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vitfil [10]
3 years ago
6

Multi celled organisms have a ______ surface area to volume ratio than single celled organisms.

Physics
1 answer:
True [87]3 years ago
6 0
A. Larger. It is larger Bc they r all larger than the other
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Because of the pole and the generator you would have to biuld
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Two astronauts on opposite ends of a spaceship are comparing lunches. One has an apple, the other has an orange. They decide to
lbvjy [14]

Answer:

Explanation:

We shall consider direction towards left as positive Let the required velocity be v and let v makes an angle φ

Applying law of conservation of momentum along direction of original motion

m₁ v₁  - m₂ v₂ = m₂v₃ - m₁ v₄

0.132 x 1.25 - .143 x 1.14 = 1.03 cos43 x .143 - v cos θ

v cos θ = .8

Applying law of conservation of momentum along direction perpendicular to direction of original motion

1.03 sin 43 x .143 = .132 x v sinθ

v sinθ = .76

squaring and adding

v² = .76 ² + .8²

v = 1.1 m /s

Tan θ = .76 / .8

θ = 44°

6 0
3 years ago
if a star 100 light years from earth is beginning to expand into a giant star how long will it take for astronomers to observe t
VikaD [51]

Answer:

100years later

Explanation:

Because the lights will arrive at world after 100 years later.

7 0
3 years ago
The diagram shows two forces of equal magnitude acting on an object. If the common magnitude of the forces is 3.6 N and the angl
Nuetrik [128]
<h3>Answer</h3>

6.6 N pointing to the right

<h3>Explanation</h3>

Given that,

two forces acting of magnitude 3.6N

angle between them = 48°

To find,

the third force that will cause the object to be in equilibrium

<h3>1)</h3>

Find the vertical and horizontal components of the two forces

vertical force1 = sin(24)(3.6)

vertical force2= -sin(24)(3.6)

<em>(negative sign since it is acting on opposite direction)</em>

vertical force3 = sin(24)(3.6) - sin(24)(3.6)

                        = 0

<h3>2)</h3>

horizontal force1 = cos(24)(3.6)

horizontal force2= cos(24)(3.6)

horizontal force3 = cos(24)(3.6) + cos(24)(3.6)

                            = 2(cos(24)(3.6))

                            = 6.5775 N

                            ≈ 6.6 N

<em />

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4 0
3 years ago
On a frictionless horizontal air table, puck A (with mass 0.254 kg ) is moving toward puck B (with mass 0.367 kg ), which is ini
irinina [24]

Answer:

v_a=0.8176 m/s

\Delta K=0.07969 J - 0.0849 J = -0.00521 J

Explanation:

According to the law of conservation of linear momentum, the total momentum of both pucks won't be changed regardless of their interaction if no external forces are acting on the system.

Being m_a and m_b the masses of pucks a and b respectively, the initial momentum of the system is

M_1=m_av_a+m_bv_b

Since b is initially at rest

M_1=m_av_a

After the collision and being v'_a and v'_b the respective velocities, the total momentum is

M_2=m_av'_a+m_bv'_b

Both momentums are equal, thus

m_av_a=m_av'_a+m_bv'_b

Solving for v_a

v_a=\frac{m_av'_a+m_bv'_b}{m_a}

v_a=\frac{0.254Kg\times (-0.123 m/s)+0.367Kg (0.651m/s)}{0.254Kg}

v_a=0.8176 m/s

The initial kinetic energy can be found as (provided puck b is at rest)

K_1=\frac{1}{2}m_av_a^2

K_1=\frac{1}{2}(0.254Kg) (0.8176m/s)^2=0.0849 J

The final kinetic energy is

K_2=\frac{1}{2}m_av_a'^2+\frac{1}{2}m_bv_b'^2

K_2=\frac{1}{2}0.254Kg (-0.123m/s)^2+\frac{1}{2}0.367Kg (0.651m/s)^2=0.07969 J

The change of kinetic energy is

\Delta K=0.07969 J - 0.0849 J = -0.00521 J

3 0
3 years ago
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