How much energy .....

Explain the causes of atmospheric pressure and why it changes with altitude?

Verdich [7]

Answer:

As altitude increases, the amount of gas molecules in the air decreases—the air becomes less dense than air nearer to sea level. This is what meteorologists and mountaineers mean by "thin air." Thin air exerts less pressure than air at a lower altitude

4 0

2 years ago

A 600 g model rocket is on a cart that is rolling to the right at a speed of 4.0 m/s. The rocket engine, when it is fired, exert

murzikaleks [220]

Answer:

The rocket should be launched when the cart is 13.48m away from a point directly below the hoop.

Explanation:

Step 1: Data given

mass of the rocket = 600 grams

speed = 4.0 m/s

Step 2: Calculate weight

Fw = mg

with Fw = the weight (in Newton)

with m = the mass (in kg)

with g = the acceleration due to gravity (9.81 m/s²).

Fw = (0.600 kg)(9.81 m/s²) = 5.886 N

Step 3: Calculate force available to provide acceleration

The rocket engine, when it is fired, exerts a 8.0 AND vertical thrust on the rocket.

5.886 N of that force will be used to counter the rocket's weight, leaving 2.114 N of force available to provide acceleration.

Step 4: Calculate the rocket's upward acceleration:

Fnet = m*a

With Fnet = the net force (the force that remains after the rocket's weight is compensated)

with a = the rocket's acceleration (in m/s²)

2.114 N = (0.600 kg)*a

a = 3.52 m/s² = the rocket's upward acceleration

Step 5: Calculate how long it will take to rise 20 meters into the air.

Δy = v0*t + 1/2 at²

with v0 = 0m/s

Δy = 1/2 at²

20 m = 1/2(3.52)t²

20 m = (1.76 m/s²)t²

11.36 = t²

t = 3.37 s

This means the rocket will take 3.37 seconds to reach the hoop. It should be launched when the cart is 3.37 seconds away from being directly beneath the hoop.

Step 6: Calculate the distance

v = Δx / t

4.0 m/s = Δx / 3.37 s

Δx = 13.48 m

The rocket should be launched when the cart is 13.48m away from a point directly below the hoop.

8 0

4 years ago

Help, due in 10 minutes. Really confused. Thank you.

alexandr402 [8]

Answer: animal 4

Explanation: I think it is correct

6 0

3 years ago

Assume it takes 10 J to stretch a spring 10 cm beyond its natural length. Find the work required (in Joules) to stretch the spri

Lady bird [3.3K]

Answer:

W = 30 J

Explanation:

given,

Work done = 10 J

Stretch of spring, x = 0.1 m

We know,

dW = F .dx

we know, F = k x

$\int dW = \int_0^{0.1} k.x dx$

$W = \int_0^{0.1} k.x dx$

$W = k[\dfrac{x^2}{2}]_0^{0.1}$

$10 = k\dfrac{0.1^2}{2}$

k = 2000

now, calculating Work done by the spring when it stretched to 0.2 m from 0.1 m.

$W = \int_{0.1}^{0.2} 2000 x dx$

$W = 2000 [\dfrac{x^2}{2}]_{0.1}^{0.2} dx$

W = 1000 x 0.03

W = 30 J

Hence, work done is equal to 30 J.

4 0

3 years ago

If the magnitude of the moment of F about line CD is 57 N·m, determine the magnitude of F.If the magnitude of the moment of F ab

bazaltina [42]

Answer:

F_ab = 260.17 N

Explanation:

Given:

- Moment of force F about CD, (M)_cd = 57 Nm

Find:

- First we will write down the position vectors of points A, B , C , D:

- We will take left and bottom most corner of cube to be the origin.

- The unit vectors i , j , k are along vertical planes and outside the plane, respectively.

- The position vectors wrt to the origin are:

Point A = 0.2 k

Point B = 0.4 i + 0.2 j

Point C = 0.2 j + 0.4 k

Point D = 0.4 i + 0.4 k

- Now we will determine the Force vector F_ab along vector AB.

vec (AB) = B - A = 0.4 i + 0.2 j - 0.2 k

unit (AB) = 0.4 i + 0.2 j - 0.2 k / sqrt ( 0.4^2 + 2*0.2^2)

= [5 / sqrt(6)] * ( 0.4 i + 0.2 j - 0.2 k )

Hence,

vec(F_ab) = Fab*[5 / sqrt(6)] * ( 0.4 i + 0.2 j - 0.2 k )

- Now, form a unit vector along the line CD:

vec(CD) = D - C = 0.4 i - 0.2 j

unit (CD) = 0.4 i - 0.2 j / sqrt ( 0.4^2 + 0.2^2)

= [sqrt(5)]*(0.4 i - 0.2 j)

- Now select a point on line CD, lets say C. Find the moment arm from line of action of force along AB and line CD. Hence, vector AC is:

vec(AC) =r_ac = C - A = 0.2 j + 0.2 k

- Now the moment about a line CD due to force is:

(M)_cd = unit(CD) . ( r_ac x vec(F_ab) )

The cross product of r_ac and vec(F_ab) is as follows:

(M)_c = ( r_ac x vec(F_ab) ) :

$\left[\begin{array}{ccc}i&j&k\\0&0.2&0.2\\0.8165&0.40824&-0.40824\end{array}\right]$

(M)_c = F_ab[- sqrt(6)/15 i + sqrt(6)/15 j - sqrt(6)/15 k]

The dot product of (M)_c and unit (CD) is as follows:

(M)_cd = unit(CD) . (M)_c :

(M)_cd = F_ab[- sqrt(6)/15 i + sqrt(6)/15 j - sqrt(6)/15 k] . [sqrt(5)]*(0.4 i - 0.2 j)

(M)_cd = F_ab*(sqrt(30) / 25)

- The given magnitude of the moment is (M)_cd. Calculate F_ab:

57 = F_ab*(sqrt(30) / 25)

F_ab = 260.17 N

7 0

4 years ago