Ask question

Ask question

All categories

3 years ago

8

Someone drops a 50 − g pebble off of a docked cruise ship, 70.0 m from the water line. A person on a dock 3.0 m from the water l

ine holds out a net to catch the pebble. (a) How much work is done on the pebble by gravity during the drop? (b) What is the change in the gravitational potential energy during the drop? If the gravitational potential energy is zero at the water line, what is the gravitational potential energy (c) when the pebble is dropped? (d) When it reaches the net? What if the gravitational potential energy was 30.0 Joules at water level? (e) Find the answers to the same questions in (c) and (d).

1 answer:

alexira [117]3 years ago

6 0

Answer:

a) The work done on the pebble is 32.9 J.

b) The change in the gravitational potential energy is -32.9 J.

c) When the pebble is dropped, the potential energy is 34.3 J.

d) When the pebble reaches the net, the potential energy is 1.47 J.

e) If the potential energy is 30.0 J at water level, it will be 31.47 J at the net and 64.3 J at 70.0 m.

Explanation:

a) The work done by the gravity force can be calculated using the following equation:

W = F · Δy

Where:

W = work

F = gravity force

Δy = vertical displacement (final height - initial height)

The gravity force can be calculated as:

F = m · g

Where:

F = gravity force

m = mass

g = acceleration due to gravity (-9.81 m/s² considering the upward direction as positive).

Then, the work done on the pebble by the gravity force will be:

W = m · g · (hf - hi) (hf = final height, hi = initial height)

W = 0.050 kg · (-9.81 m/s²) · (3.0 m - 70.0 m)

W = 32.9 J

The work done on the pebble is 32.9 J.

(Notice that the work is positive. That means that the force that does the work is in the same direction as the movement).

b) The potential energy (EP) can be calculated as follows:

EP = m · g · h (h = height)

The change in the gravitational potential energy is calculated as the difference of the potential energy between the two positions:

ΔEP = EPf - EPi

Where:

ΔEP = change in the gravitational potential energy.

EPf = final potential energy.

EPi = initial potential energy.

Then:

ΔEP = EPf - EPi

ΔEP = (0.050 kg · 9.81 m/s² · 3.0 m) - (0.050 kg · 9.81 m/s² · 70.0 m)

ΔEP = -32.9 J

c) When the pebbel is dropped, the potential energy will be:

EP = m · g · h

EP = 0.050 kg · 9.81 m/s² · 70.0 m = 34.3 J

d) When it reaches the net h = 3.0 m. Then:

EP = 0.050 kg · 9.81 m/s² · 3.0 m = 1.47 J

e) We have to add 30 J to the potential energy values calculated above:

The potential energy at 70.0 m will be: 34.3 J + 30 J = 64.3 J

The potential energy at 3.0 m will be: 1.47 J + 30 J = 31.47 J

You might be interested in

A sprinter is advised to reduced his speed slowly after completing his race.why

VARVARA [1.3K]

<span>The sprinter is advised to reduce his speed slowly after completing the race because of the power that is needed when the stoppage is down in a faster manner could be very great. This would translate to the great usage in gasoline. Also, the inertia of the vehicle is quiet high so it is hard to stop it very suddenly. </span>

5 0

4 years ago

protons in an atomic nucleus are typically 10-15 m apart. what is the electric force of repulsion between nuclear protons?

Aleonysh [2.5K]

230 Newton

Electric charge consists of two types i.e. positively electric charge and negatively electric charge.There was a famous scientist who investigated about this charges. His name is Coulomb and succeeded in formulating the force of attraction or repulsion between two charges i.e. :

F = electric force (N)

k = electric constant (N m² / C²)

q = electric charge (C)

r = distance between charges (m)

The value of k in a vacuum = 9 x 10⁹ (N m² / C²)

F = k(q1 q2)/ r^2

Distance between protons = d = 10⁻¹⁵ m

charge of proton = q = 1.6 × 10⁻¹⁹ C

Here q1=q2

electric force = F =230N

Coulomb's Law. Two protons in an atomic nucleus are typically separated by a distance of 2×10−15m. The electric repulsive force between the protons is huge, but the attractive nuclear force is even stronger and keeps the nucleus from bursting apart.

2 Nuclei and the Need for an Attractive Nuclear Force. The Coulomb force also acts within atomic nucleii, whose characteristic dimension is 10 m, which is called a fermi. There are two protons in a He nucleus, which repel each other because of the Coulomb force.

Find more about electric force of repulsion between nuclear protons

brainly.com/question/8404637

#SPJ4

5 0

1 year ago

PLEASE HELP PLEASE HELP ME PLEASE HELP ME

klemol [59]

Answer:24

Explanation:

7 0

2 years ago

As the wavelength increases, the frequency (2 points) decreases and energy decreases. increases and energy increases. decreases

frozen [14]

Bohr's equation for the change in energy is
$\Delta E= \frac{hc}{\lambda}$
where
h = Planck's constant
c == the velocity of light
λ = wavelength.

The velocity is related to wavelength and frequency, f, by
c = fλ

Let us examine the given answers on the basis of the given equations.

a. As λ increases, f decreases and ΔE decreases.
TRUE

b. As λ increases, f increases and ΔE increases.
FALSE

c. As λ increases, f increases and ΔE decreases.
FALSE

Answer:
As the wavelength increases, the frequency decreases and energy decreases.

3 0

4 years ago

Read 2 more answers

Two sound waves, from two different sources with the same frequency, 540 Hz, travel in the same direction at 330 m s . The sourc

oee [108]

Answer:

The value is $\Delta \phi = 4.12 \ rad$

Explanation:

From the question we are told that

The frequency of each sound is $f_1 = f_2 = f = 540 \ Hz$

The speed of the sounds is $v = 330 \ m/s$

The distance of the first source from the point considered is $a = 4.40 \ m$

The distance of the second source from the point considered is $b = 4.00 \ m$

Generally the phase angle made by the first sound wave at the considered point is mathematically represented as

$\phi_a = 2 \pi [\frac{a}{\lambda} + ft]$

Generally the phase angle made by the first sound wave at the considered point is mathematically represented as

$\phi_b = 2 \pi [\frac{b}{\lambda} + ft]$

Here b is the distance o f the first wave from the considered point

Gnerally the phase diffencence is mathematically represented as

$\Delta \phi= \phi_a - \phi_b = 2 \pi [\frac{ a}{\lambda} + ft ] - 2 \pi [\frac{b}{\lambda} + ft ]$

=> $\Delta \phi = \frac{2\pi [ a - b]}{ \lambda }$

Gnerally the wavelength is mathematically represented as

$\lambda = \frac{v}{f}$

=> $\lambda = \frac{330}{540}$

=> $\lambda = 0.611 \ m$

=> $\Delta \phi = \frac{2* 3.142 [ 4.40 - 4.0 ]}{ 0.611 }$

=> $\Delta \phi = 4.12 \ rad$

5 0

3 years ago