Differentiate between a derived quantity and a derived unit.

A 460 g model rocket is on a cart that is rolling to the right at a speed of 3.0 m/s. The rocket engine, when it is fired, exert

ollegr [7]

Answer:

The rocket should be launched at a horizontal distance of 6.45 m left of the loop.

Explanation:

Given:

Mass of the rocket model (m) = 460 g = 0.460 kg [1 g = 0.001 kg]

Speed of the cart (v) = 3.0 m/s

Thrust force by the rocket engine (F) = 8.5 N

Vertical height of the loop (y) = 20 m

Let the horizontal distance left to the loop for launch be 'x'. Also, let 't' be the time taken by the rocket to reach the loop.

Now, there are two types of motion associated with the rocket- one is horizontal and the other vertical.

So, we will apply kinematics of motion in the two directions separately.

Vertical motion:

Given:

Force acting in the vertical direction is given as:

$F_y=F-mg=8.5-0.46\times 9.8=3.992\ N$

So, acceleration in the vertical direction is given as:

Acceleration = Force ÷ mass

$a_y=\frac{F_y}{m}=\frac{3.992\ N}{0.46\ kg}=8.678\ m/s^2$

Vertical displacement of rocket is same as the height of loop. So, $y=20\ m$

There is no initial velocity in the vertical direction. So, $u_y=0\ m/s$

Now, applying equation of motion in vertical direction. we have:

$y=u_yt+\frac{1}{2}a_yt^2\\\\20=0+\frac{1}{2}\times 8.678t^2\\\\20=4.339t^2\\\\t^2=\frac{20}{4.339}\\\\t=\sqrt{\frac{20}{4.339}}=2.15\ s$

Now, time taken to reach the loop is 2.15 s.

Horizontal motion:

There is no acceleration in the horizontal motion. So, displacement in the horizontal direction is equal to the product of horizontal speed and time.

Also, displacement of the rocket in the horizontal direction is nothing but the horizontal distance of its launch left of the loop. So,

$x=vt\\\\x=3.0\ m/s\times 2.15\ s\\\\x=6.45\ m$

Therefore, the rocket should be launched at a horizontal distance of 6.45 m left of the loop.

5 0

3 years ago

What is the frequency of an electromagnetic wave that has a wavelength of 4,000m?

NeTakaya

Answer:

f=75000hz

Explanation:

fλ= c

f= c/λ

f = 3.0×10^8/4000

f=75000hz

7 0

3 years ago

An empty water tank has a rectangular base with side lengths of 1.2 m and 2.3 m. The weight of the tank applies a pressure of 35

LekaFEV [45]

Answer:

Weight = 966 Newton.

Explanation:

Given the following data;

Length = 1.2 m

Width = 2.3 m

Pressure = 350 Pa

To find the weight of the tank;

We know that weight is the force of gravity acting on an object multiplied by its mass.

Weight = mg = force

Hence, we would determine the force using the parameters that were given.

But we would first determine the area of the rectangular tank.

Area of rectangle, A = length * width

A = 1.2 * 2.3

A = 2.76 m²

Mathematically, pressure is given by the formula;

Pressure = force/area

Force = pressure * area

Substituting into the formula, we have;

Force = 2.76 * 350

Force = 966 Newton

Therefore, the weight of the tank is 966 Newton.

5 0

3 years ago

Space Station Suppose a space-station is designed in s shape of a torus such as the one depicted in Stanley Kubrick's "2001: A s

yaroslaw [1]

Answer:

w = 3.2 rev / min

Explanation:

For this exercise we will use the centrine acceleration equal to the acceleration of gravity

a = v² / r

Angular and linear variables are related.

v = w r

Let's replace

a = w² r = g

w = √ g / r

r = d / 2

r = 175/2 = 87.5 m

w = √( 9.8 / 87.5)

w = 0.3347 rad / s

Let's reduce to rotations per min

w = 0.3347 rad / s (1 rov / 2pi rad) (60 s / 1 min)

w = 3.2 rev / min

Suppose the space station rotates counterclockwise, we have two possibilities for the car

The first car turns counterclockwise (same direction of the station

$v_{c}$ = $w_{c}$ r

[texwv_{c}[/tex] = $v_{c}$ / r

[texwv_{c}[/tex] = 25.0 / 87.5

[texwv_{c}[/tex] = 0.286 rad / s

When the two rotate in the same direction their angular speeds are subtracted

w total = w -[texwv_{c}[/tex]

w total = 0.3347 - 0.286

w total= 0.487 rad / s

The car goes in the opposite direction of the station the speeds add up

w = 0.3347 + 0.286

w = 0.62 rad / s

From this values we can see that the person feels a variation of the acceleration of gravity, feels that he has less weight when he goes in the same direction of the season and that his weight increases when he goes in the opposite direction to the season.

3 0

4 years ago

An object is dropped from a height h. during the final second of its fall, it traverses a distance of 41.2 m. what was h?

Leya [2.2K]

41.2 = h-1/2g(t-1)^2
 {-h = -1/2gt^2-1/2g+g*t-41.2
 {h = 1/2gt^2
 summing them up
 0 = -1/2g+g*t-41.2
 41.2 +4.9 = g*t
 t = 46.1/9.8 = 4.70 sec
 h = 1/2gt^2 =4.9*(4.70^2) = 108.241 m

3 0

3 years ago