All categories

3 years ago

A 400.0 kg storage box is held 10 m above ground by a forklift. What is its gravitational potential energy? (PE = mgh)

Physics

2 answers:

castortr0y [4]3 years ago

7 0

Answer: D.) 39,200 J

Via the equation of potential energy PE = mgh where m is mass, g is the average gravity on earth and h is the height. In this case m = 400 kg, g = 9.8, h = 10 m thus:

$P.E.=(400kg)(9.8\frac{m}{s^2} )(10m)=39,200 J$

P.E.= 39,200 Joules

solmaris [256]3 years ago

3 0

<h2>Answer: D.) 39,200 J</h2>

Explanation:

for USA Testprep

You might be interested in

(a) Calculate the height of a cliff if it takes 2. 35 s for a rock to hit the ground when it is thrown straight up from the clif

Natali [406]

(a) The height of the cliff will be 8.26 meters.

(b) The time would it take to reach the ground will be 0.717 sec.

<h3>What is velocity?</h3><h3 />

The change of displacement with respect to time is defined as the velocity. Velocity is a vector quantity. it is a time-based component.

(a) The height of the cliff will be 8.26 meters.

According to Newton's second equation of motion

$\rm H =ut-\frac{1}{2} gt^2 \\\\ \rm H =8\times 2.35-\frac{1}{2} 9.81 (2.35)^2\\\\\rm H =8.16 \; m$

Hence The height of the cliff will be 8.26 meters.

(b)The time would it take to reach the ground will be 0.717 sec.

We must have the final velocity to find the time so;

$\rm v^2=u^2+2gh\\\\ \rm v^2=8^2+2\times 9.81 \times 8.6 \\\\ \rm v= \sqrt{8^2+2\times 9.81 \times 8.6}\\\\\rm v=15.03 \;m/sec$

According to Newton's third equation of motion ;

$\rm v=u-gt \\\\ \rm t=\frac{v-u}{g} \\\\ \rm t=\frac{15.03-8}{9.81} \\\\ \rm t=0.717 sec.$

Hence the time would it take to reach the ground will be 0.717 sec.

To learn more about the velocity refer to the link ;

brainly.com/question/862972

3 0

2 years ago

Water flows through a 4.50-cm inside diameter pipe with a speed of 12.5 m/s. At a later position, the pipe has a 6.25-cm inside

jek_recluse [69]

Given,

The initial inside diameter of the pipe, d₁=4.50 cm=0.045 m

The initial speed of the water, v₁=12.5 m/s

The diameter of the pipe at a later position, d₂=6.25 cm=0.065 m

From the continuity equation,

$\begin{gathered} A_1v_1=A_2v_2 \\ \pi(\frac{d_1}{2})^2v_1=\pi(\frac{d_2}{2})^2v_2 \\ \Rightarrow d^2_1v_1=d^2_2v_2 \end{gathered}$

Where A₁ is the area of the cross-section at the initial position, A₂ is the area of the cross-section of the pipe at a later position, and v₂ is the flow rate of the water at the later position.

On substituting the known values,

$\begin{gathered} 0.045^2\times12.5=0.065^2\times v_2 \\ \Rightarrow v_2=\frac{0.045^2\times12.5}{0.065^2} \\ =5.99\text{ m/s} \end{gathered}$

Thus, the flow rate of the water at the later position is 5.99 m/s

4 0

1 year ago

A particle with charge 3.20×10−19 c is placed on the x axis in a region where the electric potential due to other charges increa

lys-0071 [83]

Answer:

-5 V

Explanation:

The charged particle (which is positively charged) moves from point A to B, and its kinetic energy increases: it means that the particle is following the direction of the field, so its potential energy is decreasing (because it's been converted into potential energy), therefore it is moving from a point at higher potential (A) to a point at lower potential (B). This means that the value

vb−va

is negative.

We can calculate the potential difference between the two points by using the law of conservation of energy:

$\Delta K+ \Delta U=0\\\Delta K + q\Delta V=0$