HELP!!!

What is the G.P.E of 675 newtons on 26000 meter mountain

vagabundo [1.1K]

GPE=mgh, where m=mass, g=acceleration due to gravity, h=height
We have a force of 675N already, so mg=675N

<span>GPE=mgh=675(26000)=17.55x10^6J</span>

3 0

3 years ago

At any point P, the total electric field due to a group of point charges equals the vector sum of the electric fields of the ind

wel

The answer is
A. True

8 0

3 years ago

A mass of 4kg suspended by a light string 2m long and at rest is projected horizontally with a velocity of 1.5 m/s. find the ang

Dafna11 [192]

Answer:

19.5°

Explanation:

The energy of the mass must be conserved. The energy is given by:

1) $E=\frac{1}{2}mv^2+mgh$

where m is the mass, v is the velocity and h is the hight of the mass.

Let the height at the lowest point of the be h=0, the energy of the mass will be:

2) $E=\frac{1}{2}mv^2$

The energy when the mass comes to a stop will be:

3) $E=mgh$

Setting equations 2 and 3 equal and solving for height h will give:

4) $h=\frac{v^2}{2g}$

The angle ∅ of the string with the vertical with the mass at the highest point will be given by:

5) $cos\phi=\frac{l-h}{l}$

where l is the lenght of the string.

Combining equations 4 and 5 and solving for ∅:

6) $\phi={cos}^{-1}(\frac{l-h}{l})={cos}^{-1}(1-\frac{h}{l})={cos}^{-1}(1-\frac{v^2}{2gl})$

8 0

4 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

An object is thrown upward from the top of a 128-foot building with an initial velocity of 112 feet per second. The height h of

lorasvet [3.4K]

Answer:

Time, t = 8 seconds

Explanation:

An object is thrown upward from the top of a 128-foot building with an initial velocity of 112 feet per second. The height h as a function of time t is given by :

$h=-16t^2+112t+128$

We need to find the time when the object will hit the ground. When it will hit the ground, h = 0

So,

$-16t^2+112t+128=0$

On solving the above quadratic equation, we get the value of t = 8 seconds. So, after 8 seconds the object will hit the ground. Hence, this is the required solution.

6 0

4 years ago