I don't get this stuff!!! help me!!!

A net force of -1,000 Newtons is delivered to the object over a time of .02 seconds. Calculate the new velocity of the object.

MariettaO [177]

The change in velocity (v₂ - v₁) is

 (-20) / (the object's mass).

Call it a crazy hunch, but I can't shake the feeling that there was more
to the question before the part you copied, that mentioned the object's
mass, and its velocity before this force came along.

4 0

3 years ago

A ball of gas becomes a ____ when nuclear fusion begins in its core.

Ira Lisetskai [31]

The answer to this is Protostar.

This is a process where it is gathering mass from its parent molecular cloud. Its a very young star meaning, the star was now born.

Hope this helped :)
Have a great day

8 0

3 years ago

What are the scientific mesurments ordered from greatest to least

Sveta_85 [38]

Meters Micrometers centimeters millimeters

5 0

3 years ago

A ball is thrown vertically upward from the top of a building 112 feet tall with an initial velocity of 96 feet per second. The

nordsb [41]

Answer:

$t=6.96s$

Explanation:

From this exercise, our knowable variables are hight and initial velocity

$v_{oy}=96ft/s$

$y_{o}=112ft$

To find how much time does the ball strike the ground, we need to know that the final position of the ball is y=0ft

$y=y_{o}+v_{oy}t+\frac{1}{2}gt^{2}$

$0=112ft+(96ft/s)t-\frac{1}{2}(32.2ft/s^{2})t^{2}$

Solving for t using quadratic formula

$t=\frac{-b±\sqrt{b^{2}-4ac } }{2a}$

$a=-\frac{1}{2} (32.2)\\b=96\\c=112$

$t=-0.999s$ or $t=6.96s$

Since time can't be negative the answer is t=6.96s

7 0

3 years ago

The dragster has a mass of 1.3 Mg and a center of mass at G. A parachute is attached at C provides a horizontal braking force of

adell [148]

Answer:

The deceleration of the dragster upon releasing the parachute such that the wheels at B are on the verge of leaving the ground is 16.33 m/s²

Explanation:

The additional information to the question is embedded in the diagram attached below:

The height between the dragster and ground is considered to be 0.35 m since is not given ; thus in addition win 0.75 m between the dragster and the parachute; we have: (0.75 + 0.35) m = 1.1 m

Balancing the equilibrium about point A;

F(1.1) - mg (1.25) = $ma_a (0.35)$