An object starts from rest and accelerates at a rate of 3.0 m/s/s for 6.0 seconds. The velocity at the end of the 6.0 seconds is

Physics

1 answer:

Zepler [3.9K]3 years ago

6 0

Vf=vi+at
=0+3×6
=18m/s

the answer is D.

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A meter stick is held vertically with one end on the floor and is then allowed to fall. Find the speed of the other end when it

Tems11 [23]

Answer:

5.4 ms⁻¹

Explanation:

Here we have to use conservation of energy. Initially when the stick is held vertical, its center of mass is at some height above the ground, hence the stick has some gravitational potential energy. As the stick is allowed to fall, its rotates about one. gravitational potential energy of the stick gets converted into rotational kinetic energy.

$L$ = length of the meter stick = 1 m

$m$ = mass of the meter stick

$w$ = angular speed of the meter stick as it hits the floor

$v$ = speed of the other end of the stick

we know that, linear speed and angular speed are related as

$v = r w\\w = \frac{v}{r}$

$h$ = height of center of mass of meter stick above the floor = $\frac{L}{2} = \frac{1}{2} = 0.5 m$

$I$ = Moment of inertia of the stick about one end

For a stick, momentof inertia about one end has the formula as

$I = \frac{mL^{2} }{3}$

Using conservation of energy

Rotational kinetic energy of the stick = gravitational potential energy

$(0.5) I w^{2} = mgh\\(0.5)(\frac{mL^{2} }{3}) (\frac{v}{L} )^{2} = mgh\\(0.5)(\frac{v^{2} }{3}) = gh\\(0.5)(\frac{v^{2} }{3}) = (9.8)(0.5)\\v = 5.4 ms^{-1}$

7 0

3 years ago

Calculate the propellant mass required to launch a 2000 kg spacecraft from a 180 km circular orbit on a Hohmann transfer traject

Finger [1]

Answer:

t = 12,105.96 sec

Explanation:

Given data:

weight of spacecraft is 2000 kg

circular orbit distance to saturn = 180 km

specific impulse = 300 sec

saturn orbit around the sun R_2 = 1.43 *10^9 km

earth orbit around the sun R_1= 149.6 * 10^ 6 km

time required for the mission is given as t

$t = \frac{2\pi}{\sqrt{\mu_sun}} [\frac{1}{2}(R_1 + R_2)]^{3/2}$

where

$\mu_{sun}$ is gravitational parameter of sun = 1.32712 x 10^20 m^3 s^2. $t = \frac{2\pi}{\sqrt{ 1.32712 x 10^{20}}} [\frac{1}{2}(149.6 * 10^ 6 +1.43 *10^9 )]^{3/2}$