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3 years ago

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A car traveling 77 km/h slows down at a constant 0.45 m/s^2 just by "letting up on the gas." --Part a : Calculate the distance t

he car coasts before it stops.
--Part b: Calculate the time it takes to stop.
--Part c: Calculate the distance it travels during the first second.
--Part d: Calculate the distance it travels during the fifth second. Need help with all parts A-D, please show all work and formulas used.

1 answer:

sweet [91]3 years ago

5 0

Answer:

(a) 508.37 m

(b) 47.53 s

(c) 21.165 m

(d) 19.365 m

Explanation:

initial velocity, u = 77 km/h = 21.39 m/s

acceleration, a = - 0.45 m/s^2

(a) final velocity, v = 0

Let the distance traveled is s.

Use third equation of motion

$v^{2}=u^{2}+2as$

$0^{2}=21.39^{2}-2 \times 0.45 \times s$

s = 508.37 m

(b) Let t be the time taken to stop.

Use first equation of motion

v = u + at

0 = 21.39 - 0.45 t

t = 47.53 s

(c) Use the formula for the distance traveled in nth second

$s_{n^{th}=u+\frac{1}{2}a\left ( 2n-1 \right )}$

where n be the number of second, a be the acceleration, u be the initial velocity.

put n = 1, u = 21.39 m/s , a = - 0.45m/s^2

$s_{n^{th}=21.39-\frac{1}{2}\times 0.45\left ( 2\times 1-1 \right )}$

$s_{n^{th}=21.165m$

(d) Use the formula for the distance traveled in nth second

$s_{n^{th}=u+\frac{1}{2}a\left ( 2n-1 \right )}$

where n be the number of second, a be the acceleration, u be the initial velocity.

put n = 5, u = 21.39 m/s , a = - 0.45m/s^2

$s_{n^{th}=21.39-\frac{1}{2}\times 0.45\left ( 2\times 5-1 \right )}$

$s_{n^{th}=19.365m$

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A 4.25 kg block is sent up a ramp inclined at an angle theta=37.5° from the horizontal. It is given an initial velocity ????0=15

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Answer:

d = 11.79 m

Explanation:

Known data

m=4.25 kg : mass of the block

θ =37.5° :angle θ of the ramp with respect to the horizontal direction

μk= 0.460 : coefficient of kinetic friction

g = 9.8 m/s² : acceleration due to gravity

Newton's second law:

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∑F : algebraic sum of the forces in Newton (N)

m : mass s (kg)

a : acceleration (m/s²)

We define the x-axis in the direction parallel to the movement of the block on the ramp and the y-axis in the direction perpendicular to it.

Forces acting on the block

W: Weight of the block : In vertical direction

N : Normal force : perpendicular to the ramp

f : Friction force: parallel to the ramp

Calculated of the W

W= m*g

W= 4.25 kg* 9.8 m/s² = 41.65 N

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Wx= Wsin θ= 41.65*sin 37.5° = 25.35 N

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We apply the formula (1)

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N = Wy

N = 33.04 N

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f = μk* N= 0.460*33.04

f = 15.2 N

We apply the formula (1) to calculated acceleration of the block:

∑Fx = m*ax , ax= a : acceleration of the block

-Wx-f = m*a

-25.35-15.2 = (4.25)*a

-40.55 = (4.25)*a

a = (-40.55)/ (4.25)

a = -9.54 m/s²

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Because the block moves with uniformly accelerated movement we apply the following formula to calculate the final speed of the block :

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Where:

d:displacement (m)

v₀: initial speed (m/s)

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Data:

v₀ = 15 m/s

vf = 0

a = -9.54 m/s²

We replace data in the formula (2) to calculate the distance along the ramp the block reaches before stopping (d)

vf²=v₀²+2*a*d

0 = (15)²+2*(-9.54)*d

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Answer:

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I have attached an illustration of a solid disk with the respective forces applied, as stated in this question.

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$F_1 = 90.0N\\F_2 = 125N$

Other parameters given include:

Mass of solid disk, $M = 24.3kg$

and radius of solid disk, $r = 0.364m$

a.) The formula for determining torque ( $T$ ), is $T = r * F$

Hence the net torque produced by the two forces is given as a summation of both forces:

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b.) The angular acceleration of the disk can be found thus:

using the formula for the Moment of Inertia of a solid disk;

$I_{disk} = {\frac{1}{2}}Mr^2$

where $M$ = Mass of solid disk

and $r$ = radius of solid disk

We then relate the torque and angular acceleration ( $\alpha$ ) with the formula:

$T = I\alpha \\-12.7 = ({\frac{1}{2}}Mr^2)\alpha \\\alpha = -{\frac{12.7}{1.61}} = -7.9 rad/s^2$

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