Solve this equation using a common base

A spacecraft is traveling with a velocity of v0x = 5320 m/s along the +x direction. Two engines are turned on for a time of 739

OlgaM077 [116]

Answer:

The velocities after 739 s of firing of each engine would be 6642.81 m/s in the x direction and 5306.02 in the y direction

Step-by-step explanation:

For a constant acceleration: $v_{f}=v_{0}+at$ , where $v_{f}$ is the final velocity in a direction after the acceleration is applied, $v_{0}$ is the initial velocity in that direction before the acceleration is applied, a is the acceleration applied in such direction, and t is the amount of time during where that acceleration was applied.
<em>Then for the x direction</em> it is known that the initial velocity is $v_{0x} =$ 5320 m/s, the acceleration (the applied by the engine) in x direction is $a_{x}$ 1.79 m/s2 and, the time during the acceleration was applied (the time during the engines were fired) of the is 739 s. Then: $v_{fx}=v_{0x}+a_{x}t=5320\frac{m}{s} +1.79\frac{m}{s^{2} }*739s=6642.81\frac{m}{s}$
In the same fashion, <em>for the y direction</em>, the initial velocity is $v_{0y} =$ 0 m/s, the acceleration in y direction is $a_{y}$ 7.18 m/s2, and the time is the same that in the x direction, 739 s, then for the final velocity in the y direction: $v_{fy}=v_{0y}+a_{y}t=0\frac{m}{s} +7.18\frac{m}{s^{2} }*739s=5306.02\frac{m}{s}$

8 0

3 years ago

The shaded sector above covers 1/3 of the circle. If the radius of the circle above is 3 cm, what is the area of the sector in t

Mekhanik [1.2K]

Given:

The shaded sector above covers $\dfrac{1}{3}$ of the circle.

Radius of the circle = 3 cm

To find:

The area of the sector in terms of π.

Solution:

The area of a circle is

$A=\pi r^2$

Substituting $r=3$ , we get

$A=\pi (3)^2$

$A=9\pi$

It is given that the shaded sector above covers $\dfrac{1}{3}$ of the circle.

The area of shaded sector $=\dfrac{1}{3}A$

$=\dfrac{1}{3}(9\pi)$

$=3\pi$

Therefore, the area of shaded sector is 3π sq. cm.

5 0

3 years ago

What is the equation of a line, in general form, that passes though points (-1,2) and (5,2)?

Tom [10]

Answer:

A

Step-by-step explanation:

To write the equation of the line. first calculate the slope using the slope formula.

$m = \frac{y_2-y_1}{x_2-x_1} = \frac{2-2}{5--1}= \frac{0}{6} = 0$

Since the slope of this line is 0, it is horizontal and has the form y=b where b is the y-coordinate. So y = 2 is the equation. In general form, it would be y-2 = 0

3 0

3 years ago

Someone pls help:( I can’t

max2010maxim [7]

Answer:

x = 8 is the answer :)

Step-by-step explanation:

Slope(m) = -2/5

slope is passing through the points: (x,1) and (3,3)

$m=\frac{y_2-y_1}{x_2-x_1}$

$m = \frac{3-1}{3-x}$

$m=\frac{2}{3-x}$

$-\frac{2}{5} =\frac{2}{3-x}$

-2(3 - x) = 2 × 5

-6 + 2x = 10

2x = 10 + 6

2x = 16

x = 8

Hopefully, this answer helped :)

8 0

3 years ago

PLEASE HELP!!!!!! DUE!!!!!!!!!

ollegr [7]

Answer:

y = 1/4x - 2

Step-by-step explanation:

Line whose perpendicular which we have to find : 8x+2y=14

A line which is in the form of ax + by = c

has slope - a / b

Thus slope for 8x+2y=14 is -8/2 = -4.

_______________________________________

Slope of two lines with slope m1 and m2 which are perpendicular to each other have the slope relation as

m1*m2 = -1.

here m1 is -4. we have to fine m2

thus

-4*m2 = -1

=> m2 = -1/-4 = 1/4

_________________________________________

Equation of slope intercept form line is y = mx + c

c is the intercept of line on y axis

and m is the slope of line.

Thus required line has equation

y = mx + c

m we have calculated as 1/4

thus

y = 1/4x + c

Given that this line passes through (−4,−3). Hence, this point will satify the equation y = 1/4x + c.

Substituting value of x and y with coordinates we have (−4,−3)

-3 = (1/4) * (-4) + c

-3 = -1 + c

=> c = -3 + 1

=> c = -2

y = 1/4x + c , substituting the value c calculated

y = 1/4x - 2

Thus , equation of line perpendicular to the line 8x+2y=14 and passing through point (−4,−3) is y = 1/4x - 2.

3 0

3 years ago