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-Dominant- [34]

3 years ago

5

If a client's knees hurt after a one-week walking regimen, which one of the following suggestions is most appropriate?

1 answer:

Flura [38]3 years ago

5 0

A, it's better to take a break to let your body rejuvenate.

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The highest element in the hierarchical breakdown of the wbs is

Makovka662 [10]

Work package. Hope this helps!

5 0

3 years ago

Which of the following has the greatest kinetic energy? A. a mass of 4m at velocity v. B. a mass of 3m at velocity 2v. C. a mass

Artist 52 [7]

Answer:

Option C

Explanation:

Kinetic energy is the energy that the body possesses by virtue of its motion.

The formula for Kinetic energy is given by $\frac{1}{2} mv^2$

Using this formula let us find kinetic energy for the bodies given and find out which is the greatest

A) KE = $\frac{1}{2} (4m)(v^2) = 2mv^2$

B) KE = $\frac{1}{2} (3m)(2v)^2 = 6mv^2$

C) KE = $\frac{1}{2} (2m)(3v)^2 = 9mv^2$

D) KE = $\frac{1}{2} (3)(4v)^2 = 8mv^2$

Comparing these we find that 9mv^2 is the highest.

Hence option C is the answer.

4 0

3 years ago

Scientific way of thinking

noname [10]

Answer:

huh,? can you explain the question more please

7 0

3 years ago

Read 2 more answers

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

Julie throws a ball to her friend Sarah. The ball leaves Julie's hand a distance 1.5 meters above the ground with an initial spe

iris [78.8K]

Answer:

Explanation:

1. $V_{x}$ = $V_{0}$ * cos $\alpha$ ⇒ 16*cos32 ≈ 13.6 m/s (13.56)

2. $V_{y}$ = $V_{0}$ * sin $\alpha$ ⇒ 16* sin32 ≈ 9.4 m/s

3. $y_{max}$ = $\frac{v_{0}^2*sin^2\alpha}{2g}$ = $\frac{16^2*sin^232}{2*9.8}$ (the g (gravity) depends on the country but i'll take the average g which is 9.2m/s^2)

$y_{max}$ ≈ 3.6677+1.5 ≈ 5.2m

4. $x_{max}$ = $\frac{v_{0}^2*sin(2\alpha)}{g}$ = $\frac{16^2*sin(2*32)}{9.8}$ ≈ 23.5m (23.47)

5. -

answer 4 could be wrong, not certain about that one and i don't know 5

3 0

3 years ago