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beks73 [17]
3 years ago
8

A student builds and calibrates an accelerometer and uses it to determine the speed of her car around a certain unbanked highway

curve. The accelerometer is a plumb bob with a protractor that she attaches to the roof of her car. A friend riding in the car with the student observes that the plumb bob hangs at an angle of 15.0° from the vertical when the car has a speed of 21.5 m/s. (a) What is the centripetal acceleration of the car rounding the curve? m/s2 (b) What is the radius of the curve? m (c) What is the speed of the car if the plumb bob deflection is 7.00° while rounding the same curve? m/s
Physics
1 answer:
Sunny_sXe [5.5K]3 years ago
6 0

Answer:

a) 9.46M m/s^2

b) 48.83m

c) 68.22m/s

Explanation:

The centripetal force must be equal to the horizontal component (x component) of the gravitational force over the plum bob. Thus, we have

F_C=F_{xg}=Mgcos\theta=M\frac{v^2}{r}  ( 1 )

where M is the mass of the car, v its speed, r the radius of the curvature and g the gravity constant.

a) By replacing we obtain:

F_c=M(9.8m/s^2)(cos15\°)=9.46M\ m/s^2   ( 2 )

where it is only necessary to put the mass of the car M in (2).

b) By canceling M in (2) and taking apart r we get:

gcos\theta =\frac{v^2}{r}\\\\r=\frac{v^2}{gcos\theta}=\frac{(21.5m/s)^2}{(9.8m/s^2)cos15\°}=48.83m

c) If the angle is 7° the speed is given by:

v=\sqrt{gcos\theta r}=\sqrt{(9.8m/s^2)(cos7\°)(48.83m)}=68.22\frac{m}{s}

hope this helps!!

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andre [41]

Answer:

1.5 km/s²

Explanation:

Given that:

a car starts from rest; it means the initial velocity (u) = 0 km/hr = 0 m/s

after time (t) = 20 seconds

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The acceleration (a) of the car can be determined by using the formula:

a = \dfrac{v-u}{t}

a = \dfrac{30\  m/s -0 \ m/s}{20 \ s}

a = \dfrac{30 \  m/s}{20 \ s}

a = 1.5 km/s²

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3 years ago
Using equations, determine the temperature, pressure and density of the air for a aircraft flying at 19.5 km. Is this aircraft s
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Answer:

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d) it is a subsonic aircraft

Explanation:

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Given :

Po (atmospheric pressure at sea level )  = 101kpa

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attached below is the remaining part of the solution

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