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jeka94
2 years ago
10

A sample of tendon 3.00 cm long and 4.00 mm in diameter is found to break under a minimum force of 128 N. If instead the sample

had been 1.50 cm long and of uniform composition and cross-sectional area, what minimum force would have been required to break it
Physics
1 answer:
Hatshy [7]2 years ago
3 0

Answer:128 N

Explanation:

Sample of 3 cm and 4 mm diameter found to break under a minimum force of 128 N .

If sample is 1.5 cm long with same cross-sectional area then minimum force required to break is also 128 N because the applied force is same for any length and diameter of tendon.        

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Consider a steel guitar string of initial length L=1.00 meter and cross-sectional area A=0.500 square millimeters. The Young's m
Reil [10]

Answer:

\Delta l=0.015m

Explanation:

We have given initial length of the steel guitar l = 1 m

Cross sectional area A=0.5mm^2=0.5\times 10^{-6}m^2

Young's modulus \gamma=2\times 10^{11}Pa

Force F = 1500 N

So stress =\frac{force}{area}=\frac{1500}{0.5\times 10^{-6}}=3000\times 10^{-6}=3\times 10^{9}Pa

We know that young's modulus =\frac{stress}{strain}

So 2\times 10^{11}=\frac{3\times 10^{9}}{strain}

strain=1.5\times 10^{-2}=0.015m

Now strain =\frac{\Delta l}{l}

0.015=\frac{\Delta l}{1}

\Delta l=0.015m

6 0
3 years ago
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An electric motor rotating a workshop grinding wheel at 1.06 102 rev/min is switched off. Assume the wheel has a constant negati
kvasek [131]

Answer:

t = 106π / 30*2.1

Explanation:

w_{i} = 1.06*10^{2}    => 106

    => 106 x 2π/60

    => 106/30π

∝ = -2.1 rad/sec²

w_{f} => 0

w_{f} = w_{i} + ∝t

∴ (w_{f} - w_{i}) / ∝ = t

t = 106π / 30*2.1

6 0
3 years ago
To travel at constant speed, a car engine provides 24KW of useful power. The driving force on the car is 600N. At what speed doe
Keith_Richards [23]

Answer:

The speed traveled by the car is 40 meter per second.

Explanation:

The formula for the relation between the power and the force is as follows:

P = Fv

Where F is the force and v is the speed.

As given

To travel at constant speed, a car engine provides 24KW of useful power. The driving force on the car is 600N.

F = 600 N

Convert power from KW to W.

1 KW = 1000 W

24 KW = 24 × 1000 W

           = 24000 W

Thus

P = 24000 W

Put these values in the formula.

24000 = 600 × v

24000 = 600v

v = \frac{24000}{600}

v = 40 meter per second .

Therefore the speed of the car is 40 meter per second .

3 0
3 years ago
suppose you needed to build a raft to cross a fast-moving river. Describe the physical and chemical properties of the raft that
erastovalidia [21]
1.Density of the material building the raft must be lower than the water. 
<span>2. Material must not react with water. </span>
<span>3.Material must have high strength. </span>
<span>4.Raft must be wide in-order to avoid drawing in the river.</span>
4 0
3 years ago
For a given initial projectile speed Vo, calculate what launch angle A gives the longest range R. Show your work, don't just quo
pickupchik [31]
The optimal angle of 45° for maximum horizontal range is only valid when initial height is the same as final height. 

<span>In that particular situation, you can prove it like this: </span>

<span>initial velocity is Vo </span>
<span>launch angle is α </span>

<span>initial vertical velocity is </span>
<span>Vv = Vo×sin(α) </span>

<span>horizontal velocity is </span>
<span>Vh = Vo×cos(α) </span>

<span>total time in the air is the the time it needs to fall back to a height of 0 m, so </span>
<span>d = v×t + a×t²/2 </span>
<span>where </span>
<span>d = distance = 0 m </span>
<span>v = initial vertical velocity = Vv = Vo×sin(α) </span>
<span>t = time = ? </span>
<span>a = acceleration by gravity = g (= -9.8 m/s²) </span>
<span>so </span>
<span>0 = Vo×sin(α)×t + g×t²/2 </span>
<span>0 = (Vo×sin(α) + g×t/2)×t </span>
<span>t = 0 (obviously, the projectile is at height 0 m at time = 0s) </span>
<span>or </span>
<span>Vo×sin(α) + g×t/2 = 0 </span>
<span>t = -2×Vo×sin(α)/g </span>

<span>Now look at the horizontal range. </span>
<span>r = v × t </span>
<span>where </span>
<span>r = horizontal range = ? </span>
<span>v = horizontal velocity = Vh = Vo×cos(α) </span>
<span>t = time = -2×Vo×sin(α)/g </span>
<span>so </span>
<span>r = (Vo×cos(α)) × (-2×Vo×sin(α)/g) </span>
<span>r = -(Vo)²×sin(2α)/g </span>

<span>To find the extreme values of r (minimum or maximum) with variable α, you must find the first derivative of r with respect to α, and set it equal to 0. </span>

<span>dr/dα = d[-(Vo)²×sin(2α)/g] / dα </span>
<span>dr/dα = -(Vo)²/g × d[sin(2α)] / dα </span>
<span>dr/dα = -(Vo)²/g × cos(2α) × d(2α) / dα </span>
<span>dr/dα = -2 × (Vo)² × cos(2α) / g </span>

<span>Vo and g are constants ≠ 0, so the only way for dr/dα to become 0 is when </span>
<span>cos(2α) = 0 </span>
<span>2α = 90° </span>
<span>α = 45° </span>
4 0
2 years ago
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