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dusya [7]
3 years ago
9

(a) a light-rail commuter train accelerates at a rate of 1.15 m/s2. how long does it take it to reach its top speed of 80.0 km/h

starting from rest?
Physics
1 answer:
snow_lady [41]3 years ago
6 0
A = 1.15m/s2, Vf = 80.0km/h --> we need it in m/s, so:
Vf = 80km/h × 1000m/1km × 1h/3600s
= 22.22m/s
Top speed = Vf, initial speed = Vi
time (t) = V(Vf-Vi) ÷ a
t = (22.22-0)m/s ÷ 1.15m/s2
t = 22.22m/s × s2/1.15m
= 19.32 seconds
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In a local bar, a customer slides an empty beer mug down the counter for a refill. the height of the counter is 1.15 m. the mug
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<span>A.    </span>Let’s say the horizontal component of the velocity is vx and the vertical is vy. <span>
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x = v0x t

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<span>1.50m = v0x t    ---->  v0x= 1.50/t</span>

<span>-1.15m = -(1/2) (9.81) t^2    -----> t =0.4842 s</span>

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v0x = 3.10 m/s

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4 0
3 years ago
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The magnitude of the total friction force exerted by the road on the tires at B will be 919.46 lb.

<h3>What is the friction force?</h3>

It is a type of opposition force acting on the surface of the body that tries to oppose the motion of the body. Its unit is Newton (N).

Mathematically, it is defined as the product of the coefficient of friction and normal reaction.

The given data in the problem is;

The weight is,W= 3,220 lb

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The distance traveled is,d= 300 ft

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\rm v^2 = u^2 +2ad \\\\ \rm (45 \times \frac{5280}{3600} )^2 = (60 \times  \frac{5280}{3600}  )^2 +2a\times 300 \\\\

The tangential accelerations are;

\rm a_t = \frac{66^2 -88^2}{600} \\\\ \rm a_t =  -5.65 ft/sec^2 \\\\

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The net or the total friction force exerted by the road on the tires at B. is found as;

\rm F = \sqrt{(565)^2+(726)^2} \\\ F = 919.946 \ lb

Hence, the magnitude of the total friction force exerted by the road on the tires at B will be 919.46 lb.

To learn more about the friction force, refer to the link;

brainly.com/question/1714663

#SPJ1

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I think this is the answer:

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