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djverab [1.8K]
3 years ago
15

An airplane traveling at 201 m/s makes a turn. What is the smallest radius of the circular path (in km) that the pilot can make

and keep the centripetal acceleration under 5.0 m/s2 ?
Physics
1 answer:
egoroff_w [7]3 years ago
4 0

Answer:

the smallest radius of the circular path is 8.1 km

Explanation:

The computation of the smallest radius of the circular path is given below:

Given that

V = Velocity = 201 m/s

a_c = acceleration = 5 m/s^2

radius = ?

As we know that

a_c = V^2 ÷ r

5 = 201^2 ÷ r

r = 201^2 ÷ 5

= 8,080.2 g

= 8.1 km

Hence, the smallest radius of the circular path is 8.1 km

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Two frisky grasshoppers collide in midair at the top of their respective trajectories and grab onto each other, holding tight th
Svetlanka [38]

Answer:

The decrease in Kinetic energy is 0.0107 Joules

Explanation:

Given

Mass of grasshoppers

Let m1 = Mass of grasshopper 1

Let m2 = Mass of grasshopper 2

Let u1 = initial speed of grasshopper 1

Let u2 = initial speed of grasshopper 2

m1 = 250g = 0.25kg

m2 = 130g = 0.13kg

u1 = 15cm/s = 0.15m/s

u2 = 65cm/s = 0.65m/s

First, we calculate the final velocity of the grasshoppers after collision using conservation of momentum.

Using

m1u1 + m2u2 = (m1 + m2) * v

Where v = final velocity

By substituton

0.25 * 0.15 + 0.13 * 0.65 = (0.25 + 0.13) * v

0.0375 + 0.0845 = 380v

0.122 = 0.38v

Make v the subject of formula

v = 0.122/0.38

v = 0.321 m/s

Calculating the Kinetic energies before and after impact.

Before collision;

KE = ½m1u1²+ ½m2u2²

KE = ½(m1u1² + m2u2²)

By substituton;

KE = ½(0.25 * 0.15² + 0.13 * 0.65²)

KE = 0.030275J

After collision:

KE = ½(m1+m2)v²

KE = ½(0.25 + 0.13) * 0.321²

KE = 0.01957779 J

Change in kinetic energy = ∆KE

∆KE = 0.030275J - 0.01957779J

∆KE = 0.01069721J

∆KE = 0.0107 J --- Approximately

Hence the decrease in Kinetic energy is 0.0107 Joules

7 0
3 years ago
A 16.5-kg crate starts at rest at the top of a 60.0° incline. The coefficients of friction are μs = 0.400 and μk = 0.300. The cr
irga5000 [103]

Answer:

t = 1.62 s

Explanation:

given,

mass of the block m₁ = 16.5 Kg

m₂ = 8 Kg

angle of inclination = 60°

μs = 0.400 and μk = 0.300

time to slide 2 m = ?

a) let a is the acceleration of the block m₁ downward.

Net force acting on m₂,

F₂ = T - m₂ g

m₂a = T - m₂ g

a = \dfrac{T}{m_2} - g.......(1)

net force acting on m₁

F₁ = m₁g sin(60°) - μ_k m₁g cos (60°) - T

m₁ a = m₁g sin(60°) - μ_k m₁g cos (60°) - T

a = g sin(60^0) - \mu_k g cos (60^0) - \dfrac{T}{m_1}.........(2)

from equations 1 and 2

\dfrac{T}{m_2} - g = g sin(60^0) - \mu_k g cos (60^0) - \dfrac{T}{m_1}

\dfrac{T}{m_2} +\dfrac{T}{m_1} = g+ g sin(60^0) - \mu_k g cos (60^0)

 T\dfrac{m_1+m_2}{m_2\times m_1} = g+ g sin(60^0) - \mu_k g cos (60^0)

 T = \dfrac{g+ g sin(60^0) - \mu_k g cos (60^0)}{\dfrac{m_1+m_2}{m_2\times m_1}}

 T = {m_2\times m_1}\dfrac{g+ g sin(60^0) - \mu_k g cos (60^0)}{{m_1+m_2}}  

T = {16.5\times 8}\dfrac{9.8 + 9.8 sin(60^0) - 0.3\times 9.8 cos (60^0)}{{16.5+8}}

T = 90.61 N

from equation (1)

a = \dfrac{90.61}{8} - 9.8.......(1)

a = 1.52 m/s²

let t is the time taken

Apply,

d = ut + 0.5 a t²

2 = 0 + 0.5 x 1.52 x t²

t = \sqrt{2.63}

t = 1.62 s

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