1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Igoryamba
3 years ago
7

An 800 kg fishing boat going south at 12 m/s runs into a stopped pontoon boat (1400 kg). The boats stick together and move south

. What is the velocity of the boats after the collision?
Physics
1 answer:
kykrilka [37]3 years ago
7 0

Answer:

the velocity of the boats after the collision is 4.36 m/s.

Explanation:

Given;

mass of fish, m₁ = 800 kg

mass of boat, m₂ = 1400 kg

initial velocity of the fish, u₁ = 12 m/s

initial velocity of the boat, u₂ = 0

let the final velocity of the fish-boat after collision = v

Apply the principle of conservation of linear momentum for inelastic collision;

m₁u₁ + m₂u₂ = v(m₁ + m₂)

800 x 12    +   1400 x 0 = v(800 + 1400)

9600 = 2200v

v = 9600/2200

v = 4.36 m/s

Therefore, the velocity of the boats after the collision is 4.36 m/s.

You might be interested in
2. What process does water redeposit into a lake in the form of rain?
weqwewe [10]

The process that water redeposit into a lake in the form of rain is precipitation.

<h3>What is water runoffs?</h3>

Water runoff occurs when there is more water than land can absorb.

The water flows across the surface of the land and into nearby creeks, streams, or lakes.

Runoff can come from both natural processes and human activity.

When rain falls to the earth from clouds and runs downhill into rivers and lakes.

During evaporation, the water turns from liquid into gas, and moves from oceans and lakes into the atmosphere where it forms clouds.

<h3>What is precipitation?</h3>

Precipitation is any liquid (rain) or frozen water that forms in the atmosphere and falls back to the Earth, for example it could fall on land or into lakes and rivers.

Thus,  the process that water redeposit into a lake in the form of rain is precipitation.

Learn more about precipitation here: brainly.com/question/1783904

#SPJ1

5 0
1 year ago
As the scattering angle of the photon increases, what happens to the wavelength associated with the photon?
frez [133]

As the scattering angle of the photon increases, the wavelength associated with the photon increases.

<h3><u>Explanation:</u></h3>

The particle with quantum mechanical property is known as Compton wavelength. The wavelength of a photon increases during collision. When the scattering angle of the photon is 0 degree then the photon's wavelength increases by 0 and when the scattering angle  is 180 degree then the wavelength of  the photon will become double. This is known as Compton wavelength.

When a photon undergoes collision process, the photo loses its energy and this energy is transferred to the electrons. This causes energy of the photon to decrease and thus the frequency also decreases. Thus, the wavelength of the photon will increase.

6 0
3 years ago
The drawing shows a large cube (mass = 21.0 kg) being accelerated across a horizontal frictionless surface by a horizontal force
MaRussiya [10]

Answer:

The blocks must be pushed with a force higher than 359 Newtons horizontally in order to accomplish this friction levitation feat.

Explanation:

The first step in resolving any physics problem is to draw the given scenario (if possible), see the attached image to have an idea of the objects and forces involved.

The large cube in red is being pushed from the left by a force \vec{P} whose value is to be found. That cube has its own weight \vec{w}_1=m_1\vec{g}, and it is associated with the force of gravity which points downward. Newton's third law stipulates that the response from the floor is an upward pointing force on the cube, and it's called the normal force \vec{N}_1.

A second cube is being pushed by the first, and since the force \vec{P} is strong enough it is able to keep such block suspended as if it were glued to the first one, due to friction. As in the larger cube, the smaller one has a weight \vec{w}_2=m_2\vec{g} pointing downwards, but the normal force in this block doesn't point upwards since its 'floor' isn't below it, but in its side, therefore the normal force directs it to the right as it is shown in the picture. Normal forces are perpendicular to the surface they contact. The final force is the friction between both cubes, that sets a resistance of one moving parallel the other. In this case, the weight of the block its the force pointing parallel to the contact surface, so the friction opposes that force, and thus points upwards. Friction forces can be set as Fr=\mu~N, where \mu is the coefficient of static friction between the cubes.

Now that all forces involved are identified, the following step is to apply Newton's second law and add all the forces for each block that point in the same line, and set it as equal its mass multiplied by its acceleration. The condition over the smaller box is the relevant one so its the first one to be analyzed.

In the vertical component: \Sigma F^2_y=Fr-w_2=m_2 a_y Since the idea is that it doesn't slips downwards, the vertical acceleration should be set to zero a_y=0, and making explicit the other forces: \mu N_2-m_2g=0\quad\Rightarrow (0.710)N_2-(4.5)(10)=0\quad\Rightarrow N_2=(4.5)(10)/(0.710)\approx 63.38 [N]. In the last equation gravity's acceleration was rounded to 10 [m/s^2].

