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hodyreva [135]
3 years ago
13

A car traveling 90 km/hr is 100 m behind a truck traveling 50 km/hr. How long will it take the car to reach the truck?

Physics
2 answers:
Ber [7]3 years ago
8 0

v = speed of car = 90 km/h

u = speed of truck = 50 km/h

d = initial separation distance = 100 m = 0.1 km

They meet at time t such that

vt = d + ut

t(v - u) = d

t = d/(v - u) = (0.1 km) / [(90 - 50) km/h] = 0.0025 hours

olganol [36]3 years ago
7 0

95t - .210 = 50t

95t - 50t = 0.210

45t = 0.210

t = 0.210/45

t = 0.0047 km/hr

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3 years ago
A 20.0 kg crate sits at rest at the bottom of a 15.0-m-long ramp that is inclined at 34.0° above horizontal. A constant horizont
ankoles [38]

Answer:

987 joules, 3.01s

Explanation:

(A)

from the attached diagram

net force, Fnet, pulling the crate up the ramp is given by

Fnet = FcosФ - WsinФ - Fr

where FcosФ is the component of horizontal force 290N resolved parallel to the plane

WsinФ = mgsinФ = component of the weight of the crate resolved parallel to the plane

Fr = constant opposing frictional force

Fnet = 290cos34⁰ - 20 × 9.8 × sin34° - 65

Fnet = 240.421 - 109.602 - 65

Fnet = 65.82N

Work done on the crate up the ramp, W, is given by

W = Fnet × d (distance up the plane)

W = 65.819 × 15

W = 987.285 joules

W = 987 joules (to 3 significant Figures)

(B)

to calculate the time of travel up the ramp

we use the equation of motion

s = ut + \frac{1}{2}at^{2}

where s = distance up the plane, 15m

u = Initial velocity of the crate, which is 0 for a body that is initially at rest

a = acceleration up the plane, given by

a = \frac{Fnet}{m}

where m = mass of the crate, 20 kg

now, a = \frac{65.819}{20} \\a = 3.291\frac{m^{2} }{s}

from, s = ut + \frac{1}{2}at^{2}

15 = 0*t + \frac{1}{2}* 3.291 * t^{2}

15 = 0 + 1.645t^{2}

15 = 1.645t^{2}

t = \sqrt{\frac{15}{1.645} }

t = 3.019

t = 3.01s (to 3 sig fig)

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