All categories

3 years ago

If a marble is released from a height of 10 meters how long would it take to hit the ground?

1 answer:

3 0

S = ut + 0.5at^2

10 = 0 + 0.5(9.81)t^2 {and if g = 10 then t^2 = 2 so t ~1.414} 

t^2 ~ 2.04 

t ~ 1.43 seconds

You might be interested in

A skier leaves the horizontal end of a ramp with a velocity of 31.0 m/s and lands 156.3 m from the base of a ramp how high is th

BartSMP [9]

Answer:

The height of ramp = 124.694 m

Explanation:

Using second equation of motion,

$s = ut + \frac{1}{2}at^2$

From the question,

u = 31 m/s; s = 156.3 m, a=0

substituting values

$156.3 = 31\times t + 0$

t = $\frac{156.3}{31 }$

= 5.042 s

Similary, for the case of landing

t = 5.042 s; initial velocity, u =0

acceleration = acceleration due to gravity, g = 9.81 $m/s^2$

Substituting in $h = ut + \frac{1}{2}gt^2$

$h = 0 + \frac{1}{2} \times 9.81 \times (5.042)^2$

h = 124.694 m

So height of ramp = 124.694 m

3 0

3 years ago

A hollow cylinder of mass 2.00 kg, inner radius 0.100 m, and outer radius 0.200 m is free to rotate without friction around a ho

Aneli [31]

Answer:

h=2.86m

Explanation:

In order to give a quick response to this exercise we will use the equations of conservation of kinetic and potential energy, the equation is given by,

$\Delta PE_i + \Delta KE_i = \Delta PE_f +\Delta KE_f$

There is no kinetic energy in the initial state, nor potential energy in the end,

$mgh+0=0+KE_f$

In the final kinetic energy, the energy contributed by the Inertia must be considered, as well,

$mgh = (\frac{1}{2}mv^2+\frac{1}{2}I\omega^2)$

The inertia of the bodies is given by the equation,

$I=\frac{m(R_1^2+R^2_2)}{2}$

$I=\frac{2(0.2^2+0.1^2)}{2}$

$I=0.05Kgm^2$

On the other hand the angular velocity is given by

$\omega =\frac{v}{R_2}=\frac{4}{1/5} = 2rad/s$

Replacing these values in the equation,

$(0.5)(9.8)(h) =\frac{1}{2}*0.5*4^2+\frac{1}{2}*0.05*20^2$

Solving for h,

$h=2.86m$

5 0

3 years ago

3.9 divided by 15 gcuu

Vsevolod [243]

0.26 there you go buddy

8 0

3 years ago

Read 2 more answers

Science Will give brainless and 22 points

fenix001 [56]

Answer:

Mercury 0.39 AU, 36 million miles

57.9 million km

Venus 0.723 AU

67.2 million miles

108.2 million km

Earth 1 AU

93 million miles

149.6 million km

Mars 1.524 AU

141.6 million miles

227.9 million km

Jupiter 5.203 AU

483.6 million miles

778.3 million km

Saturn 9.539 AU

886.7 million miles

1,427.0 million km

Uranus 19.18 AU

1,784.0 million miles

2,871.0 million km

Neptune 30.06 AU

2,794.4 million miles

4,497.1 million km

Pluto (a dwarf planet) 39.53 AU

3,674.5 million miles

5,913 million km

4 0

3 years ago

Read 2 more answers

In March 1999 the Mars Global Surveyor (GS) entered its final orbit about Mars, sending data back to Earth. Assume a circular or

shusha [124]

Answer: $5.944\times 10^{23}\ kg$

Explanation:

Given

Time period $T=7.08\times 10^3\ s$

Orbital speed $v=3.40\times 10^3\ m/s$

mass of GS $m_{GS}=930\ kg$

Radius of Mars $r=3.43\times 10^6\ m$

Consider the mass of mars is M

Here, Gravitational pull will provide the centripetal force

$F_G=F_c$

$\dfrac{GMm_{GS}}{r^2}=\dfrac{m_{GS}v^2}{r}\\M=\dfrac{v^2\cdot r}{G}\\M=\dfrac{(3.43\times 10^3)^2\cdot 3.43\times 10^6}{6.67\times 10^{-11}}$

$M=5.944\times 10^{23}\ kg$

5 0

3 years ago