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marin [14]
3 years ago
5

In the previous part, you determined the maximum angle that still allows the crate to remain at rest. If the coefficient of fric

tion is less than 0.7, what happens to this angle? (Note that you can adjust the coefficient of friction by clicking on the More Features tab near the top of the window and then using the slider bar in the right panel.)
Physics
1 answer:
creativ13 [48]3 years ago
6 0

Answer:

the angle must decrease to maintain balance and that the system remains motionless

Explanation:

In a system of a block in an inclined plane, using Newton's second law

X Axis       Wₓ-fr = 0

Y Axis       N -W_{y} = 0

The equation for the force of friction is

          fr = μ N

And using trigonometry for weight components

         sin θ = Wₓ / W

         cos θ = W_{y}/ W

         Wₓ = W sin θ

         W_{y} = W cos θ

We replace

         Wₓ = fr

         W sin θ = μ Wcos θ

         tan θ = μ

If the coefficient of friction decreases the force of static friction decreases, ₓso the value of the tangent must decrease so the angle must decrease to maintain balance and that the system remains motionless or with constant speed

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The Achilles tendon connects the muscles in your calf to the back of your foot. When you are sprinting, your Achilles tendon alt
drek231 [11]

Answer:

5 mm

Explanation:

Youngs's modulus (Y) is described by the following expression:

Y=\frac{F*L}{\Delta L*A}

Where F is the force exerted on the tendon, L is its length, A is its area and ΔL is its change in length (stretching).

The force in this case is 8 times the weight of the runner:

F= 8*m*g\\F= 8*70*9.8\\F=5488 N

Therefore, the change in length of the tendon is given by:

\Delta L=\frac{F*L}{Y*A}\\\Delta L=\frac{5488*0.15}{0.15*10^{10}*1.1*10^{-4}}\\\Delta L= 0.004989 m

the runner's Achilles tendon will stretch by 0.004989 m, which is roughly 5 mm.

8 0
3 years ago
A stone is dropped from the upper observation deck of a tower, 950 m above the ground. (assume g = 9.8 m/s2.) (a) find the dista
sergey [27]
<span>d = 950 m - 4.9t^2 m The distance an object moves under constant acceleration is d = 0.5at^2 where d = distance a = acceleration t = time. Since we're falling and since we're starting at 950 m above ground, the formula becomes: d = 950 m - 0.5at^2 Substituting known values, and simplifying gives us d = 950 m - 0.5*9.8 m/s^2 * t^2 d = 950 m - 4.9 m/s^2 * t^2 Since time is in seconds, we can cancel out the seconds in the units, getting d = 950 m - 4.9t^2 m</span>
8 0
3 years ago
Read 2 more answers
A 0.540-kg bucket rests on a scale. Into this bucket you pour sand at the constant rate of 56.0 g/s. If the sand lands in the bu
romanna [79]

Answer:

a) 12.8212 N

b) 12.642 N

Explanation:

Mass of bucket = m = 0.54 kg

Rate of filling with sand  = 56.0 g/ sec = 0.056 kg/s

Speed of sand = 3.2 m/s

g= 9.8 m/sec2

<u>Condition (a);</u>

Mass of sand = Ms = 0.75 kg

So total mass becomes = bucket mass + sand mass = 0.54 +0.75=1.29 kg

== > total weight = 1.29 × 9.8 = 12.642 N

Now impact of sand = rate of filling × velocity = 0.056 × 3.2 =  0.1792 kg. m /sec2=0.1792 N

Scale reading is sum of impact of sand and weight force ;

i-e

scale reading = 12.642 N+0.1792 N = 12.8212 N

<u>Codition (b);</u>

bucket mass + sand mass = 0.54 +0.75=1.29 kg

==>weight = mg = 1.29 × 9.8 = 12.642 N (readily calculated above as well)

6 0
3 years ago
Momentum of the 2 kg mass moving with velocity 10 m/s is *
gulaghasi [49]
20 kg*m/s because there is 2 kg mass and 10 m/s so you can multiply.
8 0
4 years ago
A sample of a material has 200 radioactive particles in it today. Your grandfather measured 400 radioactive particles in it 60 y
sukhopar [10]

Answer:

Amount of radioactive particles left after 60 years = 100 particles.

Explanation:

Amount of radioactive particles before 60 years = 400

Amount of radioactive particles present today = 200

That is radio active particles reduced to half. That is 60 years is half life of this radio active material.

After 60 years this 200 radio active particles will reduce to half.

Amount of radioactive particles left after 60 years = 0.5 x 200 = 100 particles.

8 0
3 years ago
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