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LUCKY_DIMON [66]
3 years ago
11

A basketball player pushes down with a force of 34 N on a basketball that is inflated

Physics
1 answer:
zloy xaker [14]3 years ago
8 0

Answer:

2.2 cm

Explanation:

Pressure = force / area

P = F / A

A = F / P

A = 34 N / (9×10⁴ Pa)

A = 0.000378 m²

Area of a circle is:

A = πr²

0.000378 m² = πr²

r = 0.011 m

Diameter is double the radius:

d = 2r

d = 0.022 m

d = 2.2 cm

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Answer:

Six meters refers to the distance the baseball travels in one second.

Explanation:

6 m/s means that every second, the baseball travels 6m. So, 6 meters is the distance traveled in one second.

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A person is standing on a scale in an unmoving elevator. The elevator starts to move upwards at 1 m/s squared. Is the scale read
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Answer: GREATER

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HENCE

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F_grav = G * m1 * m2 / r^2

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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

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