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denpristay [2]
2 years ago
12

You catch a volleyball (mass 0.270 kg) that is moving downward at 7.50 m/s. In stopping the ball, your hands and the volleyball

descend together a distance of 0.150 m. How much work do your hands do on the volleyball in the process of stopping it?
Physics
1 answer:
Svetllana [295]2 years ago
7 0

Answer:

Explanation:

deceleration a

v² = u² - 2as

0 = 7.5² - 2 a x .15

a = 187.5 m / s²

If R be the force exerted by hand

R- mg = ma

R = mg + ma

= .27 ( 9.8 + 187.5 )

= 53.27 N

Work done by hand

= force x displacement

= 53.27 x - .15

= -  8 J  

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elena-14-01-66 [18.8K]

Answer:

Explanation:

Given the velocity of a particle modeled by the equation

v = 13-3t+8t² where t is in seconds

Given t = 0 and position s = +5m

A) To get the position as a function of time, we will integrate the function with respect to t ad shown;

v = 13-3t+8t²

S = ∫13-3t+8t² dt

S = 13t-13t²/2+8t³/3 + C

at t = 0 and S = +5m

5 = 13(0)-13(0)²/2+8(0)³/3+C

5 = 0-0+0+C

C = 5

Substituting c = 5 into the displacement function

S = 13t-13t²/2+8t³/3 + C

S = 13t-13t²/2+8t³/3 + 5

B) acceleration is the change in velocity with respect to time.

a = dv/dt

Given v = 13-3t+8t²

a= dv/dt = -3+16t

a = 16-3t

C) acceleration at t = 6s is derived by plugging in t = 6 into the resulting equations in (B)

a = 16-3t

a = 16-3(6)

a = 16-18

a = -2m/s²

D) net displacement from t = 0 to t = 6s

At t = 0:

S(0) = 13(0)-13(0)²/2+8(0)³/3 + 5

S(0) = 0+5

S(0) = 5m

At t = 6s

S(6) = 13(6)-13(6)²/2+8(6)³/3 + 5

S(6) = 78-234+576+5

S(6) = 425m

Net displacement from t = 0s to t = 6s is s(6)-s(0)

= 425-5

= 420m

E) Total distance travelled D = S(6)+S(0)

= 425+5

= 430m

F) Average velocity = ∆S/∆t

Average velocity = S(6)-S(0)/6-0

Average velocity = 425-5/6

Average velocity = 420/6

Average velocity = 70m/s

4 0
2 years ago
Have you ever chewed on a wintergreen mint in front of a mirror in the dark? If you have, you may have noticed some sparks of li
lutik1710 [3]

Answer:

Part a)

E = 3.66 eV

Part b)

\lambda = 508.5 nm

Explanation:

Part a)

change in the energy due to decay of photon is given as

E = h\nu

here we know that

\nu = 8.88 \times 10^{14} Hz

now we have

E = (6.6 \times 10^{-34})(8.88 \times 10^{14})

E = 5.86 \times 10^{-19} J

E = 3.66 eV

Part b)

While electron return to its ground state it will emit a photon of energy 2/3rd of the total energy

so we have

\Delta E = \frac{2}{3}(3.66 eV)

\Delta E = 2.44 eV

now to find the wavelength we have

\Delta E = \frac{hc}{\lambda}

2.44 = \frac{1242}{\lambda}

\lambda = 508.5 nm

3 0
2 years ago
Imagine that you replace the block in the video with a happy or sad ball identical to the one used as a pendulum, so that the sa
gizmo_the_mogwai [7]

Answer: a) The Answer to the question is option a) None of it.

Explanation:

The reason is because according to the law of conservation of energy Energy can neither be created nor destroyed but can be transformed from one form to the other. Therefore none of the kinetic energy was dissipated, rather it was transformed to another form of energy.

6 0
2 years ago
A cylinder rotating about its axis with a constant angular acceleration of 1.6 rad/s2 starts from rest at t = 0. At the instant
OverLord2011 [107]

Answer:

The magnitude of the total linear acceleration is 0.27 m/s²

b. 0.27 m/s²

Explanation:

The total linear acceleration is the vector sum of the tangential acceleration and radial acceleration.

The radial acceleration is given by;

a_t = ar

where;

a is the angular acceleration and

r is the radius of the circular path

a_t = ar\\\\a_t = 1.6 *0.13\\\\a_t = 0.208 \ m/s^2

Determine time of the rotation;

\theta = \frac{1}{2} at^2\\\\0.4 = \frac{1}{2} (1.6)t^2\\\\t^2 = 0.5\\\\t = \sqrt{0.5} \\\\t = 0.707 \ s\\\\

Determine angular velocity

ω = at

ω = 1.6 x 0.707

ω = 1.131 rad/s

Now, determine the radial acceleration

a_r = \omega ^2r\\\\a_r = 1.131^2 (0.13)\\\\a_r = 0.166 \ m/s^2

The magnitude of total linear acceleration is given by;

a = \sqrt{a_t^2 + a_r^2} \\\\a = \sqrt{0.208^2 + 0.166^2} \\\\a = 0.266 \ m/s^2\\\\a = 0.27  \ m/s^2

Therefore, the magnitude of the total linear acceleration is 0.27 m/s²

b. 0.27 m/s²

5 0
2 years ago
A 2.5 g marshmallow is placed in one end of a 40 cm pipe, as shown in the figure above. A person blows into the left end of the
frosja888 [35]

Answer:

Your answer would be

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I belive this is the answer sorry if im wrong!

3 0
2 years ago
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