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3 years ago

11

What is the acceleration of this object? The object's mass is 2 kg. Small box with dot in the middle. Arrows originate at dot. A

rrow up and Arrow down are same length. Arrow up labeled F sub norm equals 20 N and Arrow down labeled F sub grav equals 20 N. Arrow to the right is longer than arrow to the left. Arrow to the right is labeled F sub app equals 22 N and arrow to the left is labeled F sub frict equals 14 N. 11 m/s2 0 m/s2 4 m/s2 7 m/s2

2 answers:

Murrr4er [49]3 years ago

8 0

Answer:

4 m/s2

Explanation:

I took the test

weeeeeb [17]3 years ago

8 0

Answer:

4m/s2

Explanation:

i took the test

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Which phrase describes the thermal energy of a sample of matter?

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where are the phrase's?

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What is the name of the force that is a combination of all forces that act on an object?

jarptica [38.1K]

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If the combination of all the forces acting on an object sums upto be zero, then the net force is known as <u>balanced force.</u>

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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>

$\Delta t=t-t'$

$\Delta t=1.0642-1.0469$

$\Delta t=0.0173\ s$

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Answer:

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4 years ago

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Zina [86]

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4 years ago