal pot is conduction.
The sun gives you a sunburn is a radiation
3 0
3 years ago
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A merry-go-round with a rotational inertia of 600 kg m2 and a radius of 3.0 m is initially at rest. A 20 kg boy approaches the m
nekit [7.7K]

Answer:

The velocity of the merry-go-round after the boy hops on the merry-go-round is 1.5 m/s

Explanation:

The rotational inertia of the merry-go-round = 600 kg·m²

The radius of the merry-go-round = 3.0 m

The mass of the boy = 20 kg

The speed with which the boy approaches the merry-go-round = 5.0 m/s

F_T \cdot r = I \cdot \alpha  = m \cdot r^2  \cdot \alpha

Where;

F_T = The tangential force

I =  The rotational inertia

m = The mass

α = The angular acceleration

r = The radius of the merry-go-round

For the merry go round, we have;

I_m \cdot \alpha_m  = I_m \cdot \dfrac{v_m}{r \cdot t}

I_m = The rotational inertia of the merry-go-round

\alpha _m = The angular acceleration of the merry-go-round

v _m = The linear velocity of the merry-go-round

t = The time of motion

For the boy, we have;

I_b \cdot \alpha_b  = m_b \cdot r^2  \cdot \dfrac{v_b}{r \cdot t}

Where;

I_b = The rotational inertia of the boy

\alpha _b = The angular acceleration of the boy

v _b = The linear velocity of the boy

t = The time of motion

When the boy jumps on the merry-go-round, we have;

I_m \cdot \dfrac{v_m}{r \cdot t} = m_b \cdot r^2  \cdot \dfrac{v_b}{r \cdot t}

Which gives;

v_m = \dfrac{m_b \cdot r^2  \cdot \dfrac{v_b}{r \cdot t} \cdot r \cdot t}{I_m} = \dfrac{m_b \cdot r^2  \cdot v_b}{I_m}

From which we have;

v_m =  \dfrac{20 \times 3^2  \times 5}{600} =  1.5

The velocity of the merry-go-round, v_m, after the boy hops on the merry-go-round = 1.5 m/s.

5 0
3 years ago
A cannon is mounted on a cart which sits on the ground, supported by frictionless wheels. The mass of the cannon and cart is 4.6
Kitty [74]

Answer:

The velocity of the launcher after the projectile is launched is  -5.011 m/s

Explanation:

Here we have the mass of the cannon and cart, m₁ =  4.65 kg

Velocity of cannon and cart, v₁ = 2.00 m/s

Mass of projectile, m₂ = 50.0 g = 0.05 kg

Velocity of projectile, v₂ = 647 m/s

Velocity of the launcher, v₃ = Required

Mass of cannon and cart, launcher after launching projectile m₃ = 4.65-0.05

= 4.6 kg

Therefore, from the principle of the conservation of linear momentum, we have

Total initial momentum = Total final momentum

m₁ × v₁ = m₂ × v₂ + m₃ × v₃

Substituting gives

4.65 kg × 2.00 m/s = 0.05 kg × 647 m/s + 4.6 kg × v₃

4.65 kg × 2.00 m/s - 0.05 kg × 647 m/s = 4.6 kg × v₃

-23.05 kg·m/s = 4.6 kg × v₃

v_3 = \frac{-23.05 \, kg\cdot m/s}{4.6 \, kg} =  \frac{-461}{92} m/s

v₃ = -5.011 m/s.

3 0
3 years ago
Read 2 more answers
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