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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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Cardiovascular fitness can be measured by what?
12345 [234]
VO2 max is considered to be the most valid measure<span> of </span>cardio respiratory fitness<span>. It </span>measures<span> the capacity of the heart, lungs, and blood to transport oxygen to the working muscles, and </span>measures<span> the utilization of oxygen by the muscles during exercise.</span>
5 0
2 years ago
An electron in a vacuum chamber is fired with a speed of 9800 km/s toward a large, uniformly charged plate 75 cm away. The elect
melisa1 [442]

Answer:

The plate's surface charge density is -8.056\times10^{-9}\ C/m^2

Explanation:

Given that,

Speed = 9800 km/s

Distance d= 75 cm

Distance d' =15 cm

Suppose we determine the plate's surface charge density?

We need to calculate the surface charge density

Using work energy theorem

W=\Delta K.E

W=\dfrac{1}{2}mv_{f}^2-\dfrac{1}{2}mv_{i}^2

Here, final velocity is zero

W=0-\dfrac{1}{2}mv_{i}^2...(I)

We know that,

W=-Fd

W=-E\times e\times d

W=-\dfrac{\lambda}{2\epsilon_{0}}\times e\times d...(II)

From equation (I) and (II)

-\dfrac{1}{2}mv_{i}^2=-\dfrac{\lambda}{2\epsilon_{0}}\times e\times d

Charge is negative for electron

\lambda=\dfrac{mv^2\epsilon_{0}}{(-e)d}

Put the value into the formula

\lambda=-\dfrac{9.1\times10^{-31}\times(9800\times10^{3})^2\times8.85\times10^{-12}}{1.6\times10^{-19}\times(75-15)\times10^{-2}}

\lambda=-8.056\times10^{-9}\ C/m^2

Hence, The plate's surface charge density is -8.056\times10^{-9}\ C/m^2

3 0
2 years ago
A 2 kg stone is tied to a 0.5 m string and swung around a circle at a constant angular velocity of 12 rad/s. the angular momentu
Illusion [34]
Starting from the angular velocity, we can calculate the tangential velocity of the stone:
v=\omega r= (12 rad/s)(0.5 m)= 6 m/s

Then we can calculate the angular momentum of the stone about the center of the circle, given by
L=mvr
where
m is the stone mass
v its tangential velocity
r is the radius of the circle, that corresponds to the length of the string.

Substituting the data of the problem, we find
L=(2 kg)(6 m/s)(0.5 m)=6 kg m^2 s^{-1}
3 0
2 years ago
While the cart is moving along an aisle, it comes in contact with a smear of margarine that had recently been dropped on the flo
liberstina [14]

Answer:

No, the velocity of the grocery cart will not remain constant, but instead will increase continuously

Explanation:

No the grocery cart will not move at constant velocity over the spilled maragine because as the frictional force is changed to - 20 N, the net force will not be equal to zero

Net force on the cart when frictional force is changed to - 20 N is 40 - 20 N = 20 N

As the force is positive, it means that the force is acting in the direction of motion of the cart and will increase the velocity of the cart as the force is acting in the direction of motion

∴ The velocity of the cart will not remain constant instead it will increase

3 0
2 years ago
(a) A velocity selector consists of electric and magnetic fields described by the expressions E = E and B = B ĵ, with B = 10.0 m
Allushta [10]

Answer:

(a) 160000 kV/m

(b) 1336 keV

Explanation:

(a) magnetic filed, B = 10 T

energy of electron, E = 740 eV

mass of electron, m = 9.1 x 10^-31 kg

Let v be the velocity of electron.

E = 1/2 mv^2

740 x 1.6 x 10^-19 = 0.5 x 9.1 x 10^-31 x v^2

v = 1.6 x 10^7 m/s

v = E / B

E = v x B = 1.6 x 10^7 x 10 = 16 x 10^7 V/m

E = 160000 kV/m

(b) E = 16 x 10^7 V/m

B = 10 T

Let v be the velocity of protons.

v = E / B = 16 x 10^7 / 10 = 1.6 x 10^7 m/s

Kinetic energy of proton, E = 1/2 mv^2

                                          = 0.5 x 1.67 x 10^-27 x 1.6 x 1.6 x 10^14

                                          = 2.14 x 10^-13 J = 1336000 eV = 1336 keV

8 0
2 years ago
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