Multiply.<br> (2x + 4)(x

A marble is flicked off the edge of a table. The motion of the marble follows a parabolic path.

dem82 [27]

When marble is flicked off the edge of a table then the at the instant it will move off the table the velocity is horizontal while due to force of gravitation the marble will start accelerating downwards.

So here we will say that the velocity of marble will remain constant through out the motion in X direction as there is no force in that direction

While in Y direction the marble will have gravitational force due to which marble will accelerate downwards

So the path of the marble will be parabolic path

hence the correct answer will be

<em>Only one continuous force is acting on the marble.</em>

5 0

4 years ago

Which of the following statements is true? 1 mL = 1 g 1 g = 1 oz 1 mL = 1 cm3 1 g = 1 cm

Nikitich [7]

The first statement is true: 1 mL = 1 g

6 0

3 years ago

Read 2 more answers

2. An electric motor transfers 2003 in 10.0 s. If it then transfers the same amount of energy in 8.0 s, has its

Marizza181 [45]

Answer:

decreased

Explanation:

because it transfered from 10.0s

to 8.0s

5 0

2 years ago

In a jet engine, _________________ is sucked into the front of a compressor, where _____________ is injected into the high press

Kay [80]

The answers are; air, fuel, back, forward

The jet engine utilizes the principle that for every action force there is an equal reaction force in the opposite direction. When the exhaust from the combustion in the engine is ejected with great force backwards, the reaction force pushes the plane forward with an equivalent force.

4 0

3 years ago

A skateboarder is moving at 1.75 m/s when she starts going up an incline that causes an acceleration of -0.20 m/s2

Rudiy27

Answer:

Approximately $7.66\; \rm m$ .

Explanation:

<h3>Solve this question with a speed-time plot</h3>

The skateboarder started with an initial speed of $u = 1.75\; \rm m \cdot s^{-1}$ and came to a stop when her speed became $v = 0\; \rm m \cdot s^{-1}$ . How much time would that take if her acceleration is $a = -0.20\; \rm m \cdot s^{-1}$ ?

$\begin{aligned} t &= \frac{v - u}{a} \\ &= \frac{0\; \rm m \cdot s^{-1} - 1.75\; \rm m \cdot s^{-1}}{-0.20\; \rm m \cdot s^{-2}} \approx 8.75\; \rm s\end{aligned}$ .

Refer to the speed-time graph in the diagram attached. This diagram shows the velocity-time plot of this skateboarder between the time she reached the incline and the time when she came to a stop. This plot, along with the vertical speed axis and the horizontal time axis, form a triangle. The area of this triangle should be equal to the distance that the skateboarder travelled while she was moving up this incline until she came to a stop. For this particular question, that area is approximately equal to:

$\displaystyle \frac{1}{2} \times 1.75\; \rm m \cdot s^{-1} \times 8.75\; \rm s \approx 7.66\; \rm m$ .

In other words, the skateboarder travelled $15.3\; \rm m$ up the slope until she came to a stop.

<h3>Solve this question with an SUVAT equation</h3>

A more general equation for this kind of motion is:

$\displaystyle x = \frac{1}{2}\, (u + v) \, t = \frac{1}{2}\, (u + v)\cdot \frac{v - u}{a}= \frac{v^2 - u^2}{2\, a}$ ,

where:

$u$ and $v$ are the initial and final velocity of the object,
$a$ is the constant acceleration that changed the velocity of this object from $u$ to $v$ , and
$x$ is the distance that this object travelled while its velocity changed from $u$ to $v$ .

For the skateboarder in this question:

$\begin{aligned}x &= \frac{v^2 - u^2}{2\, a}\\ &= \frac{\left(0\; \rm m \cdot s^{-1}\right)^2 - \left(1.75\; \rm m \cdot s^{-1}\right)^2}{2\times \left(-0.20\; \rm m \cdot s^{-2}\right)}\approx 7.66\; \rm m \end{aligned}$ .

6 0

4 years ago

Multiply. (2x + 4)(x - 4)