Ask question

Ask question

All categories

2 years ago

11

At which position does the diver have the most gravitational potential energy

2 answers:

marta [7]2 years ago

4 0

Directly vertical. The surface area is the smallest, allowing for the water upthrust to be kept to a minimum, whilst the mass is the same, which keeps the same weight and gravitational pull.

Ray Of Light [21]2 years ago

3 0

Potantial energy formula = m.g.h
m: mass
g: gravitational acceleration
h: height

m and g are the same. So, we must to look at to the height. At the vertical position, the potantional energy increase.

When the h increase, the energy will increases.

You might be interested in

All waves transfer____. Energy weight velocity mass

Katarina [22]

I believe the answer is energy

5 0

2 years ago

Can a conductor be given limitless charge

k0ka [10]

Answer:

No

Explanation:

You could try to give it enough to fill all valence electrons in all of the atoms in the conductor, but practically this could not be achieved.

5 0

2 years ago

Given f(x)=3x-11 which of the following is f(7)<br><br> A. 10<br> B. 32<br> C. -6<br> D. 6

Crazy boy [7]

Answer:

The answer is A. 10

Explanation:

<em>Given </em>

<em>f(</em><em>x</em><em>)</em><em> </em><em>=</em><em> </em><em>3x </em><em>-</em><em> </em><em>1</em><em>1</em>

<em>So, </em><em> </em>

<em>f(</em><em>7</em><em>)</em><em> </em><em>=</em><em> </em><em>3</em><em> </em><em>*</em><em> </em><em>7</em><em> </em><em>-</em><em> </em><em>1</em><em>1</em>

<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>=</em><em> </em><em>2</em><em>1</em><em> </em><em>-</em><em> </em><em>1</em><em>1</em>

<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>=</em><em> </em><em>1</em><em>0</em>

5 0

2 years ago

Read 2 more answers

Water falls without splashing at a rate of 0.200 L/s from a height of 3.60 m into a 0.730 kg bucket on a scale. If the bucket is

dimaraw [331]

Answer:

15.106 N

Explanation:

From the given information,

The weight of the bucket can be calculated as:

$W_b = m_bg = \\ \\ W_b = (0.730 \ kg) ( 9.80 \ m/s^2) \\ \\ W_b = 7.154 \ N$

The mass of the water accumulated in the bucket after 3.20s is:

$m_w= (0.20 \ L/s) ( 3.20)s$

$m _w=0.64 \ kg$

To determine the weight of the water accumulated in the bucket, we have:

$W_w = m_w g$

$W_w = ( 0.64 \ kg )(9.80\ m \ /s^2)$

$W_w = 6.272 \ N$

For the speed of the water before hitting the bucket; we have:

$v = \sqrt{2gh}$

$v = \sqrt{2*9.80 \ m/s^2 * 3.60 \ m}$

v = 8.4 m/s

Now, the force required to stop the water later when it already hit the bucket is:

$F = v ( \dfrac {dm}{dt} )$

$F = (8.4 \ m/s)( 0.200 \ L/s)$

F = 1.68 N

Finally, the reading scale is:

$F_{scale$ = 7.154 N + 6.272 N + 1.68 N

= 15.106 N

6 0

2 years ago

Read 2 more answers

Suppose a spring has a relaxed length of 28.3 cm. The simulation refers to this as the natural length. This is the length of the

den301095 [7]

Answer:

Explanation:

Normal length of spring = 28.3 cm

stretched length of spring = 38.2 cm

length of extension = 38.2 - 28.3 = 9.9 cm

= 9.9 x 10⁻² m

force applied to stretch = .55 x 9.8 ( mg )

= 5.39 N

Force constant = force applied / extension

= 5.39 / 9.9 x 10⁻²

= .5444 x 10² N /m

= 54.44 N/m

4 0

3 years ago