A toy rocket is launched straight up by using a spring. The rocket is initially pressed down on the spring so that the spring is compressed by 9 cm. If the spring constant is 1050 N/m and the mass of the rocket is 50 g, how high will the rocket go? You may neglect the effects of air resistance.

Answers

Answer 1
Answer:

Answer:

Rocket will go to a height of 8.678 m

Explanation:

Mass of the rocket m = 50 gram = 0.05 kg

Spring constant k = 1050 N /m

Spring is stretched to 9 cm

So x = 0.09 m

Work done in stretching the spring

E=(1)/(2)kx^2=(1)/(2)* 1050* 0.09^2=4.2525J

From energy conservation this energy will convert into potential energy

Potential energy is equal to E=mgh, here m is mass, g is acceleration due to gravity and h is height

So 0.05* 9.8* h=4.2525

h=8.678m

So rocket will go to a height of 8.678 m

Answer 2
Answer:

Answer:

8.68 m

Explanation:

compression in spring, x = 9 cm = 0.09 m

Spring constant, K = 1050 N/m

mass of rocket, m = 50 g = 0.05 kg

Let it go upto height h.

Use conservation of energy

Potential energy stored in spring = potential energy of the rocket

(1)/(2)kx^(2)=mgh

0.5 x 1050 x 0.09 x 0.09 = 0.05 x 9.8 x h

h = 8.68 m

Thus, the rocket will go upto height 8.68 m.


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A 3.00N rock is thrown vertically into the air from the ground. At h=15.0m, v=25m/s upward. Use the work-energy theorem to find the initial speed of the rock.a. 3m/sb. 30.3 m/sc. None of the above
Coherent light with wavelength 598 nm passes through two very narrow slits, and the interference pattern is observed on a screen a distance of 3.00 m from the slits. The first-order bright fringe is a distance of 4.84 mm from the center of the central bright fringeFor what wavelength of light will thefirst-order dark fringe be observed at this same point on thescreen?Express your answer in micrometers(not in nanometers).
A 6.0-cm-diameter horizontal pipe gradually narrows to 4.0 cm. When water flows through this pipe at a certain rate, the gauge pressure in these two sections is 32.0 kPa and 24.0 kPa, respectively. What is the volume rate of flow?
(a) Is the velocity of car A greater than, less than, or the same as thevelocity of car B?(b) Is the initial position of car A greater than, less than, or equal to theinitial position of car B?(c) In the time period from t = 0 tot = 1 s, is car A ahead of car B,behind car B, or at the same position as car B?

Light emitted by element X passes through a diffraction grating that has 1200 slits/mm. The interference pattern is observed on a screen 75.0 cm behind the grating. First-order maxima are observed at distances of 56.2 cm, 65.9 cm, and 93.5 cm from the central maximum. What are the wavelengths of light emitted by element X?

Answers

Answer:

Explanation:

Width of slit = 10⁻³ / 1200

d  = 8.3 x 10⁻⁷ m

First order maxima will be observed at

x = λD/d

D = 75 cm = 75 x 10⁻² m

56.2 x 10⁻² = λ₁D/d

= λ₁ x 75 x 10⁻² / 8.3 x 10⁻⁷

λ₁ = 56.2 x 8.3 x 10⁻⁷ / 75

= 6.219 x  10⁻⁷ m

= 6219 A

Similarly

λ₂ = 65.9 x 8.3 x 10⁻⁷ / 75

= 7293 A

λ₃ =  93.5 x 8.3 x 10⁻⁷ / 75

= 10347 A

Convert this measurement
664.2 km=____cm

Answers

(664.2 km) · (1,000 m/km) · (100 cm/m) =

(664.2 · 1,000 · 100) (km·m·cm/km·m) =

66,420,000 cm

For metric conversion, you can remember this acronym for help:

King Henry died unusually drinking chocolate milk. Which stand for:

Kilo - unit * 1000

Hecto - unit * 100

Deca - unit * 10

Unit - unit * 1

Deci - unit * (1)/(10)

Centi - unit * (1)/(100)

Milli - unit * (1)/(100)

Kilometers and centimeters are five places apart apart, so you move the decimal point in 664.2 to the right five times, which means 664.2 km = 66420000 cm.

