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Strike441 [17]
2 years ago
11

Look at Concept Simulation 5.2 to review the concepts involved in this question. Two cars are identical, except for the type of

tread design on their tires. The cars are driven at the same speed and enter the same unbanked horizontal turn. Car A cannot negotiate the turn, but car B can. Which tread design, the one on car A or the one on car B, yields a larger coefficient of static friction between the tires and the road?
Physics
1 answer:
dimulka [17.4K]2 years ago
7 0

Answer:

Tread design on car B would yield a larger coefficient of static friction between the tires and the road

Explanation:

The car model using the coefficient of static friction doesn't work well with tires. A higher coefficient of static friction would require more force to cause a loss of attraction.

The static frictional force helps to keep the unbanked horizontal turn. This means that the frictional force is the centripetal force.

The tread design of car B ensures that the centripetal force is enough to negotiate the turn. On the other hand, the tread design of car A does not provide the necessary centripetal force, hence car A is unable to negotiate the turn.

Therefore, tread design on car B would yield a larger coefficient of static friction between the tires and the road.

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A plane flying at 70.0 m/s suddenly stalls. If the acceleration during the stall is 9.8 m/s2 directly downward, the stall lasts
tino4ka555 [31]

Answer:

v = 66.4 m/s

Explanation:

As we know that plane is moving initially at speed of

v = 70 m/s

now we have

v_x = 70 cos25

v_x = 63.44 m/s

v_y = 70 sin25

v_y = 29.6 m/s

now in Y direction we can use kinematics

v_y = v_i + at

v_y = 29.6 - (9.81 \times 5)

v_y = -19.5 m/s

since there is no acceleration in x direction so here in x direction velocity remains the same

so we will have

v = \sqrt{v_x^2 + v_y^2}

v = \sqrt{63.44^2 + 19.5^2}

v = 66.4 m/s

4 0
2 years ago
Angular and Linear Quantities: A child is riding a merry-go-round that has an instantaneous angular speed of 1.25 rad/s and an a
serious [3.7K]

To solve this problem we will use the kinematic equations of angular motion in relation to those of linear / tangential motion.

We will proceed to find the centripetal acceleration (From the ratio of the radius and angular velocity to the linear velocity) and the tangential acceleration to finally find the total acceleration of the body.

Our data is given as:

\omega = 1.25 rad/s \rightarrow The angular speed

\alpha = 0.745 rad/s2 \rightarrow The angular acceleration

r = 4.65 m \rightarrow The distance

The relation between the linear velocity and angular velocity is

v = r\omega

Where,

r = Radius

\omega = Angular velocity

At the same time we have that the centripetal acceleration is

a_c = \frac{v^2}{r}

a_c = \frac{(r\omega)^2}{r}

a_c = \frac{r^2\omega^2}{r}

a_c = r \omega^2

a_c = (4.65 )(1.25 rad/s)^2

a_c = 7.265625 m/s^2

Now the tangential acceleration is given as,

a_t = \alpha r

Here,

\alpha = Angular acceleration

r = Radius

\alpha = (0.745)(4.65)

\alpha = 3.46425 m/s^2

Finally using the properties of the vectors, we will have that the resulting component of the acceleration would be

|a| = \sqrt{a_c^2+a_t^2}

|a| = \sqrt{(7.265625)^2+(3.46425)^2}

|a| = 8.049 m/s^2 \approx 8.05 m/s2

Therefore the correct answer is C.

7 0
2 years ago
You are called as an expert witness to analyze the following auto accident: Car B, of mass 2000 kg, was stopped at a red light w
OLga [1]

Answer:

a). va=17.23 \frac{m}{s} or 38.54 mph

b). v=38.54 mph and limit is 35 mph

c). Completely inelastic

d). Eka=192.967 kJ

Ekt=76.071 kJ

Explanation:

m_{a}=1300kg\\m_{b}=2000kg\\x_{f}=7.25m\\u_{k}=0.65

The motion is an inelastic collision so

m_{a}*v_{a}+m_{b}*v_{b}=(m_{a}+m_{b})*v_{f}

The force of the motion is contrarest by the force of friction so

F-F_{uk} =0\\F=F_{uk}\\F_{uk}=u_{k}*m*g\\F=m*a\\a=\frac{F}{m}\\ a=\frac{F_{uk}}{m}\\a=\frac{u_{k}*m*g}{m}\\a=u_{k}*g\\a=0.65*9.8\frac{m}{s^{2}} \\a=6.39\frac{m}{s^{2}}

