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VMariaS [17]
3 years ago
15

#1. A car moving at the rate of 65 miles per hour passes a truck moving in the same direction at 38 miles per hour . What is the

speed of the car to the nearest mile per hour from the truck frame of reference ?
#2. The X-43 is the fastest aircraft ever. It flew a distance of 3,961 km from Los Angeles to New York at the speed of 11,265 km/h . How long did it take the x-43 to fly that distance?


#3. A mail carrier walked six blocked to the north one block to the east , and then six blocks to the south. What is the mail carriers displacement ?
Physics
1 answer:
valentina_108 [34]3 years ago
8 0
<span>#1. A car moving at the rate of 65 miles per hour passes a truck
moving in the same direction at 38 miles per hour.
The speed of the car from the truck's frame of reference is
27 miles per hour. 

#2. The X-43 is the fastest aircraft ever. It flew a distance of 3,961 km
from Los Angeles to New York at the speed of 11,265 km/h.
It took the X-43  (3,961 km) / (11,265 km/h) = 0.352 hour, or 21min 5.8sec 
to fly that distance? 

#3. A mail carrier walked six blocks to the north one block to the east ,
and then six blocks to the south.
After these three moves, the mail carrier's displacement is one block
to the East.</span>
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A baseball is hit with a bat. The direction of the ball is completely reversed and its speed is doubled. If the actual contact w
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Before the impact, let the velocity of the baseball was v m/s.

After being hit by the bat its velocity is -2v
So, change in velocity, Deltav=v-(-2v)=3v
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A horizontal 810-N merry-go-round of radius 1.60 m is started from rest by a constant horizontal force of 55 N applied tangentia
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Answer:

576 joules

Explanation:

From the question we are given the following:

weight = 810 N

radius (r) = 1.6 m

horizontal force (F) = 55 N

time (t) = 4 s

acceleration due to gravity (g) = 9.8 m/s^{2}

K.E = 0.5 x MI x ω^{2}

where MI is the moment of inertia and ω is the angular velocity

MI = 0.5 x m x r^2

mass = weight ÷ g = 810 ÷ 9.8 = 82.65 kg

MI = 0.5 x 82.65 x 1.6^{2}

MI = 105.8 kg.m^{2}

angular velocity (ω) = a x t

angular acceleration (a) = torque ÷ MI

where torque = F x r = 55 x 1.6 = 88 N.m

a= 88 ÷ 105.8 = 0.83 rad /s^{2}

therefore

angular velocity (ω) = a x t = 0.83 x 4 = 3.33 rad/s

K.E = 0.5 x MI x ω^{2}

K.E = 0.5 x 105.8 x 3.33^{2} = 576 joules

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