# 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.

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

From energy conservation this energy will convert into potential energy

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

So

So rocket will go to a height of 8.678 m

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

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.

## Related Questions

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?

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

(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 *

Centi - unit *

Milli - unit *

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?

The current flows in the second wire is

Explanation:

Given that,

Upward current = 24 A

Force per unit length

Distance = 7.0 cm

We need to calculate the current in second wire

Using formula of magnetic force

Where,

=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

Hence, The current flows in the second wire is

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

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:

Now the following formula for area calculation should be used:

= 80.99

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?

λ = 5.734 x 10⁻⁷ m = 573.4 nm

Explanation:

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

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 = = 0.00453 m

Therefore,

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?

(a):

(b):

(c):

(d):

(e):

Explanation:

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

The units of terms and should also be same as that of x, i.e., meters.

The unit of t is seconds.

(a):

Unit of

Therefore, unit of

(b):

Unit of

Therefore, unit of

(c):

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

(d):

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

(e):

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

Applying both these conditions,

For ,

Since, c is a positive constant therefore, for ,

Thus, particle does not reach its maximum value at

For ,

Here,

Thus, the particle reach its maximum x value at time