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3 years ago

Curves on some test tracks and race courses, such as the Daytona International Speedway in Florida, are very steeply banked.

Physics

1 answer:

docker41 [41]3 years ago

8 0

Answer:

Speed, v = 45.84 m/s

Explanation:

Given that,

Radius of the curve, r = 100 m

Banking of the curve, $\theta=65^{\circ}$

On the banking of curve, the speed of the vehicle is given by :

$v=\sqrt{rg\ tan\theta}$

$v=\sqrt{100\times 9.8\times \ tan(65)}$

v = 45.84 m/s

So, the speed at which a 100 m radius curve banked is 45.84 m/s. Hence, this is the required solution.

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Explain in detail what happens to the Earths surface after an earthquake due to the force of tension

irga5000 [103]

After the two plates Collide the amount of tension is what makes the two plates slip. When two plates slip the Earths surface begins to move because of the slipping of the plates thats when we feel an earth quake. After the earth quake is finshed the tension between the two plates happen again so the Earths surface may be cracked as a result of the immense slipping. But the tension will happen again and devolpe slowly over time. Intill it happens again almost like a cycle

8 0

3 years ago

if one paperclip has the mass of 1 gram and 1000 paperclips has a mass of 1 kilogram how many kilograms are 8000 paperclips ?

elixir [45]

8 kilograms. Hope I've helped you

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3 years ago

A wooden rod of negligible mass and length 80.0cm is pivoted about a horizontal axis through its center. A white rat with mass 0

algol13

Answer:

The speed of the animals is 1.64m/s.

Explanation:

Let us work with variables and call the mass of the two rats $m_1$ and $m_2$ , and the length of the rod $L$ .

Using the law of conservation of energy, which says the potential energies of the rats must equal their kinetic energies, we know that when the rod swings to the vertical position,

$m_1\frac{L}{2}g -m_2\frac{L}{2}g = \frac{1}{2}m_1v^2+\frac{1}{2}m_2v^2$

$(m_1 -m_2)g\frac{L}{2} = \frac{1}{2}(m_1+m_2)v^2$ ,

solving for $v$ , we get:

$\boxed{v = \sqrt{\frac{(m_1 -m_2)gL}{(m_1+m_2)}} }$

Putting in the values for $m_1$ , $m_2$ , $g$ , and $L$ we get:

$v = \sqrt{\frac{(0.450kg -0.220kg)(9.8m/s^2)(0.8m)}{(0.450kg+0.220kg)}}$

$\boxed{ v= 1.64m/s}$

Therefore, as the rod swings through the vertical position , the speed of the rats is 1.64 m/s.

7 0

2 years ago

The downsprue leading into the runner of a certain mold has a length of 175 mm. The cross-sectional area at the base of the spru

adoni [48]

Answer:

(a) Velocity at bottom is 1.85 m/s

(b) Volume flow rate is 7.4 x 10⁻⁴ m³/s.

Explanation:

(a)

Applying Bernoulli's Equation on both ends of the down sprue, with the assumptions that every point is at atmospheric pressure and the liquid metal at the pouring basin is at zero velocity. The equation then becomes:

V = √2gh

where,

V = velocity at bottom of down sprue

h = height of down sprue = 175 mm = 0.175 m

V = √2(9.8 m/s²)(0.175 m)

V = 1.85 m/s

(b)

The volume flow rate is given as:

Volume Flow Rate = (V)(A)

where,

V = velocity at bottom = 1.85 m/s

A = Area of bottom = 400 mm² = 0.0004 m²

Therefore,

Volume Flow Rate = (1.85 m/s)(0.0004 m²)

Volume Flow Rate = 7.4 x 10⁻⁴ m³/s = 740 cm³/s

(c)

The time required to fill the cavity is given as:

Volume Flow Rate = V/t

where,

V = Volume of mold Cavity = 0.001 m³

t = time required to fill the cavity = ?

Therefore,

t = V/Volume Flow Rate

t = 0.001 m³/7.4 x 10⁻⁴ m³/s

t = 1.35 s

5 0

3 years ago

A disk of radius 10 cm is pulled along a frictionless surface with a force of 16 N by a string wrapped around the edge. At the i

drek231 [11]

Answer:

t = 0.2845Nm (rounded to 4 decimal places)

Explanation:

The disk rotates at a distance of an arc length of 28cm

Arc length = radius × central angle × π/180

28cm = 10cm × central angle × π/180

Central angle = $\frac{28}{10}$ × 180/π ≈ 160.4°

Torque (t) = rFsin(central angle) , where F is the applied force

Radius in meters = 10/100 = 0.1m

t = 0.1m × 16N × sin160.4°

t = 0.2845Nm (rounded to 4 decimal places)

8 0

3 years ago