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2 years ago

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A stunt car driver testing the use of air bags drives a car at a constant speed of25 m/s for a total of 100m. He applies his bra

kes and accelerates uniformly to a stop just as he reaches a wall 50m away.
Required:
Sketch the a graph showing position v.s time and velocity v.s time.

1 answer:

PIT_PIT [208]2 years ago

4 0

Answer:

The graphs are attached

Explanation:

We are told that he starts with a constant speed of 25 m/s for a distance of 100 m.

At constant velocity, v = distance/time

time(t) = distance(d)/velocity(v)

t1 = 100/25

t1 = 4 s

Now, we are told that he applies his brakes and accelerates uniformly to a stop just as he reaches a wall 50m away.

It means, he decelerate and final velocity is zero.

Thus;

v² = u² + 2as

0² = 25² + 2a(50)

25² = - 100a

625 = - 100a

a = - 625/100

a = - 6.25 m/s²

v = u + at

0 = 25 + (-6.25t)

25 = 6.25t

t = 25/6.25

t = 4 s

With the values gotten, kindly find attached the distance-time and velocity-time graphs.

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Ceres: Yes!
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Eris: Yes!
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Haumea: Yes!
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1 year ago

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A child is sliding a toy block (with mass = m) down a ramp. The coefficient of static friction between the block and the ramp is

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Answer:

$F=mg(sin(\theta )-0.25 cos(\theta ))$

Explanation:

The free body diagram of the block on the slide is shown in the below figure

Since the block is in equilibrium we apply equations of statics to compute the necessary unknown forces

N is the reaction force between the block and the slide

For equilibrium along x-axis we have

$\sum F_{x}=0\\\\mgsin(\theta )-\mu N-F=0\\\therefore F=mgsin(\theta)-\mu N......(\alpha )\\Similarly\\\sum F_{y}=0\\\\N-mgcos(\theta )=0\\\therefore N=mgcos(\theta ).......(\beta )\\\\$

Using value of N from equation β in α we get value of force as

$F=mg(sin(\theta )-\mu cos(\theta ))$

Applying values we get

$F=mg(sin(\theta )-0.25 cos(\theta ))$

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2 years ago

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A penny is placed on a rotating turntable. Where on the turntable does the penny require the largest centripetal force to remain

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Answer:

m = mass of the penny

r = distance of the penny from the center of the turntable or axis of rotation

w = angular speed of rotation of turntable

F = centripetal force experienced by the penny

centripetal force "F" experienced by the penny of "m" at distance "r" from axis of rotation is given as

F = m r w²

in the above equation , mass of penny "m" and angular speed "w" of the turntable is same at all places. hence the centripetal force directly depends on the radius .

hence greater the distance from center , greater will be the centripetal force to remain in place.

So at the edge of the turntable , the penny experiences largest centripetal force to remain in place.

Explanation:

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2 years ago

During a compaction test in the lab a cylindrical mold with a diameter of 4in and a height of 4.58in was filled. The compacted s

Ray Of Light [21]

Answer:

part a : <em>The dry unit weight is 0.0616 </em> $lb/in^3$ <em />

part b : <em>The void ratio is 0.77</em>

part c : <em>Degree of Saturation is 0.43</em>

part d : <em>Additional water (in lb) needed to achieve 100% saturation in the soil sample is 0.72 lb</em>

Explanation:

Part a

Dry Unit Weight

The dry unit weight is given as

$\gamma_{d}=\frac{\gamma}{1+\frac{w}{100}}$

Here

$\gamma_d$ is the dry unit weight which is to be calculated
γ is the bulk unit weight given as

$\gamma =weight/Volume \\\gamma= 4 lb / \pi r^2 h\\\gamma= 4 lb / \pi (4/2)^2 \times 4.58\\\gamma= 4 lb / 57.55\\\gamma= 0.069 lb/in^3$

w is the moisture content in percentage, given as 12%

Substituting values

$\gamma_{d}=\frac{\gamma}{1+\frac{w}{100}}\\\gamma_{d}=\frac{0.069}{1+\frac{12}{100}} \\\gamma_{d}=\frac{0.069}{1.12}\\\gamma_{d}=0.0616 lb/in^3$

<em>The dry unit weight is 0.0616 </em> $lb/in^3$ <em />

Part b

Void Ratio

The void ratio is given as

$e=\frac{G_s \gamma_w}{\gamma_d} -1$

Here

e is the void ratio which is to be calculated

$\gamma_d$ is the dry unit weight which is calculated in part a
$\gamma_w$ is the water unit weight which is 62.4 $lb/ft^3$ or 0.04 $lb/in^3$

G is the specific gravity which is given as 2.72

Substituting values

$e=\frac{G_s \gamma_w}{\gamma_d} -1\\e=\frac{2.72 \times 0.04}{0.0616} -1\\e=1.766 -1\\e=0.766$

<em>The void ratio is 0.77</em>

Part c

Degree of Saturation

Degree of Saturation is given as

$S=\frac{G w}{e}$

Here

e is the void ratio which is calculated in part b

G is the specific gravity which is given as 2.72

w is the moisture content in percentage, given as 12% or 0.12 in fraction

Substituting values

$S=\frac{G w}{e}\\S=\frac{2.72 \times .12}{0.766}\\S=0.4261$

<em>Degree of Saturation is 0.43</em>

Part d

Additional Water needed

For this firstly the zero air unit weight with 100% Saturation is calculated and the value is further manipulated accordingly. Zero air unit weight is given as

$\gamma_{zav}=\frac{\gamma_w}{w+\frac{1}{G}}$

Here

$\gamma_{zav}$ is the zero air unit weight which is to be calculated

$\gamma_w$ is the water unit weight which is 62.4 $lb/ft^3$ or 0.04 $lb/in^3$

G is the specific gravity which is given as 2.72

w is the moisture content in percentage, given as 12% or 0.12 in fraction

$\gamma_{zav}=\frac{\gamma_w}{w+\frac{1}{G}}\\\gamma_{zav}=\frac{0.04}{0.12+\frac{1}{2.72}}\\\gamma_{zav}=\frac{0.04}{0.4876}\\\gamma_{zav}=0.08202 lb/in^3\\$

Now as the volume is known, the the overall weight is given as

$weight=\gamma_{zav} \times V\\weight=0.08202 \times 57.55\\weight=4.72 lb$

As weight of initial bulk is already given as 4 lb so additional water required is 0.72 lb.

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2 years ago

1. A2 .7-kg copper block is given an initial speed of 4.0m/s on a rough horizontal surface. Because of friction, the block final

tatyana61 [14]

Answer:

A. Increase in temperature is 0.0176 degree Celsius. b. the remaining energy will be lost.

Explanation:

The mass of copper block = 7kg

Initial speed = 4.0 m/s

Specific heat of copper = 0.385 j/g degree Celcius.

a. The increase in temperature is calculated below:

$\text{Change in kinetic energy} = \frac{1}{2}mv^{2} \\= \frac{1}{2} \times 2.7 \times 4^{2} = 21.6 J \\$

85% of energy is converted into internal energy.

$mc\DeltaT = 21.6 \times 0.85 \\2.7 \times 385 \times \Delta T = 21.6 \times 0.85 \\\Delta T = 0.0176 degree \ celsius$

b. The remaining 15 per cent of kinetic energy will be lost and it will be changed into other forms.

7 0

2 years ago