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Mrrafil [7]
3 years ago
8

Convert 1.4×10^9km^3 into cubic meters

Physics
2 answers:
storchak [24]3 years ago
7 0

Explanation: Hi! here we have 1.4×10^9km^3 and want to write this quantity in meters.

First, a kilometer is one thousand meters, so 1km = 1000m, so 1km^{3} = (1000m)^{3}= 1000000000m^{3}. = 10^9 m^3

so changin this in our quantity we get 1.4×10^9km^3  = 1.4×10^9*10^9 m^3=1.4×10^18m^3

and there you hace the original quantity written in meters.

Vadim26 [7]3 years ago
3 0
1km=10^3 m,1km^3=10^9cubic metres answer is 1.4x10^18cubic meters
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In the "before" part of Fig. 9-60, car A (mass 1100 kg) is stopped at a traffic light when it is rear-ended by car B (mass 1400
liq [111]

Complete Question:

In the "before" part of Fig. 9-60, car A (mass 1100 kg) is stopped at a traffic light when it is rear-ended by car B (mass 1400 kg). Both cars then slide with locked wheels until the frictional force from the slick road (with a low ?k of 0.15) stops them, at distances dA = 6.1 m and dB = 4.4 m. What are the speeds of (a) car A and (b) car B at the start of the sliding, just after the collision? (c) Assuming that linear momentum is conserved during the collision, find the speed of car B just before the collision.

Answer:

a) Speed of car A at the start of sliding = 4.23 m/s

b) speed of car B at the start of sliding = 3.957 m/s

c) Speed of car B before the collision = 7.28 m/s

Explanation:

NB: The figure is not provided but all the parameters needed to solve the question have been given.

Let the frictional force acting on car A, f_{ra} = \mu mg\\............(1)

Since frictional force is a type of force, we are safe to say f_{ra} = ma.......(2)

Equating (1) and (2)

ma = \mu mg\\a = \mu g\\\mu = 0.15\\a = 0.15 * 9.8 = 1.47 m/s^{2}

a) Speed of A at the start of the sliding

d_{A} = 6.1 m\\Speed of A at the start of sliding, v_{A} = \sqrt{2ad_{A} }\\ v_{A} = \sqrt{2*1.47*6.1 } \\v_{A} = \sqrt{17.934 } \\v_{A} = 4.23 m/s

b) Speed of B at the start of the sliding

d_{A} = 4.4 m\\Speed of A at the start of sliding, v_{B} = \sqrt{2ad_{B} }\\ v_{B} = \sqrt{2*1.47*4.4 } \\v_{B} = \sqrt{12.936 } \\v_{B} = 3.957 m/s

Let the speed of car B before collision = v_{B1}

Momentum of car B before collision = m_{B} v_{B1}

Momentum after collision = m_{A} v_{A} + m_{B} v_{B2}

Applying the law of conservation of momentum:

m_{B} v_{B1}  = m_{A} v_{A} +m_{B} v_{B2}

m_{A} = 1100 kg\\m_{B} = 1400 kg

(1400*v_{B1} ) = (1100 * 4.23) + ( 1400 * 3.957)\\(1400*v_{B1} ) = 10192.8\\v_{B1} = 10192.8/1400\\v_{B1 = 7.28 m/s

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3 years ago
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Why do scientist use different types of models to represent compounds
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Because they are different they all show different traits.
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3 years ago
A 120 kg box is on the verge of slipping down an inclined plane with an angle of inclination of 47º. What is the coefficient of
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Given :

A 120 kg box is on the verge of slipping down an inclined plane with an angle of inclination of 47º.

To Find :

The coefficient of static friction between the box and the plane.

Solution :

Vertical component of force :

mg\ sin\ \theta =  120\times 10 \times sin\ 47^\circ{}=877.62 \ N

Horizontal component of force(Normal reaction) :

mg\ cos\ \theta =  120\times 10 \times cos\ 47^\circ{}=818.40 \ N

Since, box is on the verge of slipping :

mg\ sin\ \theta= \mu(mg \ cos\ \theta)\\\\\mu = tan \ \theta\\\\\mu = tan\ 47^o\\\\\mu = 1.07

Therefore, the coefficient of static friction between the box and the plane is 1.07.

Hence, this is the required solution.

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Different between progressive wave and stationary wave​
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2) In Progressive waves the energies are equally and efficiently transferred along the travelling waves. Every particle are transferring some kind of energy to a next further particle on the same path, basically most of the energies are lost because of which there's no energy acquired by it.

2) In Stationary or standing waves there's no absolute transfer of any significant amount of energies which are not transferred along a path of the wave. Particles in stationery waves are giving and contributing in energy submission and also acquire some of the energy back because of which the net transferring of energies between the particles in a specific period as nullified.

3) Phases of the progressive waves of the particles in these waves are varying in a continuous manner and have changing values between them.

3) Phases of the Stationary or standing waves of the particles in these waves are not changing and always same to the contrary opposite when placed between the consecutively running sets of nodes (Between two nodes of particles).

4) Progressive waves have no particles which show they're having a rest phase or a permanent rest phase in a medium (particle medium).

4) Stationary waves have significant amount of particles of the medium to show that there having a rest phase or a permanent rest phase at the nodes of those particles.

5) Amplitudes of Progressive waves are totally and completely different particles are neutral and are having same values.

5) Amplitudes of Stationary or Standing waves of the particles in between those tow consecutively sets of nodes in between them and antinodes provided are having different values and vary much more progressively.

6) All of the particles in Progressive waves containing it don't specially cross their given mean positions in a simultaneous manner.

6) All of the particles in Stationary waves containing it frequently and steadily cross their given mean positions in a simultaneous manner.

7) In Progressive waves the particles don't show any attainment of a displacement provided in a maximum amount in a simulations manner.

7) In Stationary waves the particles are showing and exhibiting the attainments of various displacements in a maximum amount in a simultaneous manner.

8) Maximum velocities achieved by Progressive waves are indeed same or similar for all the given particles when they're showing a passing of those given mean positions.

8) Maximum velocities achieved by Stationary waves of those particles when they're crossing their given mean positions are in a continuity of increasement for those particles between those "nodes" and of course the consecutively set "antinodes" further which it's showing a significant decreasement after it corresponds and reaches the second or usually the next nearest node.

9) Progressive waves have crest and troughs in their waves that're moving into a forward direction.

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Illusion [34]

Answer:

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Explanation:

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