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Ulleksa [173]
3 years ago
7

A motor-cycle rider going 90km/hr around a curve with a radius of 100m must lean at an angle to the vertical. The angle at which

he leans is *​
Physics
1 answer:
BartSMP [9]3 years ago
3 0

Answer:

The angle of inclination of the rider to the vertical is 32.5⁰.

Explanation:

Given;

linear speed of the motor cycle, v = 90 km/hr = 25 m/s

radius of the circle, r = 100 m

The angle of inclination of the rider to the vertical is calculated using banking angle formula;

\theta = tan^{-1}(\frac{v^2}{rg} )\\\\\theta = tan^{-1} (\frac{25^2}{100 \times 9.8} )\\\\\theta = 32.5 ^0

Therefore, the angle of inclination of the rider to the vertical is 32.5⁰

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An antelope moving with constant acceleration covers the distance 68.0 m between two points in time 7.50 s. Its speed as it pass
Darya [45]

Answer:

A)The speed of the antelope at the first point is 2.43 m/s.

B) The acceleration of the antelope is 1.77 m/s²

Explanation:

The equations of the position and velocity of the antelope is given by the following expressions:

x = x0 + v0 · t + 1/2 · a ·t²

v = v0 + a · t

Where:

x = position of the antelope at time t

x0 = initial position

v0 = initial velocity

t = time

a = acceleration

v = velocity at time t

A) Let´s place the center of the frame of reference at the first point. The equation of position at t = 7.50 s will be:

x = x0 + v0 · t + 1/2 · a ·t²

68.0 m = 0 m + v0 · 7.50 s + 1/2 · a · (7.50 s)²

We also know that at the second point the velocity is 15.7 m/s. Then at t = 7.50 the velocity will be 15.7 m/s.

v = v0 + a · t

15.7 m/s = v0 + a · 7.50 s

We can solve this equation for "a" and replace it in the equation of height to obtain "v0". Then:

a = (15.7 m/s - v0) / 7.50 s

Replacing it in the equation for position:

68.0 m = 0 m + v0 · 7.50 s + 1/2 · a · (7.50 s)²

68.0 m = v0 · 7.50 s + 1/2 · (15.7 m/s - v0) / 7.50 s · (7.50 s)²

68.0 m = v0 · 7.50 s + 7.85 m/s · 7.50 s - 3.75 s · v0

68.0 m - 7.85 m/s · 7.50 s = 3.75 s · v0

(68.0 m - 7.85 m/s · 7.50 s) / 3.75 s = v0

v0 = 2.43 m/s

The speed of the antelope at the first point is 2.43 m/s.

B) The acceleration of the antelope will be:

a = (15.7 m/s - v0) / 7.50 s

a = (15.7 m/s - 2.43 m/s) / 7.50 s

a = 1.77 m/s²

The acceleration of the antelope is 1.77 m/s²

5 0
3 years ago
Calculate the Work done if the force is 2000 Newtons<br> and the distance is 5 km.
kodGreya [7K]

Answer:

10 million joules or 10,000 KJ

Explanation:

Work= Force x Displacement

convert 5km into meters -5km=5000m

W= 2000N x 5000m

w=10,000,000 Joules

or 10,000KJ

4 0
2 years ago
Two asteroids collide and stick together. The first asteroid has a mass of 15\times 10^3\,\mathrm{kg}15×10 3 kg and is initially
statuscvo [17]

Answer:

Final speed is 900.06 m/s at 0.2215^{\circ}  

Solution:

As per the question:

Mass of the first asteroid, m = 15\times 10^{3}\kg

Mass of the second asteroid, m' = 20\times 10^{3}\kg

Initial velocity of the first asteroid, v = 770 m/s

Initial velocity of the second asteroid, v' = 1020 m/s

Angle between the two initial velocities, \theta = 20^{\circ}

Now,

Since, the velocities and hence momentum are vector quantities, then by the triangle law of vector addition of 2 vectors A and B, the resultant is given by:

\vec{R} = \sqrt{A^{2} + 2ABcos\theta + B^{2}}

Thus applying vector addition and momentum conservation, the final velocity is given by:

(m + m')v_{final} = \sqrt{(mv)^{2} + 2(mv)(m'v')cos20^{\circ} + (m'v')^{2}}                               (1)

Now,

(m +m')v_{final} = (35\times 10^{3})v_{final}

(mv)^{2} = (15\times 10^{3}\times 770)^{2} = 1.334\times 10^{14}

(m'v')^{2} = (20\times 10^{3}\times 1020)^{2} = 4.16\times 10^{14}

2(mv)(m'v')cos20^{\circ} = 2(15\times 10^{3}\times 770)(20\times 10^{3}\times 1020)cos20^{\circ} = 4.43\times 10^{14}

Now, substituting the suitable values in eqn (1), we get:

v_{final} = 900.06\ m/s

Now, the direction for the two vectors is given by:

\theta = sin^{- 1} \frac{m'v'sin20^{\circ}}{(m + m')v_{final}}

\theta = sin^{- 1} \frac{20\times 10^{3}\times 1020sin20^{\circ}}{(35\times 10^{3})\times 900.06} = 0.2215^{\circ}

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