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Anastaziya [24]
3 years ago
12

How much work in joules must be done to stop a 920-kg car traveling at 90 km/h?

Physics
1 answer:
tankabanditka [31]3 years ago
3 0

Answer:

Work done, W = −287500 Joules

Explanation:

It is given that,

Mass of the car, m = 920 kg

The car is travelling at a speed of, u = 90 km/h = 25 m/s

We need to find the amount of work must be done to stop this car. The final velocity of the car, v = 0

Work done is also defined as the change in kinetic energy of an object i.e.

W=\Delta K

W=\dfrac{1}{2}m(v^2-u^2)

W=\dfrac{1}{2}\times 920\ kg(0-(25\ m/s)^2)

W = −287500 Joules

Negative sign shows the work done is done is opposite direction.

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The correct answer is  1.4285714.

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Velocity = distance/ time  

Thus time = distance/velocity

Here velocity = 350m/s

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With thermodynamics, one cannot determine ________. Question 6 options: the speed of a reaction the direction of a spontaneous r
pantera1 [17]

Answer:

A.  the speed of a reaction

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The thermodynamic aspect of a reaction will show you the energy needed for a reaction to occur. If the energy difference(ΔG)  is positive, which means the reaction is absorbing energy and it called endothermically. The opposite will be an exothermic reaction that will release energy, which means it doesn't need energy and the energy difference (ΔG) will be negative.  

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4 years ago
A 6-in-wide polyamide F-1 flat belt is used to connect a 2-in-diameter pulley to drive a larger pulley with an angular velocity
Likurg_2 [28]

Answer:

a) Fc = 4.15 N, Fi = 435.65 N, (F1)a = 640 N, and F2  = 239.6 N,

b) Ha = 1863.75 N, nfs = 1 , length = 11.8 mm

Explanation:

Given that:

γ= 9.5 kN/m³ = 9500N/m3

b = 6 inches = 0.1524 m

t = 0.0013 mm

d = 2 inches  = 0.0508 m

n = 1750 rpm

H_{nom}=2hp=1491.4W

L = 9 ft = 2.7432 m

Ks = 1.25

g = 9.81 m/s²

a)

w=\gamma b t = 9500* 0.1524*0.0013=1.88N/m

V=\frac{\pi d n}{60} =\pi *0.0508*1750/60=4.65 m/s

F_c=\frac{wV^2}{g}=1.88*4.65^2/9.81=4.15N

(F_1)_a=bF_aC_pC_v=0.1524*6000*0.7*1=640N

T=\frac{H_{nom}n_dK_s}{2\pi n}= \frac{1491*1.25*1}{2*\pi*1750/60}=10.17Nm

F_2=(F_1)_a-\frac{2T}{D}= 640-\frac{2*10.17}{0.0508} =239.6N

F_i=\frac{(F_1)_a+F_2}{2} -F_c=435.65N

b)

H_a=1491*1.25=1863.75W

n_f_s=\frac{H_a}{H_{nom}K_S }=1

dip = \frac{L^2w}{8F_i} =\frac{2.7432*1.88}{435.65}=11.8mm

7 0
3 years ago
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