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Eduardwww [97]
2 years ago
15

13–82. the 8-kg sack slides down the smooth ramp. if it has a speed of 1.5 m> s when y = 0.2 m, determine the normal reaction

the ramp exerts on the sack and the rate of increase in the speed of sack at this instant.
y
*13–84. the 2-lb block is released from rest at a and slides down along the smooth cylindrical surface. if the attached spring has a stiffness k = 2 lb> ft, determine its unstretched length so that it does not allow the block to leave the surface until u = 60°.
a
prob. 13–82
prob. 13–84
2,fuuu 2,frur
Physics
1 answer:
marissa [1.9K]2 years ago
4 0
The second problem requires a figure to be answered. For the first problem
The acceleration of the sack is
1.5² - 0² = 2a(0.2)
a = 5.63 m/s2
The reaction of the ramp is
F = 8 kg (5.63 m/s2)
F = 45 N
Differentiate the kinematic equation involving time to get the rate of increase of the velocity.
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A radioactive substance decays exponentially. A scientist begins with 200 milligrams of a radioactive substance. After 17 hours,
lianna [129]

75.17 mg of the radioactive substance will remain after 24 hours.

Answer:

Explanation:

Any radioactive substance will obey the exponential decay behavior. So according to this behavior, any radioactive substance will be decaying in terms of exponential form of disintegration constant and Time.

Disintegration constant is the rate of decay of radioactive elements. It can be measured using the half life time of the radioactive element .While half life time is the time taken by any radioactive element to decay half of its concentration. Like in this case, at first the scientist took 200 mg then after 17 hours, it got reduced to 100 g. So the half life time of this element is 17 hours.

Then Disintegration constant = 0.6932/Half Life time

Disintegration constant = 0.6932/17=0.041

Then as per the law of disintegration constant:

N = N_{0}e^{-xt}

Here N is the amount of radioactive element remaining at time t and N_{0} is the initial amount of sample, x is the disintegration constant.

So here, N_{0} = 200 mg, x = 0.041 and t = 24 hrs.

N = 200 ×e^{-24*0.041} =75.17 mg.

So 75.17 mg of the radioactive substance will remain after 24 hours.

3 0
1 year ago
Global Precipitation Measurement (GPM) is a tool scientists use to forecast weather. Which statements describe GPM? Select three
lyudmila [28]

Answer:

B.It is a satellite that collects data about rain and snow

C.Its orbit covers 90 percent of Earth’s surface

F.The sensors measure microwaves

5 0
2 years ago
The drawing shows a person (weight W = 588 N, L1 = 0.838 m, L2 = 0.398 m) doing push-ups. Find the normal force exerted by the f
zhenek [66]

Complete Question

The complete question is shown on the first uploaded image

Answer:

Force on each hand is 196.22 N

Force on each foot is 95.8 N

Explanation:

In order to get a better understanding of this question let us explain some concepts

Normal Force:

We can define normal force Fn as that type of force which makes a 90 degree angle with the surface on which it is exerted.

Torque:

We can define torque as the moment of forces that tends to produce or cause rotation

From the question we are given that

Weight of body is (W) = 584 N

The normal force on both hands (Ha) = ?

The normal force on both legs (Lg) = ?

Looking at the diagram the person is at equilibrium so

                 584 = Ha + Lg

an also this mean that torques acting on the body is balanced

         So,   0.410 Ha  = 0.840 Lg

    Making Lg the subject of formula in the equation above we

   Lg = 0.4881 Ha

 Considering the first equation and replacing Lg with this recent equation we have

                      584 = Ha + 0.4881 Ha

          Therefore Ha = 392.44 N

This value obtained is  for both hands for each hand we divide by 2

Therefore we have for each hand = 392.44/2 =196.55 N

Since we have been able to get the force on both hands we can substitute it in to the equation where we made Lg the subject of formula and we have

             Lg = 0.4881 ×  392.44

                  = 191.22 N

The value above is the force on both legs to obtain the force on each leg we have

                  191.22/2 = 95.8 N.

8 0
2 years ago
A certain signal molecule S in heart tissue is degraded by two different biochemical pathways: when only Path 1 is active, the h
Misha Larkins [42]

Answer:

Half life of S = 3.76secs

Explanation:

The concept of half life in radioactivity is applied. Half life is the time taken for a radioactive material to decay to half of its initial size.

For part 1 - How much signal will be degraded in 1secs = 1/3.9 = 0.2564

for part 2 - How much signal will be degraded in 1secs = 1/104 = 0.009615

Simply say = 1/3.9 + 1/104 = 0.266015

So both part 1 and part 2 took 1/0.266015 = 3.76secs is the half life of S when both pathways are active

6 0
2 years ago
The position of a particle moving along the x axis may be determined from the expression x(t) = btu + ctv, where x will be in me
KIM [24]

As per given equation we have

x = bt^u + ct^v

now as per the dimensional analysis we can say that dimension of right side of equation must be equal to left side of the equation

now as per left side of equation its dimension is same as length or meter

now we can say it should be meter on right side also

bt^u = M^0L^1T^0

b*T^8 = M^0L^1T^0

b = M^0L^1T^{-8}

similarly for other term we have

ct^v = M^0L^1T^0

c*T^7 = M^0L^1T^0

c = M^0L^1T^{-7}

<em>so above are the dimensions of b and c</em>

8 0
1 year ago
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