Ask question

Ask question

All categories

3 years ago

13

Dump Tower is 96 stories tall. A small, 1.2-kg object is dropped over the side of the roof of the tower and accelerates toward t

he ground. The mass of the planet housing Dump Tower is 2.7 times the mass of Ganymede and the radius of the body is 1.7 times the radius of Makemake. You will track the object for its entire fall. Each story of this tower is 3.05 meters tall.
How many seconds will it take the object to reach the ground and what will be its impact speed with the ground?

The ball will not drop until your submit your answers.

1 answer:

sattari [20]3 years ago

8 0

Answer:

The time taken by the object to reach the ground, t = 25.04 / √gₓ

The final velocity of the object, v = 23.39 √gₓ

Explanation:

Given data,

The height of the Dump Tower, h = 96 x 3.05

= 292.8 m

The mass of the planet housing Dump Tower, M = 2.7 mₓ

The radius of the planet housing Dump Tower, R = 1.7 rₐ

The acceleration due to the gravity of the planet is,

g = GM/R²

= 2.7 Gmₓ / (1.7 rₐ)²

= 0.934 Gmₓ/rₐ²

g = 0.934 gₓ

Using the II equations of motion,

S = ut + ½ gt²

= 0 + ½ (0.934 gₓ) t²

t = √(2S/ 0.934 gₓ)

= √(2 x 292.8/ 0.934 gₓ)

t = 25.04 / √gₓ

Hence, the time taken by the object to reach the ground, t = 25.04 / √gₓ

Using the I equations of motion

v = u + gt

= 0 + 0.934 gₓ (25.04 / √gₓ)

v = 23.39 √gₓ

Hence, the final velocity of the object, v = 23.39 √gₓ

You might be interested in

2. A 20 cm object is placed 10cm in front of a convex lens of focal length 5cm. Calculate

adoni [48]

Answer:

<u> </u><u>»</u><u> </u><u>Image</u><u> </u><u>distance</u><u> </u><u>:</u>

${ \tt{ \frac{1}{v} + \frac{1}{u} = \frac{1}{f} }} \\$

v is image distance
u is object distance, u is 10 cm
f is focal length, f is 5 cm

${ \tt{ \frac{1}{v} + \frac{1}{10} = \frac{1}{5} }} \\ \\ { \tt{ \frac{1}{v} = \frac{1}{10} }} \\ \\ { \tt{v = 10}} \\ \\ { \underline{ \underline{ \pmb{ \red{ \: image \: distance \: is \: 10 \: cm \: \: }}}}}$

<u> </u><u>»</u><u> </u><u>Magnification</u><u> </u><u>:</u>

• Let's derive this formula from the lens formula:

${ \tt{ \frac{1}{v} + \frac{1}{u} = \frac{1}{f} }} \\$

» Multiply throughout by fv

${ \tt{fv( \frac{1}{v} + \frac{1}{u} ) = fv( \frac{1}{f} )}} \\ \\ { \tt{ \frac{fv}{v} + \frac{fv}{u} = \frac{fv}{f} }} \\ \\ { \tt{f + f( \frac{v}{u} ) = v}}$

• But we know that, v/u is M

${ \tt{f + fM = v}} \\ { \tt{f(1 +M) = v }} \\ { \tt{1 +M = \frac{v}{f} }} \\ \\ { \boxed{ \mathfrak{formular : } \: { \tt{ M = \frac{v}{f} - 1 }}}}$

v is image distance, v is 10 cm
f is focal length, f is 5 cm
M is magnification.

${ \tt{M = \frac{10}{5} - 1 }} \\ \\ { \tt{M = 5 - 1}} \\ \\ { \underline{ \underline{ \pmb{ \red{ \: magnification \: is \: 4}}}}}$

<u> </u><u>»</u><u> </u><u>Nature</u><u> </u><u>of</u><u> </u><u>Image</u><u> </u><u>:</u>

Image is magnified
Image is erect or upright
Image is inverted
Image distance is identical to object distance.

4 0

2 years ago

In a 5000 m race, the athletes run 12 1/2 laps; each lap is 400 m.Kara runs the race at a constant pace and finishes in 17.9 min

Ksju [112]

Answer:

No. of laps of Hannah are 7 (approx).

