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1 year ago

8

When a driver presses the brake pedal, his car stops with an acceleration of -2.1 m/s2. How far will the car travel while coming

to a complete stop if its initial speed was 13 m/s?

1 answer:

Burka [1]1 year ago

8 0

The kinematic equation that will be used for this problem is:
Vf^2 = V1^2 + 2 * a * d, where,
Vf = final velocity = 0 m/s
V1 = initial velocity = 13 m/s
a = acceleration = -2.1m/s^2
d = distance traveled = ?
Inserting the given values, the equation becomes
0^2 = 13^2 + 2 * -2.1 * d
0 = 169 + - 4.2*d
169 = 4.2 * d
169/4.2 = d = 40.24
Therefore, d = 40.24 M.

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Derive an expression for the gravitational potential energy of a system consisting of Earth and a brick of mass m placed at Eart

Arlecino [84]

Answer:

The gravitational potential energy of a system is -3/2 (GmE)(m)/RE

Explanation:

Given

mE = Mass of Earth

RE = Radius of Earth

G = Gravitational Constant

Let p = The mass density of the earth is

p = M/(4/3πRE³)

p = 3M/4πRE³

Taking for instance,a very thin spherical shell in the earth;

Let r = radius

dr = thickness

Its volume is given by;

dV = 4πr²dr

Since mass = density* volume;

It's mass would be

dm = p * 4πr²dr

The gravitational potential at the center due would equal;

dV = -Gdm/r

Substitute (p * 4πr²dr) for dm

dV = -G(p * 4πr²dr)/r

dV = -G(p * 4πrdr)

The gravitational potential at the center of the earth would equal;

V = ∫dV

V = ∫ -G(p * 4πrdr) {RE,0}

V = -4πGp∫rdr {RE,0}

V = -4πGp (r²/2) {RE,0}

V = -4πGp{RE²/2)

V = -4Gπ * 3M/4πRE³ * RE²/2

V = -3/2 GmE/RE

The gravitational potential energy of the system of the earth and the brick at the center equals

U = Vm

U = -3/2 GmE/RE * m

U = -3/2 (GmE)(m)/RE

5 0

1 year ago

the flow energy of 124 L/min of a fluid passing a boundary to a system is 108.5 kJ/min. Determine the pressure at this point

Andreyy89

Answer:

The pressure at this point is 0.875 mPa

Explanation:

Given that,

Flow energy = 124 L/min

Boundary to system P= 108.5 kJ/min

$P=1.81\ kW$

We need to calculate the pressure at this point

Using formula of pressure

$P=F\times v$

$P=A_{1}P_{1}\times v_{1}$

Here, $A_{1}v_{1}=Q_{1}$

Where, v = velocity

Put the value into the formula

$1.81 =P_{1}\times0.124\times\dfrac{1}{60}$

$P_{1}=\dfrac{1.81\times60}{0.124}$

$P_{1}=875.80\ kPa$

$P_{1}=0.875\ mPa$

Hence, The pressure at this point is 0.875 mPa

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1 year ago

Paintball guns were originally developed to mark trees for logging. A forester aims his gun directly at a knothole in a tree tha

emmasim [6.3K]

Answer:

The distance between knothole and the paint ball is 0.483 m.

Explanation:

Given that,

Height = 4.0 m

Distance = 15 m

Speed = 50 m/s

The angle at which the forester aims his gun are,

$\tan\theta=\dfrac{4}{15}$

$\tan\theta=0.266$

$\cos\theta=\dfrac{15}{\sqrt{15^2+4^2}}$

$\cos\theta=0.966$

Using the equation of motion of the trajectory

The horizontal displacement of the paint ball is

$x=(u\cos\theta)t$

$t=\dfrac{x}{u\cos\theta}$

Using the equation of motion of the trajectory

The vertical displacement of the paint ball is

$y=u\sin\theta(t)-\dfrac{1}{2}gt^2$

$y=u\sin\theta(\dfrac{x}{u\cos\theta})-\dfrac{1}{2}g(\dfrac{x}{u\cos\theta})^2$

$y=x\tan\theta-\dfrac{gx^3}{2u^2(\cos\theta)^2}$

Put the value into the formula

$y=(15\times0.266)-(\dfrac{9.8\times(15)^2}{2\times(50)^2\times(0.966)^2})$

$y=3.517\ m$

We need to calculate the distance between knothole and the paint ball

$d=h-y$

$d=4-3.517$

$d=0.483\ m$

Hence, The distance between knothole and the paint ball is 0.483 m.

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2 years ago

A bag of potato chips contains 2.00 L of air when it is sealed at sea level at a pressure of 1.00 atm and a temperature of 20.0°

Genrish500 [490]

Answer:

The volume at mountains is 2.766 L.

Explanation:

Given that,

Volume $V_{1} = 2.00\ L$

Pressure $P_{1}= 1.00\ atm$

Pressure $P_{2}= 70.0\ kPa$

Temperature $T_{1}= 20.0°C = 293\ K$

Temperature $T_{2}= 7.00°C = 280\ K$

We need to calculate the volume at mountains

Using gas law

$\dfrac{PV}{T}=\ Constant$

For both temperature,

$\dfrac{P_{1}V_{1}}{T_{1}}=\dfrac{P_{2}V_{2}}{T_{2}}$

Put the value into the formula

$\dfrac{101.325\times2}{293}=\dfrac{70\times V_{2}}{280}$

$V_{2}=\dfrac{101.325\times2\times280}{293\times70}$

$V_{2}=2.766\ litre$

Hence, The volume at mountains is 2.766 L.

5 0

1 year ago

A length of 20-gauge copper wire (of diameter 0.8118 mm) is formed into a circular loop with a radius of 30.0 cm. A magnetic fie

IrinaK [193]

Answer: 3 x 10^-24 watt

Explanation:

P ( resistivity) = 1.72e-8 (from the chart).

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r= 30 cm.

R= pL/A

A= pi* r1^2

r1= 0.8118/2 * 10^-3 m

R= 1.68 x 10^-8 x (2x3.142x0.3)

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A= pi* 0.3^2

N=1

E = 1 x (14 x 3.142x 0.09) = 3.95

I=v/R

v=E,

I = 3.95 / 3.24 x 10^-8 = 1.22 x 10^8

P=I^2 x R.

= 3 x 10^-24 watt

7 0

1 year ago