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Ierofanga [76]
3 years ago
12

The horizontal distance that each thrown object covers increases as

Physics
1 answer:
GarryVolchara [31]3 years ago
6 0

Answer:

A

Explanation:

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orbit the sun have very elliptical orbits

are made of frozen gas and ice

Explanation:

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Comets move around the sun in an elliptical orbits and follow Kepler's Laws saying that they move faster when closer to the Sun.

Comets have tails even when they are far from the sun and the tails that point away from the sun.

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Pani-rosa [81]

Answer:

Vacculoes, vesicles

Explanation:

Organelles and other structures within the cells have vesicles collectively referred to as the endo-membrane system. The Lysosomes pack up the substances, and the vesicles go through the porsous membrame and secrete the substances.

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What is measurement ?...​
mariarad [96]

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The gravitational self potential energy of a solid ball of mass density ρ and radius R is E. What is the gravitational self pote
Alenkasestr [34]
It will be
E = mgh.
where h and g are constant thus
m can be written as 4/3πr^3*density
E = 4/3πr^3* density
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8 0
3 years ago
A trick shot archer shoots an arrow with a velocity of 30.0 m/s at an angle of 20.0 degrees with respect to the horizontal. An a
svlad2 [7]

Answer:

u'=10.259\ m.s^{-1} is the initial velocity of tossing the apple.

the apple should be tossed after \Delta t=0.0173\ s

Explanation:

Given:

  • velocity of arrow in projectile, v=30\ m.s^{-1}
  • angle of projectile from the horizontal, \theta=20^{\circ}
  • distance of the point of tossing up of an apple, d=30\ m

<u>Now the horizontal component of velocity:</u>

v_x=v\ cos\ \theta

v_x=30\times cos\ 20^{\circ}

v_x=28.191\ m.s^{-1}

<u>The vertical component of the velocity:</u>

v_y=v.sin\ \theta

v_y=30\times sin\ 20^{\circ}

v_y=10.261\ m.s^{-1}

<u>Time taken by the projectile to travel the distance of 30 m:</u>

t=\frac{d}{v_x}

t=\frac{30}{28.191}

t=1.0642\ s

<u>Vertical position of the projectile at this time:</u>

h=v_y.t-\frac{1}{2}g.t^2

h=10.261\times 1.0642-\frac{1}{2} \times 9.8\times 1.0642^2

h=5.3701\ m

<u>Now this height should be the maximum height of the tossed apple where its velocity becomes zero.</u>

v'^2=u'^2-2g.h

0^2=u'^2-2\times 9.8\times 5.3701

u'=10.259\ m.s^{-1} is the initial velocity of tossing the apple.

<u>Time taken to reach this height:</u>

v'=u'-g.t'

0=10.259-9.8\times t'

t'=1.0469\ s

<u>We observe that </u>t>t'<u> hence the time after the launch of the projectile after which the apple should be tossed is:</u>

\Delta t=t-t'

\Delta t=1.0642-1.0469

\Delta t=0.0173\ s

8 0
3 years ago
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