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Vlad [161]
3 years ago
6

Two blocks of masses 3 kg and 5 kg approach each other with initial velocities 4 m/s and -6 m/s respectively. The two blocks col

lide in a totally inelastic collision. What is their common final velocity after the inelastic collision?​
Physics
1 answer:
Kipish [7]3 years ago
6 0

Answer:

Explanation:

After the collision velocity of the particle is (4î - 3ĵ)m/s . ... A particle of mass 1 kg moving with a velocity of (4i^−3j^​)m/s collides with a fixed surface. ... Perfectly inelastic. D ... The common velocity of the blocks after collision is: ... A ball falls from a height of 5 m and strikes the roof of a lift. ... Stay upto date with our Newsletter! i know this is not right but just here for points see ya loser

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Marking brainliest help pls the formula are there to help ^
natta225 [31]

Answer:

Explanation:

Look at the equation for Potential Energy. PE = mass times gravity times the height. Filling in and solving for h:

34.3 = .5(9.8)h so

34.3 = 4.9h so

h = 7 meters

6 0
3 years ago
What method of separation would work best on a homogeneous mixture salt water?
Lera25 [3.4K]
The appropriate answer is c. evaporation. Evaporation is the process by which a liquid changes to a gas and in this case is the best way to remove the water from the homogenous mixture. The reaction can be sped up by heating the mixture. All the water will eventually evaporate leaving behind the sodium chloride crystals as a precipitate.
Filtration works best on mixtures that have precipitates and distillation is for separating liquids with different boiling points.
5 0
4 years ago
Read 2 more answers
(1 point) At noon, ship A is 10 nautical miles due west of ship B. Ship A is sailing west at 22 knots and ship B is sailing nort
Veronika [31]

Answer:28.8 knots

Explanation:

The ships are moving as the sides of a right triangle. Thus, Pyhogorean theorem will be useful in the following steps. Next, we have to know that the rate of change in distance, which is called velocity, can be described in terms of derivatives.

First, we have to calculate the distances covered by the ships from noon to 6 PM. In 6 hours, ship A moved 22*6=132 nautical mile. However, their first distance was 10 nautical miles, so 132+10=142 miles is the equivalent of A's displacement. For B, the distance travelled is 19*6=114 miles. From now on, A=142 miles and B=114 miles.

The distance between them is described with Pythogorean theorem, which is D=\sqrt{A^{2} +B^{2} } and when we replace the values A and D, we find Distance (D) to be 182 miles.

Now, let's make the notations clear. The velocity of A and B is notated as \frac{dA}{dt} and \frac{dB}{dt}. The rate of change of distance is also notated as \frac{dD}{dt}. Now, we have to find \frac{dD}{dt} from the Pythogorean theorem. If we derive the Pythogorean expression D=\sqrt{A^{2} +B^{2} } , we would have:

\frac{dD}{dt} =\frac{1}{2} *(A^{2} +B^{2} )^{-1/2} *(2*A*\frac{dA}{dt} + 2*B*\frac{dB}{dt} )

The derivation here includes chain rule and derives the interior parts of the parenthesis. When we insert distances for A and B and velocities for derivation notations, the formula becomes:

\frac{dC}{dt} =\frac{1}{2}*(142^{2}   +114^{2})^{-\frac{1}{2} }*(2*142*22 + 2*114*19) and the answer is 28.6 knots.

6 0
3 years ago
A horizontal 791-N merry-go-round is a solid disk of radius 1.56 m and is started from rest by a constant horizontal force of 49
ICE Princess25 [194]

Answer:

K.E=273.5J

Explanation:

Given data

F_{g}=791N\\F_{h}=49N\\r=1.56m\\t=3.03s

To find

Kinetic Energy

Solution

The moment of inertia is given as:

I=(\frac{1}{2} )MR^2\\I=\frac{1}{2}(F_{g}/g)R^2\\ I=\frac{1}{2}(\frac{791N}{9.8m/s^2} )(1.56m)^2\\ I=98.21kg.m^2

The angular acceleration is given as:

\alpha =\frac{T}{I}\\\alpha  =\frac{F_{y}R}{I}\\ \alpha =\frac{(49N)(1.56m)}{98.2kg.m^2}\\\alpha  =0.778rad/s^2

Now the angular velocity is given by:

w=\alpha t\\w=(0.778rad/s^2)(3.03s)\\w=2.36rad/s

So the kinetic energy given as:

K.E=(\frac{1}{2} )Iw^2\\K.E=\frac{1}{2}(98.21kg.m^2)(2.36rad/s)^2\\ K.E=273.5J

8 0
3 years ago
When a ball increases in speed by the same amount each second its acceleration?
MAXImum [283]
Its acceleration is constant. 
5 0
4 years ago
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