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2 years ago

An object that is speeding up is accelerating. True False

Physics

1 answer:

Stels [109]2 years ago

3 0

True. The speed of any object will have faster acceleration, and eventually slow down due to gravity.

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Douglas has a segment with endpoints I(5, 2) and J(9, 10) that is divided by a point K such that IK and KJ form a 2:3 ratio. He

Rufina [12.5K]

For the answer to the question above,
the distance from i to j is 5 parts
(2 parts from i to k and 3 parts from k to j)

The y distance from i to j is
10 - 2 = 8

Each part is 8/5 = 1.6
Therefore the distance between the 2 parts from i to k is 3.2

From the y coordinate of I which is 2 plus the 3.2 to point k
2 + 3.2 = 5.2

Answer y =5.2

Now just convert that to fraction and that will be the answer

3 0

2 years ago

Trick question (really easy) just for fun! If u get this right u get brainliest!

Maurinko [17]

Answer:

piper duh

Explanation:

6 0

2 years ago

Read 2 more answers

The train passes point A with a speed of 30 m/s and begins to decrease its speed at a constant rate of at = - 0.25 m/s^2. Determ

prisoha [69]

Explanation:

At point B, the velocity speed of the train is as follows.

$\nu^{2}_{B} = \nu^{2}_{A} + 2a_{t} (s_{B} - s_{A})$

= $(30)^{2} + 2(-0.25(412 - 0))$

= 26.34 m/s

Now, we will calculate the first derivative of the equation of train.

y = $200 e^{\frac{x}{1000}}$

$\frac{dy}{dx} = 0.2 e^{\frac{x}{1000}}$

Now, second derivative of the train is calculated as follows.

$\frac{dy}{dx} = 0.2 e^{\frac{x}{1000}}$

$\frac{d^{2}y}{dx^{2}} = 0.2 (10^{-3}) e^{\frac{x}{1000}}$

Radius of curvature of the train is as follows.

$\rho = \frac{[1 + (\frac{dy}{dx})^{2}]^{\frac{3}{2}}}{\frac{d^{2}y}{dx^{2}}}$

= $\frac{[1 + 0.2e^{\frac{400}{1000}}^{2}]^{\frac{3}{2}}}{0.2(10^{-3})e^{\frac{400}{1000}}}$

= 3808.96 m

Now, we will calculate the normal component of the train as follows.

$a_{n} = \frac{\nu^{2}_{B}}{\rho}$

= $\frac{(26.34)^{2}}{3808.96}$

= 0.1822 $m/s^{2}$

The magnitude of acceleration of train is calculated as follows.

a = $\sqrt{(a_{t})^{2} + (a_{n})^{2}}$

= $\sqrt{(-0.25)^{2} + (0.1822)^{2}}$

= $0.309 m/s^{2}$

Thus, we can conclude that magnitude of the acceleration of the train when it reaches point B, where sAB = 412 m is $0.309 m/s^{2}$ .

6 0

3 years ago

A vector has an x-component of length 10 and a y-component of length 3. What is the angle of the vector? (Hint: Use the inverse

Sonja [21]

The vector, the x-component and the y-component form a rectangle triangle where the vector is the hypothenuse and the x and y components are the two sides.
Calling $\alpha$ the angle between the vector and the horizontal direction (x), the two sides are related to $\alpha$ by
$\tan \alpha = \frac{v_y}{v_x}$
where vy and vx are the two components on the y- and x-axis. Using vx=10 and vy=3 we find
$\tan \alpha = \frac{3}{10} =0.3$
And so the angle is
$\alpha = \arctan (0.3)=16.7^{\circ}$

5 0

3 years ago

What is an element that needs to lose or gain<br> 4 valence electrons in order to become<br> stable?

tiny-mole [99]

Answer:

Elements in Group 14 could lose four, or gain four electrons to achieve a noble gas structure. In fact, if they are going to form ions, Group 14 elements form positive ions. Carbon and silicon form covalent bonds. Carbon's millions of organic compounds are all based on shared electrons in covalent bonds.

Explanation:

4 0

3 years ago