Now with the acceleration can find the time and the velocity final that make the distance 7.25m being united

x_{f}=x_{o}+v_{o}*t+2*a*t^{2}\\x_{o}=0\\v_{o}=0\\x_{f}=2*a*t^{2}\\t^{2}=\frac{x_{f}}{2*a}\\t=\sqrt{\frac{7.25m}{6.37\frac{m}{s^{2} } } } \\t=1.06s

So the velocity final can be find using this time

v_{f}=v_{o}+a*t\\v_{o}=0\\v_{f}=6.37\frac{m}{s^{2} } *1.06s\\v_{f}=6.79 \frac{m}{s}

a).

Replacing in the first equation the final velocity can find the initial velocity

m_{a}*v_{a}+m_{b}*v_{b}=(m_{a}+m_{b})*v_{f}

v_{b}=0

v_{a}= \frac{(m_{a}+m_{b)*v_{f}}}{m_{a}}\\v_{a}= \frac{(1300+2000)*6.37}{1300}\\v_{a}=17.23 \frac{m}{s}

b).

35mph*\frac{1m}{0.000621371mi} *\frac{1h}{3600s}=15.646\frac{m}{s}

Velocity limit in m/s is 15.646 m/s and the initial velocity is 17.23 m/s

so is exceeding the speed limit in about 1.58 m/s

or in miles per hour

3.5 mph

c).

The collision is complete inelastic because any mass can be returned to the original mass, so even they are no the same mass however in the moment they move the distance 7.25m as a same mass the motion is considered completely inelastic

d).

Ek=\frac{1}{2}*m*(v)^{2}\\  Eka=\frac{1}{2}*1300kg*(17.23\frac{m}{s})^{2}\\Eka=192.967 kJ\\Ekt=\frac{1}{2}*m*(v)^{2}\\Ekt=\frac{1}{2}*3300kg*(6.79\frac{m}{s})^{2}\\Ekt=76.071 kJ

8 0
2 years ago
a projectile is launched straight up at 141 m/s . How fast is it moving at the top of its trajectory? suppose it is launched upw
BARSIC [14]

The velocity of projectile has 2 components, horizontal component vcosθ and vertical component vsinθ, where v is the velocity of projection and θ is the angle between +ve X-axis and projectile motion.

In case 1, θ = 90⁰

    So horizontal component is vcos90 = 0

         Vertical component at maximum height = 0

So velocity at maximum height = 0 m/s


In case 2, θ = 45⁰

    So horizontal component is 141cos45 = 100m/s

         Vertical component at maximum height = 0

So velocity at maximum height = 100 m/s


7 0
2 years ago
A. Why is the stratosphere considered a "stable" layer in the atmosphere?
vovangra [49]

Answer:

A. Stratosphere is said to be stable layer of the atmosphere when cool air sinks and warm air rises.Due to the fact that cool air has tendency to sink ,the air is not going fluctuating up and down in the stratosphere. This means that the air remains stationary or particles remains there for a very long duration.

B. If the lifted index is negative then the parcel temperature is warmer than the actual temperature. In addition, the parcel that is less warm than the surrounding will be less dense and will rise.

C. The water vapor come from different kinds of fronts; gust fronts from existing storms as their downdraft hits the surface, spreads and lifts air in front, upper air disturbances and surface heating by solar radiation making an unbalanced vertical profile .

D. the threshold used by storm chasers to assess if the dew point temperature is high enough to produce large thunderstorms is moisture ,the surface dew point needs to be 55 degrees fahrenheit or greater for a surface based thunderstorm to occur.

E. Wind shear is the change in wind direction or speed with height in an atmosphere.

Explanation:

7 0
2 years ago
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