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anygoal [31]
3 years ago
10

A scooter is traveling at a constant speed v when it encounters a circular hill of radius r = 480 m. The driver and scooter toge

ther have mass m = 159 kg.
(a) What speed in m/s does the scooter have if the driver feels weightlessness (i.e., has an apparent weight of zero) at the top of the hill?
(b) If the driver is traveling at the speed above and encounters a hill with a radius 2r,
Physics
1 answer:
Grace [21]3 years ago
6 0

Answer:

68.585m/sec , 779.1 N

Explanation:

To feel weightless, centripetal acceleration must equal g (9.8m/sec^2). The accelerations then cancel.

From centripetal motion.

F =( mv^2)/2

But since we are dealing with weightlessness

r = 480m

g = 9.8m/s^2

M also cancels, so forget M.

V^2 = Fr

V = √ Fr

V =√ (9.8 x 480) = 4704

= 68.585m/sec.

b) Centripetal acceleration = (v^2/2r) = (68.585^2/960) = 4704/960

= 4.9m/sec^2.

Weight (force) = (mass x acceleration) = 159kg x (g - 4.9)

159kg × ( 9.8-4.9)

159kg × 4.9

= 779.1N

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What are the rules for setting up an integral of rotation?
Angelina_Jolie [31]

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<span>First the solid of revolution should be defined. The general  function is y=f(x), on an interval [a,b].</span>

Then the curve is rotated about a given axis to get the surface of the solid of revolution. That is the integral of the function.

<span>It all depends of the function f(x), which must be known in order to calculate  the integral.</span>

3 0
3 years ago
Suppose a certain jet plane creates an intensity level of 124 dB at a distance of 5.01 m. What intensity level does it create on
Zolol [24]

Answer:71 dB

Explanation:

Given

sound Level \beta _1=124 dB

distance r_1=5.01 m

From sound Intensity

\beta =10dB\log (\frac{I_1}{I_0})

124=10dB\log (\frac{I_1}{I_0})

12.4=\log (\frac{I_1}{I_0})

I_1=(1\times 10^{-12})\times 10^{12.4}

I_1=2.51 W/m^2

we know Intensity I\propto ^\frac{1}{r^2}

I_1r_1^2=I_2r_2^2

I_2=I_1(\frac{r_1}{r_2})^2

I_2=2.51\cdot (\frac{5.01}{2.25\times 10^3})^2

I_2=1.24\times 10^{-5} W/m^2

Sound level corresponding to I_2

\beta _2=10\log (\frac{I_2}{I_0})

\beta _2=10\log (\frac{1.24\times 10^{-5}}{1\times 10^{-12}})

\beta _2=70.93\approx 71 dB

6 0
3 years ago
If a person climbed Mt. Everest has a mass of 105 kg and a weight of 625 N what would be the acceleration due to gravity?
DanielleElmas [232]

The acceleration due to gravity would be 5.95 m/s²

A force is known to be a push or pull and it is the change in momentum per time. It can be expressed by using the relation.

  • Force = mass × acceleration.

From the parameters given:

  • Mass = 105 kg
  • Force = 625 N

By replacing the given values into the above equation, we can determine the acceleration.

∴

625 N = 105 kg × acceleration.

\mathbf{acceleration = \dfrac{625 \ N}{105 \ kg}}

acceleration = 5.95 N/kg

  • Since 1 N/kg = 1 m/²

acceleration = 5.95 m/s²

Learn more about acceleration(a) here:

brainly.com/question/14344386

3 0
3 years ago
A typical neutron star may have a mass equal to that of the Sun but a radius of only 20 km.(a) What is the gravitational acceler
Pepsi [2]

Answer:

331665750000\ m/s^2

3257806.62409 m/s

Explanation:

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

M = Mass of Sun = 1.989\times 10^{30}\ kg

r = Radius of Star = 20 km

u = Initial velocity = 0

v = Final velocity

s = Displacement = 16 m

a = Acceleration

Gravitational acceleration is given by

g=\dfrac{GM}{r^2}\\\Rightarrow g=\dfrac{6.67\times 10^{-11}\times 1.989\times 10^{30}}{20000^2}\\\Rightarrow g=331665750000\ m/s^2

The gravitational acceleration at the surface of such a star is 331665750000\ m/s^2

v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 331665750000\times 16+0^2}\\\Rightarrow v=3257806.62409\ m/s

The velocity of the object would be 3257806.62409 m/s

8 0
3 years ago
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