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Len [333]
3 years ago
9

You are an industrial engineer with a shipping company. As part of the package- handling system, a small box with mass 1.60 kg i

s placed against a light spring that is compressed 0.280 m. The spring has force constant 48.0 N/m . The spring and box are released from rest, and the box travels along a horizontal surface for which the coefficient of kinetic friction with the box is μk = 0.300. When the box has traveled 0.280 m and the spring has reached its equilibrium length, the box loses contact with the spring. (a) What is the speed of the box at the instant when it leaves the spring? (b) What is the maximum speed of the box during its motion?
Physics
1 answer:
wlad13 [49]3 years ago
6 0

Answer:

a) V = 0.82m/s

b) Vmax = 0.985 m/s

Explanation:

By conservation of energy we know that:

Eo = Ef    \frac{m*V^{2}}{2}-\frac{K*Xmax^{2}}{2}=-Ff*Xmax

Solving for V we get:

V = 0.82 m/s

To find the maximum speed we will do the same to an intermediate point where the compression is X and the distance for the work donde by frictions is given by (Xmax - X) = (0.28m - X):

\frac{m*Vmax^{2}}{2}+\frac{K*X^{2}}{2}-\frac{K*Xmax^{2}}{2}=-Ff*(Xmax-X)

Then we have to solve for V, derive and equal zero in order to find position X. After solving the derivative we get:

X = 0.1m    Replacing this value into the equation for Vmax:

Vmax = 0.985m/s

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As shown in the figure below, a bullet is fired at and passes through a piece of target paper suspended by a massless string. Th
NikAS [45]

Answer:

M = 0.730*m

V = 0.663*v

Explanation:

Data Given:

v_{bullet, initial} = v\\v_{bullet, final} = 0.516*v\\v_{paper, initial} = 0\\v_{paper, final} = V\\mass_{bullet} = m\\mass_{paper} = M\\Loss Ek = 0.413 Ek

Conservation of Momentum:

P_{initial} = P_{final}\\m*v_{i} = m*0.516v_{i} + M*V\\0.484m*v_{i} = M*V .... Eq1

Energy Balance:

\frac{1}{2}*m*v^2_{i} = \frac{1}{2}*m*(0.516v_{i})^2 + \frac{1}{2}*M*V^2 + 0.413*\frac{1}{2}*m*v^2_{i}\\\\0.320744*m*v^2_{i} = M*V^2\\\\M = \frac{0.320744*m*v^2_{i} }{V^2}  ....... Eq 2

Substitute Eq 2 into Eq 1

0.484*m*v_{i} = \frac{0.320744*m*v^2_{i} }{V^2} *V  \\0.484 = 0.320744*\frac{v_{i} }{V} \\\\V = 0.663*v_{i}

Using Eq 1

0.484m*v_{i} = M* 0.663v_{i}\\\\M = 0.730*m

7 0
3 years ago
Is there a link between the number of bulbs and current drawn from the power pack?
LuckyWell [14K]

Answer:

Yes there is if number of bulb is high the bulbs wouldn't glow much brighter

Explanation:

5 0
3 years ago
An athlete swings a ball, connected to the end of a chain, in a horizontal circle. The athlete is able to rotate the ball at the
Andrei [34K]

Answer:

a) 2nd case rate of rotation gives the greater speed for the ball

b) 1534.98 m/s^2

c) 1515.04 m/s^2

Explanation:

(a) v = ωR

when R = 0.60, ω = 8.05×2π

v = 0.60×8.05×2π = 30.34 m/s

Now in 2nd case

when R = 0.90, ω = 6.53×2π

v = 0.90×6.53×2π = 36.92 m/s

6.35 rev/s gives greater speed for the ball.

(b) a = ω^2 R = (8.05×2π)^2 )(0.60) = 1534.98 m/s^2

(c) a = ω^2 R = (6.53×2π)^2 )(0.90) = 1515.05 m/s^2

7 0
3 years ago
If a baseball has a zero velocity at some instant, is the acceleration of the baseball necessarily zero at that time? Explain -
ipn [44]

Answer:

No, not necessarily

Explanation:

If an object is moving with an acceleration that causes its speed to be reduced, there will be a moment in which it reaches v = 0, but this doesn't necessarily mean that the acceleration isn't acting anymore. If the object continues its movement with the same acceleration, it's velocity will become negative.

An example of an object that has zero velocity but non-zero acceleration:

If you throw an object in the air with a certain velocity, it will move vertically, reducing its velocity in a 9,8 m/s^{2} rate (which is the acceleration caused by gravity). At a certain point, the object will reach its maximum height, and will start to fall. In the exact moment that it reaches the maximum height, before it starts falling, its velocity is zero, but gravity is still acting on the object (this is the reason why it starts falling instead of just being stopped at that point). Therefore, at that point, the object has zero velocity but an acceleration of 9,8 m/s^{2}.

3 0
3 years ago
LOTS OF POINTSA rocket of mass 40 000 kg takes off and flies to a height of 2.5 km as its engines produce 500 000 N of thrust.
Svetradugi [14.3K]

Answer:

i) E = 269 [MJ]    ii)v = 116 [m/s]

Explanation:

This is a problem that encompasses the work and principle of energy conservation.

In this way, we establish the equation for the principle of conservation and energy.

i)

E_{k1}+W_{1-2}=E_{k2}\\where:\\E_{k1}= kinetic energy at moment 1\\W_{1-2}= work between moments 1 and 2.\\E_{k2}= kinetic energy at moment 2.

W_{1-2}= (F*d) - (m*g*h)\\W_{1-2}=(500000*2.5*10^3)-(40000*9.81*2.5*10^3)\\W_{1-2}= 269*10^6[J] or 269 [MJ]

At that point the speed 1 is equal to zero, since the maximum height achieved was 2.5 [km]. So this calculated work corresponds to the energy of the rocket.

Er = 269*10^6[J]

ii ) With the energy calculated at the previous point, we can calculate the speed developed.

E_{k2}=0.5*m*v^2\\269*10^6=0.5*40000*v^2\\v=\sqrt{\frac{269*10^6}{0.5*40000} }\\ v=116[m/s]

8 0
2 years ago
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