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aivan3 [116]
3 years ago
9

A 2-kg bowling ball sits on top of a building that is 40 meters tall.

Physics
2 answers:
Vesna [10]3 years ago
7 0
A 2-kg bowling ball sits on top of a building that is 40 meters tall possess gravitational potential energy. It <span> is </span>energy<span> an object possesses because of its position in a </span>gravitational <span>field. We calculate as follows:

GPE = mgh = 2(9.8)(40) = 784 J</span>
Dahasolnce [82]3 years ago
6 0
The bowling ball is at rest, so it only has gravitational potential energy.

Ug = mgy
Ug = (2)(9.8)(40) = 784 J

Need any more help?
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Clay Matthews, a linebacker for the Green Bay Packers, can reach a speed of 10.0 m/s. At the start of a play, Matthews runs down
MA_775_DIABLO [31]

Answer:

a)   D_ total = 18.54 m,   b)        v = 6.55 m / s

Explanation:

In this exercise we must find the displacement of the player.

a) Let's start with the initial displacement, d = 8 m at a 45º angle, use trigonometry to find the components

           sin 45 = y₁ / d

           cos 45 = x₁ / d

           y₁ = d sin 45

           x₁ = d sin 45

           y₁ = 8 sin 45 = 5,657 m

           x₁ = 8 cos 45 = 5,657 m

The second offset is d₂ = 12m at 90 of the 50 yard

            y₂ = 12 m

            x₂ = 0

total displacement

          y_total = y₁ + y₂

          y_total = 5,657 + 12

          y_total = 17,657 m

          x_total = x₁ + x₂

          x_total = 5,657 + 0

          x_total = 5,657 m

          D_total =   17.657 i^+ 5.657 j^  m

          D_total = Ra (17.657 2 + 5.657 2)

          D_ total = 18.54 m

b) the average speed is requested, which is the offset carried out in the time used

           v = Δx /Δt

the distance traveled using the pythagorean theorem is

         r = √ (d1² + d2²)

          r = √ (8² + 12²)

          r = 14.42 m

The time used for this shredding is

         t = t1 + t2

         t = 1 + 1.2

         t = 2.2 s

let's calculate the average speed

         v = 14.42 / 2.2

         v = 6.55 m / s

8 0
3 years ago
A child throws a baseball upward with an initial velocity of 20 m/s. The child wants to throw the baseball at least as high as t
Umnica [9.8K]
When the ball starts its motion from the ground, its potential energy is zero, so all its mechanical energy is kinetic energy of the motion:
E= \frac{1}{2}mv^2
where m is the ball's mass and v its initial velocity, 20 m/s.

When the ball reaches its maximum height, h, its velocity is zero, so its mechanical energy is just gravitational potential energy:
E=mgh

for the law of conservation of energy, the initial mechanical energy must be equal to the final mechanical energy, so we have
\frac{1}{2}mv^2 = mgh
From which we find the maximum height of the ball:
h= \frac{v^2}{2g}= \frac{(20 m/s)^2}{2 \cdot 9.81 m/s^2}=20.4 m

Therefore, the answer is yes, the ball will reach the top of the tree.

5 0
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Sloan [31]

Answer:

5 newtons

Explanation:

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tigry1 [53]

Answer:

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Explanation:

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Pachacha [2.7K]

Answer:

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n= m/M = 10/ 4 = 2.5

R=0.08206 L.atm /mol/k

760mmHg = 1 atm therefore

600mmHg = X atm

760 X = 600mmHg

X = 600/760 = 0.789 atm

P = 0.789 atm

V= ?

PV= nRT

0.789 V = 2.5 × 0.08206 × 298

V= 2.5 × 0.08206 ×298 / 0.789

V= 77.48 L

I hope I helped you ^_^

8 0
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