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Y_Kistochka [10]
4 years ago
8

A car is initially moving at 35 km/h along a straight highway. To pass another car, it speeds up to 135 km/h in 10.5 seconds at

a constant acceleration.
(a) how large was the acceleration in m/s ^2
(b)how large was the acceleration, in units go g= 9.80 m/s ^2
Physics
1 answer:
Aleks04 [339]4 years ago
6 0
Acceleration = (velocity final-velocity initial)/ time
where
velocity final = 135 km/hr x 1 hr /3600 s x 1000m/1km
                     = 37.5 m/s
velocity initial = 35 km/hr x  1hr /3600 s x 1000 m/1 km
                      =  9.72 m/s
a) acceleration = 2.646 m/s^2
b) acceleration in g units  = (2.646m/s^2)/(9.8m/s^2)
                                              = 0.27 units

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Answer:

v=12.65\ m.s^{-1}

Explanation:

Given:

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  • radius of curvature, r=71\ m
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<u>During the turn to prevent the skidding of the vehicle its centripetal force must be equal to the opposite balancing frictional force:</u>

m.\frac{v^2}{r} =\mu.N

where:

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N= normal reaction force due to weight of the car

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1350\times \frac{v^2}{71} =0.23\times (1350\times 9.8)

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3 years ago
"When fire stopping material is used where more than ____________________ nonmetallic sheathed cables pass through wood framing
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3 years ago
A car moves in a straight line at 22.0 m/s for 10.0miles, then at 30.0 m/s for another 10.0miles. Calculate the car’s average sp
maw [93]

Answer: 25.38 m/s

Explanation:

We have a straight line where the car travels a total distance D, which is divided into two segments d=10 miles:

D=d+d=2d (1)

Where d=10mi \frac{1609.34 m}{1 mi}=16093.4 m

On the other hand, we know speed is defined as:

S=\frac{d}{t} (2)

Where t is the time, which can be isolated from (2):

t=\frac{d}{S} (3)

Now, for the first segment d=16093.4 m the car has a speed S_{1}=22m/s, using equation (3):

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t_{1}=\frac{16093.4 m}{22m/s} (5)

t_{1}=731.518 s (6) This is the time it takes to travel the first segment

For the second segment d=16093.4 m the car has a speed S_{1}=30m/s,  hence:

t_{2}=\frac{d}{S_{2}} (7)

t_{2}=\frac{16093.4 m}{30m/s} (8)

t_{2}=536.44 s (9) This is the time it takes to travel the secons segment

Having these values we can calculate the car's average speed S_{ave}:

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S_{ave}=\frac{2(16093.4 m)}{731.518 s +536.44 s} (11)

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Complete Question

The complete question is shown on the first uploaded image  

Answer:

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