To avoid confusion on which direction to move the decimal point, imagine two shapes on each end of a scale. On each end, there is one large shape and one small shape. There has to be one of each on either side for it to balance. For this problem, a kilometer is a larger unit than a centimeter, so this means that the blank space needs to have a number greater than 664.2, or else the scale won't balance. Hope this helped.

A vertical straight wire carrying an upward 24-A current exerts an attractive force per unit length of 88 X 104N/m on a second parallel wire 7.0 cm away. What current (magnitude and direction) flows in the second wire?

Answers

Answer:

The current flows in the second wire is 1.3*10^(10)\ A

Explanation:

Given that,

Upward current = 24 A

Force per unit length(F)/(l) =88*10^(4)\ N/m

Distance = 7.0 cm

We need to calculate the current in second wire

Using formula of magnetic force

F=ILB

(F)/(l)=(\mu I_(1)I_(2))/(2\pi r)

Where,

(F)/(l)=force per unit length

I₁= current in first wire

I₂=current in second wire

r = distance between the wires

Put the value into the formula

88*10^(4)=(4\pi*10^(-7)*24* I_(2))/(2\pi *7*10^(-2))

I_(2)=(88*10^(4)*7*10^(-2))/(2**10^(-7)*24)

I_(2)=1.3*10^(10)\ A

Hence, The current flows in the second wire is 1.3*10^(10)\ A

A Venturi tube may be used as the inlet to an automobile carburetor. If an inlet pipe with a diameter of 2.0 cm diameter narrows to diameter of 1.0 cm, determine the pressure drop in the constricted section for an initial airflow of 3.0 m/s in the 2-cm section? (Assume air density is 1.25 kg/m

Answers

The pressure drop is equal to 80.99 Pa

Given information:

d1 = 2 cm = 0.02 m

d2 = 1 cm = 0.01 m

v = 3 m/s

p = 1.25 kg/m^3

Here we use Bernoulli's principle for the Venturi Tube:

Calculation of pressure drop:

P1 - P2 = ((v^2* p)/ 2)* ((A1^2/ A2^2)-1)\n\nP1 - P2 = \Delta P = ((v1^2* p)/ 2)* ((A1^2/ A2^2)-1)

Now the following formula for area calculation should be used:

A1 = (\pi* d1^2)/ 4 = (\pi* (0.02 m)^2)/ 4 = 0.00031 m^2\n\nA2 = (\pi* (0.01 m)^2)/ 4 = 0.000079 m^2\n\n\Delta P = ((3 m/s)^2 *1.25 kg/m^3)/ 2) * ((0.00031 m^2)^2/(0.000079 m^2)^2)-1)

= 80.99

Find out more information about the  Pressure here: brainly.com/question/356585?referrer=searchResults

Answer:

the pressure drop is equal to 80.99 Pa

Explanation:

we have the following data:

d1 = 2 cm = 0.02 m

d2 = 1 cm = 0.01 m

v = 3 m/s

p = 1.25 kg/m^3

ΔP = ?

For the calculation of the pressure drop we will use Bernoulli's principle for the Venturi Tube:

P1 - P2 = ((v^2*p)/2)*((A1^2/A2^2)-1)

where A = area

P1 - P2 = ΔP = ((v1^2*p)/2)*((A1^2/A2^2)-1)

for the calculation of the areas we will use the following formula:

A1 = (pi*d1^2)/4 = (pi*(0.02 m)^2)/4 = 0.00031 m^2

A2 = (pi*(0.01 m)^2)/4 = 0.000079 m^2

ΔP = ((3 m/s)^2*1.25 kg/m^3)/2)*((0.00031 m^2)^2/(0.000079 m^2)^2)-1) = 80.99 N/m^2 = Pa

You perform a double‑slit experiment in order to measure the wavelength of the new laser that you received for your birthday. You set your slit spacing at 1.09 mm and place your screen 8.61 m from the slits. Then, you illuminate the slits with your new toy and find on the screen that the tenth bright fringe is 4.53 cm away from the central bright fringe (counted as the zeroth bright fringe). What is your laser's wavelength ???? expressed in nanometers?

