Ask question

Ask question

All categories

3 years ago

9

A rocket is fired straight up. It contains two stages (Stage 1 and Stage 2) of solid rocket fuel that are designed to burn for 1

0.0 and 5.00 s, respectively, with no time interval between them. In Stage 1, the rocket fuel provides an upward acceleration of 15.0 m/s2. In Stage 2, the acceleration is 10.0 m/s2. Neglecting air resistance, calculate the maximum altitude above the surface of Earth of the payload and the time required for it to return to the surface. Assume the acceleration due to gravity is constant.1) Calculate the maximum altitude.
2) Calculate time required to return to the surface (i.e. the total time of flight).

1 answer:

insens350 [35]3 years ago

6 0

Answer: a) 1875m. (b) 8.66s

Explanation:

You might be interested in

Recall that the spring constant is inversely proportional to the number of coils in the spring, or that shorter springs equate t

ruslelena [56]

Answer:

$x_1= 0.0425m$

Explanation:

Using the tension in the spring and the force of the tension can by describe by

T = kx

, T = mg

Therefore:

$m*g = k*x$

With two springs, let, T1 be the tension in each spring, x1 be the extension of each spring. The spring constant of each spring is 2k so:

$T_1 = 2k*x_1$

$2T_1 = m*g=4k x_1$

Solve to x1

$x_1=\frac{m*g}{4k}$

$x_1=\frac{k*x}{4*k}$

$x_1=\frac{x}{4}$

$x_1 = 0.170 / 4$

$x_1= 0.0425m$

7 0

3 years ago

Saved While paddling a canoe up the river, Jan saw some beautiful flowers along the river bank. The canoe is 35 yards lower than

Veronika [31]

To solve this problem we must use the trigonometric relations, in this case as we have the ramp and the height the concentric trigonometric property to the $sin\theta$ will allow us to find the angle.

The sine of angle A, is defined as

$Sin A = \frac{BC}{AC}$

$Sin \theta = \frac{35}{225}$

$Sin \theta = 8.9\°$

Therefore the angle of elevation is 8.9°

6 0

4 years ago

A 12.0 V battery has as 80.0 Ohm

Vesna [10]

Potential difference=V=12V
Resistance=80ohm=R
Current be I

Using ohms law

$\\ \sf\longmapsto \dfrac{V}{I}=R$

$\\ \sf\longmapsto I=\dfrac{V}{R}$

$\\ \sf\longmapsto I=\dfrac{12}{80}$

$\\ \sf\longmapsto I=0.15A$

6 0

3 years ago

A flea jumps straight up to a maximum height of 0.490 m . What is its initial velocity v0 as it leaves the ground?

zzz [600]

Answer:

$\huge\boxed{\sf v_o=3.1\ m/s}$

Explanation:

<u>Given Data:</u>

Acceleration due to gravity = g = -9.8 m/s²

Maximum Height = h = 0.490 m

At h, $v_f$ = 0

<u>Required:</u>

$v_o=?$

<u>Formula:</u>

$2gh=v_f^2-v_o^2$

<u>Solution:</u>

Put the givens

$2 (-9.8) (0.490) = (0)\² - v_o^2\\\\-9.604=-v_o^2\\\\9.604=v_o^2\\\\Take \ sqrt\ on \ both \ sides\\\\\sqrt{9.604}=v_o^2\\\\3.1 \ m/s=v_o\\\\v_o=3.1\ m/s\\\\\rule[225]{225}{2}$

3 0

2 years ago

A cell has an internal resistance of 0.02ohms and e.m.f of 2.0v calculate it's terminal p.d if it's delivers (a)5A ( b)50A

Aleksandr-060686 [28]

Answer:

(a) The terminal voltage of the cell is <u>1.9 V.</u>

(b) The terminal voltage of the cell is <u>1.0 V.</u>

Explanation:

Given:

(a)

E.M.F of the cell (E) = 2.0 V

Internal resistance of the cell (r) = 0.02 ohms

Current passing through the cell (I) = 5 A

Now, the potential difference across the terminals of the cell is given as:

$V=E-Ir$

Plug in the given values and solve for 'V'. This gives,

$V=2.0-(5\times 0.02)\\\\V=2.0-0.1=1.9\ V$

Therefore, the terminal voltage of the cell is 1.9 V.

(b)

E.M.F of the cell (E) = 2.0 V

Internal resistance of the cell (r) = 0.02 ohms

Current passing through the cell (I) = 50 A

Now, the potential difference across the terminals of the cell is given as:

$V=E-Ir$

Plug in the given values and solve for 'V'. This gives,

$V=2.0-(50\times 0.02)\\\\V=2.0-1.0=1.0\ V$

Therefore, the terminal voltage of the cell is 1.0 V.

4 0

3 years ago