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PtichkaEL [24]
3 years ago
12

Help me with this question please

Physics
1 answer:
Salsk061 [2.6K]3 years ago
6 0

Answer:

5.en

6.ex

7.ex

8.en

Explanation:

<h3>#CARRY ON LEARNING</h3><h3>#BRAINLITS </h3>
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How would seasons on earth be different if the earth was not tilted on its axis
MakcuM [25]

If the Earth's axis were 'straight' ... pefectly perpendicular to the ecliptic
plane ... then:

-- Day and night would be the same length ... every day of the year,
everywhere on Earth !

-- There wouldn't be any seasons, anywhere.  There might still be some
'weather' ... cloudy days, sunny days, occasional rain, wind etc.  But
there would be no average change during the year.  No hot months or
cold months.  In any one place, the weather would always be generally
the same, every day, all year.  Everywhere all around the equator would be
generally the hottest on Earth, and the local climates would generally get
cooler as you moved away from the equator and toward the poles.


5 0
4 years ago
A 75 kg football player is gliding forward across very smooth ice at 4.6 m/s. He throws a 0.47 kg football straight forward. A)
lord [1]

Answer:

4.53482 m/s

4.506 m/s

Explanation:

m_1 = Mass of player = 75 kg

v_1 = Initial velocity of player = 4.6 m/s

m_2 = Mass of ball = 0.47 kg

v_1 = Initial velocity of ball = 15 m/s

The linear momentum of the system is conserved

(m_1+m_2)v_1=m_1v+m_2v_2\\\Rightarrow v=\dfrac{(m_1+m_2)v_1-m_2v_2}{m_1}\\\Rightarrow v=\dfrac{(75+0.47)4.6-0.47\times 15}{75}\\\Rightarrow v=4.53482\ m/s

The player's speed is 4.53482 m/s

In the second case the equation of momentum is

(m_1+m_2)v_1=m_1v+m_2(v_2+v_1)\\\Rightarrow v=\dfrac{(m_1+m_2)v_1-m_2(v_2+v_1)}{m_1}\\\Rightarrow v=\dfrac{(75+0.47)4.6-0.47\times (15+4.6)}{75}\\\Rightarrow v=4.506\ m/s

The player's speed is 4.506 m/s

4 0
4 years ago
A ___________ is a collection of dust and gas held together by gravity.
Genrish500 [490]
Galaxy-a collection of dust, gas, AND STARS. But I'm pretty positive this should be the answer.
4 0
3 years ago
Read 2 more answers
How do you do this problem?
kvasek [131]

Explanation:

First, find the velocity of the projectile needed to reach a height h when fired straight up.

Given:

Δy = h

v = 0

a = -g

Find: v₀

v² = v₀² + 2aΔy

(0)² = v₀² + 2(-g)(h)

v₀ = √(2gh)

Now find the height reached if the projectile is launched at a 45° angle.

Given:

v₀ = √(2gh) sin 45° = √(2gh) / √2 = √(gh)

v = 0

a = -g

Find: Δy

v² = v₀² + 2aΔy

(0)² = √(gh)² + 2(-g)Δy

2gΔy = gh

Δy = h/2

5 0
3 years ago
5. A little boy tries to push a heavy cupboard. He uses a force of 400 N, but he is unable to move
ad-work [718]

Answer:

NO HAY TRABAJO REALIZADO YA QUE EL NIÑO NO PUDO MOVER

EL ARMARIO, PARA QUE HAYA TRABAJO DEBE HABER DESPLAZAMIENTO.

Explanation:

8 0
3 years ago
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