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8090 [49]
3 years ago
12

If a plane is flying level at 910 km/h and the banking angle is not to exceed 50 ∘, what's the minimum curvature radius for the

turn?
Physics
1 answer:
hoa [83]3 years ago
4 0

Answer:

5.5 km

Explanation:

First, we convert the distance from km/h to m/s

910 * 1000/3600

= 252.78 m/s

Now, we use the formula v²/r = gtanθ to get our needed radius

making r the subject of the formula, we have

r = v²/gtanθ, where

r = radius of curvature needed

g = acceleration due to gravity

θ = angle of banking

r = 252.78² / (9.8 * tan 50)

r = 63897.73 / (9.8 * 1.19)

r = 63897.73 / 11.662

r = 5479 m or 5.5 km

Thus, we conclude that the minimum curvature radius needed for the turn is 5.5 km

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a 3 kg piece of putty that is moving with a velocity of 10 m/s collides and sticks to an 8 kg bowling ball that was at rest. wha
defon

The final velocity is 2.7 m/s

Explanation:

We can solve this problem by using the principle of conservation of momentum: in fact, in absence of external forces, the total momentum of the system must be conserved before and after the collision.

Therefore we can write:

p_i = p_f\\m_1 u_1 + m_2 u_2 = (m_1+m_2)v  

where:  

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v=\frac{m_1 u_1 + m_2 u_2}{m_1+m_2}=\frac{(3)(10)+0}{3+8}=2.7 m/s

And the positive sign indicates their final direction is the same as the initial direction of the putty.

Learn more about momentum here:

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3 years ago
The Sun has a mass of 1.99x10^30 kg and a radius of 6.96x10^8 m. Calculate the acceleration due to gravity, in meters per second
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Answer:

g=274\ m/s^2

Explanation:

Mass of the Sun, M=1.99\times 10^{30}\ kg

The radius of the Sun, r=6.96\times 10^8\ m

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g=\dfrac{GM}{r^2}\\\\g=\dfrac{6.67\times 10^{-11}\times 1.99\times 10^{30}}{(6.96\times 10^8)^2}\\\\g=274\ m/s^2

So, the value of acceleration due to gravity on the Sun is 274\ m/s^2.

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3 years ago
What is the SI unit for intensity?
Allisa [31]

Answer:

The SI unit of intensity is the watt per square meter/metre (W/m^2.)

Explanation:

Intensity is equal to the power transferred per unit area. Since power is measured in watts (W) and 1 W = 1 J/s, then intensity can be viewed as how fast energy goes through a certain area.

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