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insens350 [35]
3 years ago
6

Pages 64 to 65 of your reading material trace an extended problem that involes lifting a bag of sugar up to a shelf at first it

is lifted to one shelf then it is lifted again to a second higher shelf.based on the numbers that are obtained from the calculations, which of the following is a correct conclusion?
a) the job gets easier the higher you lift the object.
b) the weight changes as the object is lifted
c) it is less work to lift the object all the way to it's final destination in one step.
d) the amount of work is the same whether the bag is moved all at once or in two stages, provided the total height lifted is the same in either case.
Physics
1 answer:
nadezda [96]3 years ago
7 0

Answer:

d) the amount of work is the same whether the bag is moved all at once or in two stages, provided the total height lifted is the same in either case.

Explanation:

While moving the bag to the shelf in one shot we can say that the total work done is given as

W = mg(2H)

here we know that

2H = total height raised by the bag

now when we raise the bag to first shelf and then move it to next shelf

then we will have

W = W_1 + W_2[tex][tex]W = mgH + mgH

W = 2mgH

so the correct answer will be

d) the amount of work is the same whether the bag is moved all at once or in two stages, provided the total height lifted is the same in either case.

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Assume that the radius ????r of a sphere is expanding at a rate of 70 cm/min.70 cm/min. The volume of a sphere is ????=43???????
rodikova [14]

Answer:

the rate of change in volume with time is 280πr² cm³/min

Explanation:

Data provided in the question:

Radius of the sphere as 'r'

\frac{d\textup{r}}{\textup{dt}}  = 70 cm/min

Volume of the sphere, V = \frac{\textup{4}}{\textup{3}}\pi r^3

Surface area of the sphere as 4πr²

Now,

Rate of change in volume with time, \frac{d\textup{V}}{\textup{dt}}

 = \frac{d(\frac{\textup{4}}{\textup{3}}\pi r^3)}{dt}

= 3\times\frac{\textup{4}}{\textup{3}}\pi r^2}\times\frac{dr}{dt}

Substituting the value of \frac{dr}{dt}

= 3\times\frac{\textup{4}}{\textup{3}}\pi r^2}\times70

= 280πr² cm³/min

Hence, the rate of change in volume with time is 280πr² cm³/min

4 0
3 years ago
A dolphin in an aquatic show jumps straight up out of the water at a velocity of 13.0 m/s.
SVEN [57.7K]

Answer:

a) Knowns, initial speed v_{i}=13.0 m/s, final speed v_{f}=0 m/s and gravity due it is a constant g=9.8m/s^{2}

b) The maximum high reached by the dolphin is y_{max}=8.62 m

c) Total time is t=2.65s

Explanation:

a) First of all the initial speed is given at the start of the problem, gravity is constant and final speed is known any object thrown straight up reaches its max high at 0m/s speed.

b) Second, now that we know final speed we use v_{f} =v_{i}-gt, as we clear for t=\frac{v_{i}-v_{f} }{g}=\frac{20.0m/s}{9.8m/s^{2} }=1.32 s.

Then we use y=v_{i}t-\frac{1}{2} gt^{2}=(20.0m/s)(1.32s)-\frac{1}{2} (9.8m/s^{2} ) (1.32s)^{2}  =8.62m

c)Third, finally we can use y=v_{i}t-\frac{1}{2} gt^{2}, as we know y=0m when the dolphin fall into the water again and v_{i} =13.0m/s, then we have 0=(13m/s)t-\frac{1}{2} (9.8m/s^{2}  )t^{2} is a quadratic form 0=t(13.0-4.9t) so we have t_{i}=0s and t_{f}=\frac{13}{4.90}  =2.65s

6 0
3 years ago
Suppose that during any period of ¼ second there is one instant at which the crests or troughs of component waves are exactly in
kolezko [41]
<h3><u>Answer;</u></h3>

A. 4

<h3><u>Explanation;</u></h3>
  • <em><u>The period of a wave or periodic time is the time taken for a complete oscillation to occur. </u></em>For example its is the time taken between two successive crests or troughs.
  • <em><u>The beats or oscillation that occur in one second represents the frequency. Frequency is the number of complete oscillations or beats in one second in a wave.</u></em>
  • Frequency, measured in Hertz is given by the reciprocal of the periodic time.
  • Thus; <u><em>Frequency or beats per second = 1/(1/4) = 4</em></u>
  • <u><em>Hence , 4 beats per second</em></u>

6 0
3 years ago
A space vehicle is traveling at 5425 km/h relative to the Earth when the exhausted rocket motor (mass 4m) is disengaged and sent
lana66690 [7]

Answer:

V_{cE}=1489m/s

Explanation:

Given data

Space vehicle speed=5425 km/h relative to earth

The rocket motor speed=81 km/h  and mass 4m

The command has mass m

From the conservation of momentum as the system isolated

p_{i}=p_{f}\\

Since the motion in on direction we can drop the unit vector direction

MV_{i}=4mV_{mE}+mV_{CE}

Where M is the mass of space vehicle which equals to sum of the motors mass and command mass.

The velocity of the motor relative to the earth equals the velocity of the motor relative to command plus the velocity of the command relative to earth

V_{mE}=V_{mc}+V_{cE}

Where Vmc is the velocity of motor relative to command

This yields

5mV_{i}=4m(V_{mc}+V_{cE})+mV_{cE}\\5V_{i}=4V_{mc}+5V_{cE}

Substitute the given values

V_{cE}=\frac{5V_{i}-4V_{mc}}{5}\\ V_{cE}=5425*\frac{1000}{60*60}(m/s)-\frac{4}{5}*81\frac{1000}{60*60}(m/s)\\  V_{cE}=1489m/s

5 0
3 years ago
Which of the following is true regarding transverse waves?
DIA [1.3K]

Answer:

in a transverse wave, The directions that energy and matter travel in are perpendicular to one another.

these waves are types of mechanical waves that doesn't require any material medium to transfer.

4 0
3 years ago
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