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igomit [66]
3 years ago
10

How do I solve using the formula: Vf^2=vo^2+2gh ^2 means square

Physics
1 answer:
PtichkaEL [24]3 years ago
5 0

your second line ... minus 2gh ...

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A racquet ball with mass m = 0.256 kg is moving toward the wall at v = 11.8 m/s and at an angle of θ = 29° with respect to the h
icang [17]

Answer:

Part a)

P = 5.72 kg m/s

Part b)

\Delta P = 2.93 kg m/s

Part c)

F = 44.4 N

Part d)

\Delta P = 5.02 kg m/s

Part e)

\Delta t = 0.113 s

Part f)

\Delta K = 0

Explanation:

As we know that initial velocity of the ball is given as

v = 11.8 cos29 \hat i + 11.8 sin29 \hat j

v_i = 10.3 \hat i + 5.72 \hat j

Now final velocity of the system is given as

v_f = 10.3\hat i - 5.72\hat j

Part a)

now magnitude of initial momentum is given as

P = mv

P = 0.256(11.8)

P = 5.72 kg m/s

Part b)

Change in momentum is given as

\Delta P = m(v_f - v_i)

\Delta P = 0.256(5.72 + 5.72)

\Delta P = 2.93 kg m/s

Part c)

As we know that average force is defined as the rate of change in momentum

so here we have

F = \frac{\Delta P}{\Delta t}

F = \frac{2.93}{0.066}

F = 44.4 N

Part d)

Magnitude of change in momentum is given as

\Delta P = m(v_f - v_i)

\Delta P = 0.256(7.8 + 11.8)

\Delta P = 5.02 kg m/s

Part e)

As we know that in 2nd case the force is same as the initial force

so we will have

\frac{\Delta P}{\Delta t} = F

\frac{5.02}{\Delta t} = 44.4

\Delta t = 0.113 s

Part f)

Since this is elastic collision so change in kinetic energy must be ZERO

\Delta K = 0

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3 years ago
What is the main function of chloroplasts?
dangina [55]

Answer:

C

To convert sunlight into usable sugars

(

Explanation:

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Neglecting air resistance, if a ball is thrown 4.5 m/s horizontally from a 94-m cliff, how far has the ball fallen after 2.7 s?
kati45 [8]
Check the attached file for the answer.

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3 years ago
A simple motor converts
erastovalidia [21]

Answer:

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Electric motors may be classified by considerations such as power source type, internal construction, application and type of motion output. In addition to AC versus DC types, motors may be brushed or brushless, may be of various phase (see single-phase, two-phase, or three-phase), and may be either air-cooled or liquid-cooled. General-purpose motors with standard dimensions and characteristics provide convenient mechanical power for industrial use. The largest electric motors are used for ship propulsion, pipeline compression and pumped-storage applications with ratings reaching 100 megawatts. Electric motors are found in industrial fans, blowers and pumps, machine tools, household appliances, power tools and disk drives. Small motors may be found in electric watches.

In certain applications, such as in regenerative braking with traction motors, electric motors can be used in reverse as generators to recover energy that might otherwise be lost as heat and friction.

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5 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
4 years ago
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