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3 years ago

What is the final velocity, in m/s, of a hoop that rolls without slipping down a 8.20 m high hill, starting from rest?

Physics

1 answer:

pantera1 [17]3 years ago

8 0

Answer:

12.68 m/s.

Explanation:

Equations of motion:

i. S = vi*t + 1/2 * a*(t^2)

ii. vf = vi + a*t

iii. vf^2 = vi^2 + 2a*S

Where, vf = final velocity

vi = initial velocity

S = distance travelled

a = acceleration due to gravity

t = time taken

Given:

vi = 0 m/s

S = 8.2 m

vf = ?

a = 9.81 m/s^2

Using iii. Equation of motion,

vf^2 = vi^2 + 2a*S

= 2 * 9.81 * 8.2

= 160.884

vf = sqrt (160.884)

= 12.68 m/s.

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3 years ago

Your spaceship lands on an unknown planet. To determine the characteristics of this planet, you drop a wrench from 4.50 m above

Virty [35]

$8.98×10^6\:\text{m}$

Explanation:

First we need to find the acceleration due to gravity on the planet. The wrench took 0.809 s to fall from a height of 4.50 m so we can use the equation

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Solving for g, we get

$g = -\dfrac{2y}{t^2} = -\dfrac{2(-4.50\:\text{m})}{(0.809\:\text{s})} = 13.8\:\text{m/s}^2$

Recall that the acceleration due to gravity on a planet's surface can be written as

$g = G\dfrac{M_p}{R_p^2}$

We can express the mass of the planet $M_p$ in terms of its density $\rho$ as follows:

$M_p = \rho \left(\dfrac{4\pi}{3}R_p^3\right) = \dfrac{4\pi}{3}\rho R_p^3$

The expression for g then becomes

$g = \dfrac{G}{R_p^2} \left(\dfrac{4\pi}{3}\rho R_p^3\right) = \dfrac{4\pi G}{3}\rho R_p$

Solving for $R_p,$ we get

$R_p = \dfrac{3g}{4\pi G\rho}$

$\:\:\:\:\:\:\:= \left[\dfrac{3(13.8\:\text{m/s}^2)}{4\pi (6.674×10^{-11}\:\text{Nm}^2\text{/kg}^2)(5500\:\text{kg/m}^3)}\right]$

$\:\:\:\:\:\:\:= 8.98×10^6\:\text{m}$

3 0

3 years ago

A fan is to accelerate quiescent air to a velocity of 8 m/s at a rate of 9 m3 /s. determine the minimum power that must be suppl

ELEN [110]

Given:
ρ = 1.18 kg/m³, density of air
v = 8 m/s, flow velocity
Q = 9 m³/s, volumetric flow rate

The minimum power required (at 100% efficiency) is
$\frac{1}{2} (8 \, \frac{m}{s} )^{2}*(9 \, \frac{m^{3}}{s} )*(1.18 \, \frac{kg}{m^{3}}) = 339.84 \, W$

The actual power will be higher because 100% efficiency is not possible.

Answer: 339.8 W (nearest tenth)

8 0

4 years ago