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yaroslaw [1]
2 years ago
14

At what angle should the roadway on a curve with a 50m radius be banked to allow cars to negotiate the curve at 12 m/s even if t

he road way is icy (and the frictional force is zero)
Physics
1 answer:
just olya [345]2 years ago
3 0

Answer:

The banking angle of the road is 16.38 degrees.

Explanation:

Given:

Radius of the roadway on curve, R = 50 m

velocity of the moving car, V = 12 m/s

The banking angle can be calculated by using the formula below:

If there's no frictional force, then net force is equal to centripetal force.

MgTanΦ = (MV^2)/ R

TanΦ = V^2 / gR

Where;

Φ is the banking angle

g is acceleration due to gravity

TanΦ = (12 x 12) / (9.8 x 50)

TanΦ = 0.2939

Φ = arctan (0.2939)

Φ = 16.38 degrees

Therefore, the banking angle of the road is 16.38 degrees.

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On a snowy day, when the coefficient of friction μs between a car’s tires and the road is 0.50, the maximum speed that the car c
Nana76 [90]

Answer:

28.1 mph

Explanation:

The force of friction acting on the car provides the centripetal force that keeps the car in circular motion around the curve, so we can write:

F=\mu mg = m \frac{v^2}{r} (1)

where

\mu is the coefficient of friction

m is the mass of the car

g = 9.8 m/s^2 is the acceleration due to gravity

v is the maximum speed of the car

r is the radius of the trajectory

On the snowy day,

\mu=0.50\\v = 20 mph = 8.9 m/s

So the radius of the curve is

r=\frac{v^2}{\mu g}=\frac{(8.9)^2}{(0.50)(9.8)}=16.1 m

Now we can use this value and re-arrange again the eq. (1) to find the maximum speed of the car on a sunny day, when \mu=1.0. We find:

v=\sqrt{\mu g r}=\sqrt{(1.0)(9.8)(8.9)}=12.6 m/s=28.1 mph

7 0
2 years ago
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you are hiking along a river and see a tall tree on thhe opposite bank. You measure the angle of elevation of the top of the tre
Sidana [21]

Answer:

Explanation:

Let the height of the tree is y and the distance of tree from point B is x.

According to the diagram

tan61 = \frac{y}{x}

x = 0.55 y ..... (1)

tan49.5 = \frac{y}{50+x}

(50 + 0.55y) 1.17 = y ..... from equation (1)

58.5 + 0.644 y = y

0.356 y = 58.5

y = 164.3 ft

3 0
2 years ago
A 63.0 kg astronaut is on a spacewalk when the tether line to the shuttle breaks. the astronaut is able to throw a spare 10.0 kg
Llana [10]

There are other forces at work here nevertheless we will imagine it is just a conservation of momentum exercise. Also the given mass of the astronaut is light astronaut.

The solution for this problem is using the formula: m1V1=m2V2 but we need to get V1:

V1= (m2/m1) V2


V1= (10/63) 12 = 1.9 m/s will be the final speed of the astronaut after throwing the tank. 

6 0
2 years ago
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A 50.-kilogram rock rolls off the edge of a cliff. if it is traveling at a speed of 24.2 m/s when it hits the ground, what is th
ElenaW [278]

The correct answer to the question is : 29.88 m.

EXPLANATION :

As per the question, the mass of the rock m = 50 Kg.

The rock is rolling off the edges of the cliff.

The final velocity of the rock when it hits the ground v = 24 .2 m/s.

Let the height of the cliff is h.

The potential energy gained by the rock at the top of the cliff = mgh.

Here, g is known as acceleration due to gravity, and g = 9.8\ m/s^2

When the rock rolls off the edge of the cliff, the potential energy is converted into kinetic energy.

When the rock hits the ground, whole of its potential energy is converted into its kinetic energy.

The kinetic energy of the rock when it touches the ground is given as -

                Kinetic energy K.E = \frac{1}{2}mv^2.

From above we know that -

   Kinetic energy at the bottom of the cliff = potential energy at a height h

                 \frac{1}{2}mv^2=\ mgh

                ⇒ v^2=\ 2gh

                ⇒ h=\ \frac{v^2}{2g}

                ⇒ h=\ \frac{(24.2)^2}{2\times 9.8}

                ⇒ h=\ 29.88\ m

Hence, the height of the cliff is 29.88 m

             


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In photosynthesis, the water is being used to create food for the plant (Glucose). In transpiration the water is going from a liquid to a gas that's being released.
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