In its horizontal component: \Sigma F^2_x=N_2=m_2 a_x, this time the horizontal acceleration is not zero, because it is constantly being pushed. However, the value of the normal force and the mass of the block are known, so its horizontal acceleration can be determined: 63.38=(4.5) a_x \quad \Rightarrow a_x=(63.38)/(4.5)\approx 14.08 [m/s^2]. Notice that this acceleration is higher than the one of gravity, and it is understandable since you should be able to push it harder than gravity in order for it to not slip.

Now the attention is switched to the larger cube. The vertical forces are not relevant here, since the normal force balances its weight so that there isn't vertical acceleration. The unknown force comes up in the horizontal forces analysis: \Sigma F_x=P=m a_x, since the force \vec{P} is not only pushing the first block but both, the mass involved in this equation is the combined masses of the blocks, the acceleration is the same for both blocks since they move together; P=(21.0+4.5) 14.08\approx 359.04 [N]. The resulting force is quite high but not impossible to make by a human being, this indicates that this feat of friction suspension is difficult but feasable.

4 0
3 years ago
Drag each tile to the correct location.
scoray [572]

Answer:

Relative: meredith's birth, the planting of jasmine's tree, and the extinction of the stegosaurus

Absolute: harrison's birth, the building of jasmine house, and the extinction of the triceratops

Explanation: plato

7 0
2 years ago
A thin spherical shell with radius R1 = 2.00 cm is concentric with a larger thin spherical shell with radius R2 = 6.00 cm. Both
Dafna11 [192]

Answer:

a. i. 1350 V ii 0 V iii -450 V b. 6.75 kV. The inner shell is at a higher potential.

Explanation:

The formula for electric potential is given by V = Σkq/r, where k = 9 × 10⁹ Nm²/C², q = charge and r = distance.

q₁ = charge on smaller shell = +6.00 nC = +6.00 × 10⁻⁹ C, r₁ = radius of smaller shell = 2.00 cm = 2.00 × 10⁻² m.

q₂ = charge on larger shell = -9.00 nC = -9.00 × 10⁻⁹ C, r₂ = radius of larger shell = 6.00 cm = 6.00 × 10⁻² m.

a. At r = 0, inside both spheres V = kq₁/r₁ + kq₂/r₂. = k(q₁/r₁ + q₂/r₂) = 9 × 10⁹ [+6.00 × 10⁻⁹/2.00 × 10⁻² + (-9.00 × 10⁻⁹/6.00 × 10⁻²)] = 1350 V

ii. At r = 4.00 cm, the point outside of smaller shell but inside larger shell. r₁ = 4.00 cm = 4.00 × 10⁻² m and r₂ = 6.00 cm = 6.00 × 10⁻². So, V = kq₁/r₁ + kq₂/r₂. = k(q₁/r₁ + q₂/r₂) = 9 × 10⁹ [+6.00 × 10⁻⁹/4.00 × 10⁻² + (-9.00 × 10⁻⁹/6.00 × 10⁻²)] = 0 V.

iii. At r = 6.00 cm, the point outside both shells. r₁ = r₂ = r = 6.00 cm = 6.00 × 10⁻². So, V = kq₁/r₁ + kq₂/r₂. = k(q₁ + q₂)/r = 9 × 10⁹ [+6.00 × 10⁻⁹+ (-9.00 × 10⁻⁹)]/6.00 × 10⁻² = -450 V.

b. The potential of the surface of the smaller shell is V₁ = 9 × 10⁹ [+6.00 × 10⁻⁹/2.00 × 10⁻²] = 2700 V = 2.7 kV.

The potential of the surface of the larger shell is V₂ = 9 × 10⁹ [-9.00 × 10⁻⁹/2.00 × 10⁻²] = -4050 V = -4.050 kV. The potential difference V₁ - V₂ = 2700 - (-4050) V = 6750 V = 6.75 kV. Since the potential difference is positive, V₁ is higher. So, the inner shell is at a higher potential.

8 0
3 years ago
Other questions:
  • The picture above shows salt (NaCI) dissolved in water (H2O) which statement is true?
    6·1 answer
  • Radium-226 emits alpha (α), beta (β) and gamma (γ) radiation.
    12·1 answer
  • 28. the maximum distance that the particles of a medium move from the rest position is the a. amplitude of the wave. b. waveleng
    12·2 answers
  • if the vapor's volume were to be incorrectly recorded as 125ml, how will this error affect the calculated molar mass of the unkn
    9·1 answer
  • Help me with the question b.​
    8·1 answer
  • Extra CreditA particle is directed along the axis of the instrument in the gure. Aparallel plate capacitor sets up an electric e
    13·1 answer
  • Someone please help​
    7·1 answer
  • You serve a volleyball with a mass of 2.1 kg. The ball leaves your hand with a speed of 2.1 m/s. The
    14·1 answer
  • It is made up of small particles
    5·2 answers
  • Dbv-qyag-dsc.join this now on meet<br>​
    9·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!