Answers

Answer:

λ = 5.734 x 10⁻⁷ m = 573.4 nm

Explanation:

The formula of the Young's Double Slit experiment is given as follows:

\Delta x = (\lambda L)/(d)\n\n\lambda = (\Delta x d)/(L)

where,

λ = wavelength = ?

L = distance between screen and slits = 8.61 m

d = slit spacing = 1.09 mm = 0.00109 m

Δx = distance between consecutive bright fringes = (4.53\ cm)/(10) = 0.00453 m

Therefore,

\lambda = ((0.00453\ m)(0.00109\ m))/(8.61\ m)

λ = 5.734 x 10⁻⁷ m = 573.4 nm

The position of a particle moving along the x-axis depends on the time according to the equation x = ct2 - bt3, where x is in meters, t in seconds, and c and b are positive constants. What are the units of (a) constant c and (b) constant b? Find a formula in terms of c, b, and t of the (c) velocity v and (d) acceleration a. (e) At what time t ≥ 0 does the particle reach its maximum x value?

Answers

Answer:

(a):  \rm meter/ second^2.

(b):  \rm meter/ second^3.

(c):  \rm 2ct-3bt^2.

(d):  \rm 2c-6bt.

(e):  \rm t=(2c)/(3b).

Explanation:

Given, the position of the particle along the x axis is

\rm x=ct^2-bt^3.

The units of terms \rm ct^2 and \rm bt^3 should also be same as that of x, i.e., meters.

The unit of t is seconds.

(a):

Unit of \rm ct^2=meter

Therefore, unit of \rm c= meter/ second^2.

(b):

Unit of \rm bt^3=meter

Therefore, unit of \rm b= meter/ second^3.

(c):

The velocity v and the position x of a particle are related as

\rm v=(dx)/(dt)\n=(d)/(dx)(ct^2-bt^3)\n=2ct-3bt^2.

(d):

The acceleration a and the velocity v of the particle is related as

\rm a = (dv)/(dt)\n=(d)/(dt)(2ct-3bt^2)\n=2c-6bt.

(e):

The particle attains maximum x at, let's say, \rm t_o, when the following two conditions are fulfilled:

  1. \rm \left ((dx)/(dt)\right )_(t=t_o)=0.
  2. \rm \left ( (d^2x)/(dt^2)\right )_(t=t_o)<0.

Applying both these conditions,

\rm \left ( (dx)/(dt)\right )_(t=t_o)=0\n2ct_o-3bt_o^2=0\nt_o(2c-3bt_o)=0\nt_o=0\ \ \ \ \ or\ \ \ \ \ 2c=3bt_o\Rightarrow t_o = (2c)/(3b).

For \rm t_o = 0,

\rm \left ( (d^2x)/(dt^2)\right )_(t=t_o)=2c-6bt_o = 2c-6\cdot 0=2c

Since, c is a positive constant therefore, for \rm t_o = 0,

\rm \left ( (d^2x)/(dt^2)\right )_(t=t_o)>0

Thus, particle does not reach its maximum value at \rm t = 0\ s.

For \rm t_o = (2c)/(3b),

\rm \left ( (d^2x)/(dt^2)\right )_(t=t_o)=2c-6bt_o = 2c-6b\cdot (2c)/(3b)=2c-4c=-2c.

Here,

\rm \left ( (d^2x)/(dt^2)\right )_(t=t_o)<0.

Thus, the particle reach its maximum x value at time \rm t_o = (2c)/(3b).