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tresset_1 [31]
3 years ago
10

Ed biked home from school in 400 seconds. His home is located 2000 m south of the school. What was his velocity?

Physics
1 answer:
Naily [24]3 years ago
7 0

<em>answer =  5 \:  \: metre \: per \: second\\ distance = 2000 \: m \\ time = 400 \: seconds \\ velocity =  \:  \frac{distance \:t ravelled}{time \: taken}  \\  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  =  \frac{2000}{400}  \\  \:  \:  \:  \:  \:  \:  \:  \:  \:  = 5 \: metre \: per \: second \\ hope \: it \: helps</em>

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Test populations are studied. Population one is found to obey the differential equation dy1/da=o.2y1 and the population two obey
oksano4ka [1.4K]

Answer:

Population 1 indicates growth while Population 2 indicates a declining population

Explanation:

Here, using the given rate of change of the population, we want to determine which of the two is growing and which is declining

From the rate of change of both, we can determine this. Looking at the differential equation for the first one, we can see that it is of positive value. Looking at the differential equation for the second one. we can see it is of negative value

While a positive change rate indicates growth, a negative change rate will indicate otherwise

Hence, we can conclude that the one with a negative rate change will indicate a declining population

8 0
3 years ago
Which of the following statements is FALSE about collisions? (Consider Newton's Laws an the Conservation of Momentum.)
pantera1 [17]

<u>The following statements are false about collisions: </u>

  • The velocity change of two respective objects involved in a collision will always be equal.
  • Total momentum is always conserved between any two objects involved in a collision.

Answer: Option B, and D

<u>Explanation: </u>

In any collisions, equal amount of net force will be acted upon the colliding objects due to the third law of Newton, irrespective of the significance difference in mass of the objects. Similarly, they can also have different acceleration values during collision of two objects if the masses are identical.

But the statements regarding the equal change in velocity of two objects respectively involved in collision always is false, as the conservation of momentum is applicable for isolated system only. So it is true for only isolated system and not in all the systems.

The same reason goes for falsifying the fourth statement which states that total momentum is always conserved between two objects involved in a collision as this statement is only true for isolated system where the conservation of momentum can be applied. Thus the second and fourth statement is false regarding collision.

8 0
3 years ago
Thad needs to seek mental health care because he is experiencing intense anxiety which of the following providers is Least quali
Kisachek [45]

Answer:

psychologist

Explanation:

8 0
3 years ago
The World-War II battleship U.S.S Massachusetts used 16-inch guns whose barrel lengths were 15 m long. The shells each of mass 1
Vaselesa [24]

Answer:

The explosive force experienced by the shell inside the barrel is 23437500 newtons.

Explanation:

Let suppose that shells are not experiencing any effect from non-conservative forces (i.e. friction, air viscosity) and changes in gravitational potential energy are negligible. The explosive force experienced by the shell inside the barrel can be estimated by Work-Energy Theorem, represented by the following formula:

F\cdot \Delta s = \frac{1}{2}\cdot m \cdot (v_{f}^{2}-v_{o}^{2}) (1)

Where:

F - Explosive force, measured in newtons.

\Delta s - Barrel length, measured in meters.

m - Mass of the shell, measured in kilograms.

v_{o}, v_{f} - Initial and final speeds of the shell, measured in meters per second.

If we know that m = 1250\,kg, v_{o} = 0\,\frac{m}{s}, v_{f} = 750\,\frac{m}{s} and \Delta s = 15\,m, then the explosive force experienced by the shell inside the barrel is:

F = \frac{m\cdot (v_{f}^{2}-v_{o}^{2})}{2\cdot \Delta s}

F = \frac{(1250\,kg)\cdot \left[\left(750\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}\right]}{2\cdot (15\,m)}

F = 23437500\,N

The explosive force experienced by the shell inside the barrel is 23437500 newtons.

6 0
3 years ago
ASAP pls answer will mark brainiest If right
Taya2010 [7]

1. false.

2. true.

3. false.

4. true.

5. true.

6. false.

7. false. (I think).

8. true.

9. false.

10. true.

6 0
4 years ago
Read 2 more answers
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