Solution:

According to the question:

The total distance to be covered, D = 5000 m

The distance for each lap, x = 400 m

Time taken by Kara, $t_{K} = 17.9 min = 17.9\times 60 = 1074 s$

Time taken by Hannah, $t_{H} = 15.3 min = 15.3\times 60 = 918 s$

Now, the speed of Kara and Hannah can be calculated respectively as:

$v_{K} = \frac{D}{t_{K}} = \frac{5000}{1074} = 4.65 m/s$

$v_{H} = \frac{D}{t_{H}} = \frac{5000}{918} = 5.45 m/s$

Time taken in each lap is given by:

$(v_{H} - v_{K})t = x$

$(5.45 - 4.65)\times t = 400$

$t = \frac{400}{0.8}$

t = 500 s

So, Distance covered by Hannah in 't' sec is given by:

$d_{H} = v_{H}\times t$

$d_{H} = 5.45\times 500 = 2725 m$

No. of laps taken by Hannah when she passes Kara:

$n_{H} = \frac{d_{H}}{x}$

$n_{H} = \frac{2725}{400} = 6.8$ ≈ 7 laps

3 0

3 years ago

The Slingshot is a ride for two people. It consists of a single passenger cage, two towers, and two elastic bands. Potential ene

Vitek1552 [10]

my bad I was in a herie last time can you please answer my question , I am going to give you the 5 points back for this question and extra 30 points ,

Explanation:

first its going to say 10 points for my question but after that I well make answer a small question and give you 30points. like whats your favorite color . stay tuned .

8 0

3 years ago

When a metal rod is heated, its resistance changes both because of a change in resistivity and because of a change in the length

monitta

Answer:

$3.34\Omega$

Explanation:

The resistance of a metal rod is given by

$R=\frac{\rho L}{A}$

where

$\rho$ is the resistivity

L is the length of the rod

A is the cross-sectional area

The resistivity changes with the temperature as:

$\rho(T)=\rho_0 (1+\alpha (T-T_0))$

where in this case:

$\rho_0$ is the resistivity of silver at $T_0=21.0^{\circ}C$

$\alpha=6.1\cdot 10^{-3} ^{\circ}C^{-1}$ is the temperature coefficient for silver

$T=180.0^{\circ}C$ is the current temperature

Substituting,

$\rho(180^{\circ}C)=\rho_0 (1+6.1\cdot 10^{-3}(180-21))=1.970\rho_0$

The length of the rod changes as

$L(T)=L_0 (1+\alpha_L(T-T_0))$

where:

$L_0$ is the initial length at $21.0^{\circ}C$

$\alpha_L = 18\cdot 10^{-6} ^{\circ}C^{-1}$ is the coefficient of linear expansion

Substituting,

$L(180^{\circ}C)=L_0(1+18\cdot 10^{-6}(180-21))=1.00286L_0$

The cross-sectional area of the rod changes as

$A(T)=A_0(1+2\alpha_L(T-T_0))$

So, substituting,

$A(180^{\circ}C)=A_0(1+2\cdot 18\cdot 10^{-6}(180-21))=1.00572A_0$

Therefore, if the initial resistance at 21.0°C is

$R_0 = \frac{\rho_0 L_0}{A_0}=1.70\Omega$

Then the resistance at 180.0°C is:

$R(180^{\circ}C)=\frac{\rho(180)L(180)}{A(180)}=\frac{(1.970\rho_0)(1.00285L_0)}{1.00572A_0}=1.9644\frac{\rho_0 L_0}{A_0}=1.9644 R_0=\\=(1.9644)(1.70\Omega)=3.34\Omega$

7 0

3 years ago

Earthquakes and volcanic eruptions are examples of rapid events that are caused by the movement of Earth's _______. Mountain bui

andrew11 [14]

The correct answer is - C. Crustal plates, crustal plates.

The movement of the crustal plates of the Earth is the reason why the geologic processes are occurring on our planet. With their movement, the crustal plates manage to create lot of pressure, open up gaps for the magma from the mantle, as well as make adjustments deeper into the ground.

Because of the adjustments deep into the ground, or rather the crust, lot of force is released, manifested as strong and very quick vibrations, better known as earthquakes.

The gaps that are opening between the plates let the magma reach the surface, thus enabling the volcanic activities on the planet.

The pressure from the plates' collision makes the land lift up slowly, and over time a mountain range starts to form because of the continuous lifting up of the area.

5 0

3 years ago

Read 2 more answers