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dimaraw [331]
2 years ago
12

A 25 kg box of textbooks rests on a loading ramp that makes an angle α with the horizontal. The coefficient of kinetic friction

is 0.25, and the coefficient of static friction is 0.35. As the angle is increased, find the minimum angle at which the box starts to slip.
24.5°


18.5°


21.3°


19.3°
Physics
1 answer:
Lunna [17]2 years ago
8 0

Up until the moment the box starts to slip, the static friction is maximized with magnitude <em>f</em>, so that by Newton's second law,

• the net force acting on the box parallel to the ramp is

∑ <em>F</em> = <em>mg</em> sin(<em>α</em>) - <em>f</em> = 0

where <em>mg</em> sin(<em>α</em>) is the magnitude of the parallel component of the box's weight; and

• the net force acting perpendicular to the ramp is

∑ <em>F</em> = <em>n</em> - <em>mg</em> cos(<em>α</em>) = 0

where <em>n</em> is the magnitude of the normal force and <em>mg</em> cos(<em>α</em>) is the magnitude of the perpendicular component of weight.

From the second equation we have

<em>n</em> = <em>mg</em> cos(<em>α</em>)

and <em>f</em> = <em>µn</em> = <em>µmg</em> cos(<em>α</em>), where <em>µ</em> is the coefficient of static friction. Substituting these into the first equation gives us

<em>mg</em> sin(<em>α</em>) = <em>µmg</em> cos(<em>α</em>)   ==>   <em>µ</em> = tan(<em>α</em>)   ==>   <em>α</em> = arctan(0.35) ≈ 19.3°

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An airplane is moving at 350 km/hr. If a bomb is
Molodets [167]

Answers:

a) -171.402 m/s

b) 17.49 s

c) 1700.99 m

Explanation:

We can solve this problem with the following equations:

y=y_{o}+V_{oy}t-\frac{1}{2}gt^{2} (1)

x=V_{ox}t (2)

V_{f}=V_{oy}-gt (3)

Where:

y=0 m is the bomb's final jeight

y_{o}=1.5 km \frac{1000 m}{1 km}=1500 m is the bomb'e initial height

V_{oy}=0 m/s is the bomb's initial vertical velocity, since the airplane was moving horizontally

t is the time

g=9.8 m/s^{2} is the acceleration due gravity

x is the bomb's range

V_{ox}=350 \frac{km}{h} \frac{1000 m}{1 km} \frac{1 h}{3600 s}=97.22 m/s is the bomb's initial horizontal velocity

V_{f} is the bomb's fina velocity

Knowing this, let's begin with the answers:

<h3>b) Time</h3>

With the conditions given above, equation (1) is now written as:

y_{o}=\frac{1}{2}gt^{2} (4)

Isolating t:

t=\sqrt{\frac{2 y_{o}}{g}} (5)

t=\sqrt{\frac{2 (1500 m)}{9.8 m/s^{2}}} (6)

t=17.49 s (7)

<h3>a) Final velocity</h3>

Since V_{oy}=0 m/s, equation (3) is written as:

V_{f}=-gt (8)

V_{f}=-(97.22)(17.49 s) (9)

V_{f}=-171.402 m/s (10) The negative sign ony indicates the direction is downwards

<h3>c) Range</h3>

Substituting (7) in (2):

x=(97.22 m/s)(17.49 s) (11)

x=1700.99 m (12)

5 0
3 years ago
Please help on this one?
bezimeni [28]

Using the given equation you get:

E = 1.99x10^-25 / 9.0x10^-6

Divide 1.99 by 9.0: 1.99/9.0 = 0.22

For the scientific notation, when dividing subtract the two exponents:

25 -6 = 19

So you now have 0.22 x 10^-19

Now you need to change the 0.22 to be in scientific notation form:

2.2 x 10^-20

The answer is B.

3 0
3 years ago
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Speed equals distance divided by time, so 350 divided by 2.5 equals 140 kilometers per hour.
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Well I can't see the following physical properties you talked about in the question.

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explanation
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