What class of lever is shown below?

pochemuha

B

HOPE IT HELPS LET ME KNOW IF U NEED EXPLANATION

4 0

3 years ago

Suppose a rocket ship accelerates upwards with acceleration equal in magnitude to twice the magnitude of g (we say that the rock

pashok25 [27]

Answer:

a) $s_a=98100\ m$ is the height where the rocket stops accelerating and its fuel is finished and starts decelerating while it still continues to move in the upward direction.

b) $v_a=1962\ m.s^{-1}$ is speed of the rocket going when it stops accelerating.

c) $H=294300\ m$

d) $t_T=544.95\ s$

e) Zero, since the average velocity is the net displacement per unit time and when the rocket strikes back the earth surface the net displacement is zero.

Explanation:

Given:

acceleration of rocket, $a=2g=2\times 9.81=19.62\ m.s^{-2}$

time for which the rocket accelerates, $t_a=100\ s$

For the course of upward acceleration:

using eq. of motion,

$s_a=ut+\frac{1}{2}at_a^2$

where:

$u=$ initial velocity of the rocket at the launch $=0$

$s_a=$ height the rocket travels just before its fuel finishes off

so,

$s_a=0+\frac{1}{2}\times 19.62\times 100^2$

a) $s_a=98100\ m$ is the height where the rocket stops accelerating and its fuel is finished and starts decelerating while it still continues to move in the upward direction.

Now the velocity of the rocket just after the fuel is finished:

$v_a=u+at_a$

$v_a=0+19.62\times 100$

b) $v_a=1962\ m.s^{-1}$ is speed of the rocket going when it stops accelerating.

After the fuel is finished the rocket starts to decelerates. So, we find the height of the rocket before it begins to fall back towards the earth.

Now the additional height the rocket ascends before it begins to fall back on the earth after the fuel is consumed completely, at this point its instantaneous velocity is zero:

using equation of motion,

$v^2=v_a^2-2gh$

where:

$g=$ acceleration due to gravity

$v=$ final velocity of the rocket at the top height

$0^2=1962^2-2\times 9.81\times h$

$h=196200\ m$

c) So the total height at which the rocket gets:

$H=h+s$

$H=196200+98100$

$H=294300\ m$

d)

Time taken by the rocket to reach the top height after the fuel is over:

$v=v_a+g.t$

$0=1962-9.81t$

$t=200\ s$

Now the time taken to fall from the total height:

$H=v.t'+\frac{1}{2}\times gt'^2$

$294300=0+0.5\times 9.81\times t'^2$

$t'=244.95\ s$

Hence the total time taken by the rocket to strike back on the earth:

$t_T=t_a+t+t'$

$t_T=100+200+244.95$

$t_T=544.95\ s$

e)

Zero, since the average velocity is the net displacement per unit time and when the rocket strikes back the earth surface the net displacement is zero.

8 0

3 years ago

I'M GIVING 95POINTS PLZ HELP!!

Travka [436]

#1

initial height = 25000 ft

as we know that

1 ft = 0.3048 m

so we have

H = 25000(0.3048) = 7620 m

#2

If mass of the man = 90 kg

now the initial potential energy is given as

$U = mgh$

$U = (90)(9.8)(7620)$

$U = 6.72 \times 10^6 J$

#3

Gravitational force that is acting on it is given by

$F = mg$

$F = 90(9.8) = 882 N$

so this is the force by which earth is attracting him towards it as we can see due to this force the man is accelerating towards the earth

#4

As we know that gravitational force is given by

$F = \frac{GMm}{r^2}$

here we know that

r = R + h

R = radius of earth ( $6.37 \times 10^6 m$ )

h = 7620 m

now we have acceleration at that point is

$a = \frac{F}{m}$

$a = \frac{GM}{r^2}$

$a = \frac{(6.67\times 10^{-11})(5.98 \times 10^{24})}{(6.37 \times 10^6 + 7620)^}$

$a = 9.806 m/s^2$

#5

by energy conservation we have

$KE = PE$

$\frac{1}{2}mv^2 = mgh$

$v^2 = 2gh$

$v^2 = 2(9.8)(7620)$

$v = 386.5 m/s$

#6

final speed due to wind resistance is 150 mph

now we know that

1 mile = 1609 meter

so this speed in m/s is given as

$150 mph = 150(\frac{1609 m}{3600 s})$

$v = 67 m/s$

#7

No Luke is not accelerating when his speed is 150 mph

because at this speed his velocity will become constant

and since there is no change in velocity so his acceleration will become zero

#8

since at 150 mph the acceleration of Luke is zero

so net force on him must be zero

so we will have

wind force = weight or force due to earth

$F_{wind} = 882 N$

#9

If there is no wind resistance then there will be energy conservation

so KE = PE

$KE = 6.72 \times 10^6 J$

#10

Net will reduce the velocity of luke to be zero by taking long time

As we know that force on a system is given by

$F = \frac{\Delta P}{\Delta t}$

so if we increase the time interval here then we will have less force

so Luke will unhurt due to more time to decrease his speed to be zero

5 0

3 years ago

A sledgehammer is an example of what kind of simple machine?

alisha [4.7K]

I am 95 percent sure the answer in C

3 0

3 years ago

A sample of hydrogen gas will behave most like an ideal gas under the conditions of(1) low pressure and low temperature(2) low p

docker41 [41]

Ideal gas has several important properties that must be followed by real gases to behave like an ideal gas

so here we can say

1. all gas molecules moves at very fast speed in random direction

2. there should not be any interaction force between molecules of gas

3. all gas molecules must have to follow Newton's II law

4. There is no effect of gravity on the molecules of gas

so here if all above conditions are followed by gas then

the condition must be followed only at low pressure and high temperature

so correct answer is

(2) low pressure and high temperature

5 0

3 years ago