Yall have helped a lot i just need help on this then ill be done for a while

Physics

1 answer:

Lubov Fominskaja [6]3 years ago

6 0

Answer:

f(x)=a(x - h)2 + k

Much like a linear function, k works like b in the slope-intercept formula. Like where add or subtract b would determine where the line crosses, in the linear, k determines the vertex of the parabola. If you're going to go up 2, then you need to add 2.

The h determines the movement horizontally. what you put in h determines if it moves left or right. To adjust this, you need to find the number to make the parentheses equal 0 when x equals -2 (because moving the vertex point to the left means subtraction/negatives):

x - h = 0

-2 - h = 0

-h = 2

h = -2

So the function ends up looking like:

f(x)=a(x - (-2))2 + 2

Subtracting a negative cancels the signs out to make a positive:

f(x)=a(x + 2)2 + 2Explanation:

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A space probe is fired as a projectile from the Earth's surface with an initial speed of 2.05 104 m/s. What will its speed be wh

Elanso [62]

Answer:

The value is $v = 2.3359 *10^{4} \ m/s$

Explanation:

From the question we are told that

The initial speed is $u = 2.05 *10^{4} \ m/s$

Generally the total energy possessed by the space probe when on earth is mathematically represented as

$T__{E}} = KE__{i}} + KE__{e}}$

Here $KE_i$ is the kinetic energy of the space probe due to its initial speed which is mathematically represented as

$KE_i = \frac{1}{2} * m * u^2$

=> $KE_i = \frac{1}{2} * m * (2.05 *10^{4})^2$

=> $KE_i = 2.101 *10^{8} \ \ m \ \ J$

And $KE_e$ is the kinetic energy that the space probe requires to escape the Earth's gravitational pull , this is mathematically represented as

$KE_e = \frac{1}{2} * m * v_e^2$

Here $v_e$ is the escape velocity from earth which has a value $v_e = 11.2 *10^{3} \ m/s$

=> $KE_e = \frac{1}{2} * m * (11.3 *10^{3})^2$

=> $KE_e = 6.272 *10^{7} \ \ m \ \ J$

Generally given that at a position that is very far from the earth that the is Zero, the kinetic energy at that position is mathematically represented as

$KE_p = \frac{1}{2} * m * v^2$