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igor_vitrenko [27]
2 years ago
10

How long does it take a car traveling at 50 mph to travel 75 miles? Use one of the following to find the answer.

Physics
1 answer:
galben [10]2 years ago
5 0

Answer:

1.25 hours

Explanation:

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A

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Can the velocity of a body revese the direction when acceleration is constant?
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Yes, the velocity of the object can reverse direction when its acceleration is constant. For example consider that the velocity of any object at any time t is given as: ... At At t = 0 sec, the magnitude of velocity is 2m/s and is moving in the forward direction i.e.v (t) = -2.

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1. explain why mid-sized cars and SUVs have different rollover fatality rates.
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Answer:

1.  They generally have poorer fuel efficiency and require more resources to manufacture than smaller vehicles, thus contributing more to climate change and environmental degradation. Their higher center of gravity increases their risk of rollovers.

2. Mid-sized cars and SUVs both have a fixed object fatality rate of 2.6/BVM. This is most likely because they have a similar mass and inertia. The force of the crash is likely to have about the same deceleration in both vehicles.

Explanation:

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8 0
2 years ago
Water is flowing in a pipe with a circular cross section but with varying cross-sectional area, and at all points the water comp
slamgirl [31]

(a) 5.66 m/s

The flow rate of the water in the pipe is given by

Q=Av

where

Q is the flow rate

A is the cross-sectional area of the pipe

v is the speed of the water

Here we have

Q=1.20 m^3/s

the radius of the pipe is

r = 0.260 m

So the cross-sectional area is

A=\pi r^2 = \pi (0.260 m)^2=0.212 m^2

So we can re-arrange the equation to find the speed of the water:

v=\frac{Q}{A}=\frac{1.20 m^3/s}{0.212 m^2}=5.66 m/s

(b) 0.326 m

The flow rate along the pipe is conserved, so we can write:

Q_1 = Q_2\\A_1 v_1 = A_2 v_2

where we have

A_1 = 0.212 m^2\\v_1 = 5.66 m/s\\v_2 = 3.60 m/s

and where A_2 is the cross-sectional area of the pipe at the second point.

Solving for A2,

A_2 = \frac{A_1 v_1}{v_2}=\frac{(0.212 m^2)(5.66 m/s)}{3.60 m/s}=0.333 m^2

And finally we can find the radius of the pipe at that point:

A_2 = \pi r_2^2\\r_2 = \sqrt{\frac{A_2}{\pi}}=\sqrt{\frac{0.333 m^2}{\pi}}=0.326 m

6 0
3 years ago
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