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3 years ago

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A race car starts from rest on a circular track of radius 378 m. The car's speed increases at the constant rate of 0.580 m/s2. A

t the point where the magnitudes of the centripetal and tangential accelerations are equal, find the following.
a. The speed of the race car
b. The distance traveled
c. the elapsed time

1 answer:

insens350 [35]3 years ago

3 0

Answer:

a) $V=14.904m/s$

b) $d = 191.49 m$

c) $t= 25.696 s$

Explanation:

From the question we are told that:

Radius $r =378m$

Acceleration $a=0.580$

a)

Generally the equation for speed of the car is mathematically given by

$a=\frac{v^2}{r}$

$V=\sqrt{a*r}$

$V=\sqrt{0.58*383}$

$V=14.904m/s$

b)

Generally the equation for distance traveled of the car is mathematically given by

$V^2=u^2+2ad$

$d=\frac{V^2}{2a}$

$d=\frac{14.904^2}{2*0.58}$

$d = 191.49 m$

c)

Generally the equation for time of the car is mathematically given by

$V=u+at$

$t=\frac{V}{a}$

$t=\frac{14.904}{0.58}$

$t= 25.696 s$

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snow_lady [41]

Here we know that

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now from above all values we have

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photoshop1234 [79]

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Explanation:

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3 years ago

Residential building codes typically require the use of 12-gauge copper wire (diameter 0.205 cm) for wiring receptacles. Such ci

Alecsey [184]

Given Information:

Current = I = 20 A

Diameter = d = 0.205 cm = 0.00205 m

Length of wire = L = 1 m

Required Information:

Energy produced = P = ?

Answer:

P = 2.03 J/s

Explanation:

We know that power required in a wire is

P = I²R

and R = ρL/A

Where ρ is the resistivity of the copper wire 1.68x10⁻⁸ Ω.m

L is the length of the wire and A is the area of the cross-section and is given by

A = πr²

A = π(d/2)²

A = π(0.00205/2)²

A = 3.3x10⁻⁶ m²

R = ρL/A

R = 1.68x10⁻⁸*(1)/3.3x10⁻⁶

R = 5.09x10⁻³ Ω

P = I²R

P = (20)²*5.09x10⁻³

P = 2.03 Watts or P = 2.03 J/s

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3 years ago

A 1900 kg car rounds a curve of 55 m banked at an angle of 11 degrees? . If the car is traveling at 98 km/h, How much friction f

Nat2105 [25]

Answer:

22000 N

Explanation:

Convert velocity to SI units:

98 km/h × (1000 m / km) × (1 h / 3600 s) = 27.2 m/s

Draw a free body diagram. There are three forces acting on the car. Normal force perpendicular to the bank, gravity downwards, and friction parallel to the bank.

I'm going to assume the friction force is pointed down the bank. If I get a negative answer, that'll just mean it's actually pointed up the bank.

Sum of the forces in the radial direction (+x):

∑F = ma

N sin θ + F cos θ = m v² / r

Sum of the forces in the y direction:

∑F = ma

N cos θ - F sin θ - W = 0

To solve the system of equations for F, first solve for N and substitute.

N = (W + F sin θ) / cos θ

Substituting:

((W + F sin θ) / cos θ) sin θ + F cos θ = m v² / r

(W + F sin θ) tan θ + F cos θ = m v² / r

W tan θ + F sin θ tan θ + F cos θ = m v² / r

W tan θ + F (sin θ tan θ + cos θ) = m v² / r

W tan θ + F sec θ = m v² / r

F sec θ = m v² / r - W tan θ

F = m v² cos θ / r - W sin θ

F = m (v² cos θ / r - g sin θ)

Given that m = 1900 kg, θ = 11°, v = 27.2 m/s, and r = 55 m:

F = 1900 ((27.2)² cos 11 / 55 - 9.8 sin 11)

F = 21577 N

Rounding to two sig-figs, you need at least 22000 N of friction force.

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3 years ago

A solid sphere of radius 40.0 cm has a total positive charge of 16.2 μC uniformly distributed throughout its volume. Calculate t

Jobisdone [24]

Answer:

(a) E=0 : 0 cm from the center of the sphere

(b) E= 227.8*10³ N/C : 10.0 cm from the center of the sphere

(c)E= 911.25*10³ N/C : 40.0 cm from the center of the sphere

(d)E= 411.84 * 10³ N/C : 59.5 cm from the center of the sphere

Explanation:

If we have a uniform charge sphere we can use the following formulas to calculate the Electric field due to the charge of the sphere

$E=\frac{K*Q}{r^{2} }$ : Formula (1) To calculate the electric field in the region outside the sphere r ≥ a

$E=k*\frac{Q}{a^{3} } *r$ :Formula (2) To calculate the electric field in the inner region of the sphere. r ≤ a

Where:

K: coulomb constant

a: sphere radius

Q: Total sphere charge

r : Distance from the center of the sphere to the region where the electric field is calculated

Equivalences

1μC=10⁻⁶C

1cm= 10⁻²m

Data

k= 9*10⁹ N*m²/C²

Q=16.2 μC=16.2 *10⁻⁶C

a= 40 cm = 40*10⁻²m = 0.4m

Problem development

(a)Magnitude of the electric field at 0 cm :

We replace r=0 in the formula (2) , then, E=0

(b) Magnitude of the electric field at 10.0 cm from the center of the sphere

r<a , We apply the Formula (2):

$E=9*10^{9} *\frac{16.2*10^{-6} }{0.4^{3} } *0.1$

E= 227.8*10³ N/C

(c) Magnitude of the electric field at 40.0 cm from the center of the sphere

r=a, We apply the Formula (1) :

$E=\frac{9*10^{9}*16.2*10^{-6} }{0.4^{2} }$

E= 911.25*10³ N/C

(d) Magnitude of the electric field at 59.5 cm from the center of the sphere

r>a , We apply the Formula (1) :

$E=\frac{9*10^{9}*16.2*10^{-6} }{0.595^{2} }$

E= 411.84 * 10³ N/C

4 0

3 years ago