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

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

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

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.

Which of the following organisms has an adaptation that will allow it to survive in tundra biome? *A.)Plants with roots that are short and grows sideways with hairy stems and small leaves.B.)Plants that have broad leaves to capture sunlight and long roots to penetrate the soil.C.)Animals with thin fur that allows them to get rid of heat efficiently.D.)Animals with long tongues for capturing prey and sticky pads for climbing trees.

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?

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?

**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

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:

**K**ing **H**enry **d**ied** u**nusually **d**rinking **c**hocolate **m**ilk. 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.

**Answer:**

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 **

The **pressure** drop is equal to** 80.99 Pa**

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:**

Now the following formula for **area** calculation should be used:

= 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

**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:

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**

**Answer:**

(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 **