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2 years ago

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An airplane travels 2100 km at a speed of 1000 km/h. It then encounters a headwind which slows it to 800 km/h for the next 1300

km. What is the average speed for this plane trip?

1 answer:

Sidana [21]2 years ago

5 0

I don’t get it I need help

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1.How is the law of conservation of energy demonstrated by the movement of the pendulum?

Snezhnost [94]

1. Law of conservation of energy states that energy cannot be created, nor destroyed, for example, windmills take kinetic energy(movement energy) and convert it into electrical energy using gears and a generator as well as the blades.

so this supports it because the pendulum never reaches the same height twice unless you reset it so the energy is always getting less and less and not randomly getting back onto the pendulum.

2.Gravity, friction and air resistance slow it down as well

3. at the top, potential energy is the amount of energy something has relative to the amount it can disperse before stopping, for example, a book on a shelf has more potential energy than that of a book on a table, this is because when the shelf book falls it will create more energy than the table book.

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2 years ago

A(n) 7.7-kg object is sliding across the ice at 2.34 m/s in the positive x direction. An internal explosion occurs, splitting th

-BARSIC- [3]

Answer:

Average acceleration on first part of the chunk is given as

$a_1 = 13.125 m/s^2$

Average acceleration on second part of the chunk is given as

$a_2 = -13.125 m/s^2$

Explanation:

By momentum conservation along x direction we will have

$mv_i = \frac{m}{2}v_1 + \frac{m}{2}v_2$

so we have

$v_1 + v_2 = 2v$

$v_1 + v_2 = 4.68$

also by energy conservation

$\frac{1}{2}(\frac{m}{2})v_1^2 + \frac{1}{2}(\frac{m}{2})v_2^2 - \frac{1}{2}mv^2 = 17 J$

$\frac{1}{4}m(v_1^2 + v_2^2) - \frac{1}{2}mv^2 = 17$

$(v_1^2 + v_2^2) - 2v^2 = \frac{4}{7.7}(17)$

$(4.68 - v_2)^2 + v_2^2 - 2v^2 = 8.83$

$21.9 + 2v_2^2 - 9.36 v_2 - 10.95 = 8.83$

$2v_2^2 - 9.36v_2 + 2.12 = 0$

by solving above equation we will have

$v_1 = 4.44 m/s$

$v_2 = 0.24 m/s$

Average acceleration on first part of the chunk is given as

$a_1 = \frac{4.44 - 2.34}{0.16}$

$a_1 = 13.125 m/s^2$

Average acceleration on second part of the chunk is given as

$a_2 = \frac{0.24 - 2.34}{0.16}$

$a_2 = -13.125 m/s^2$

5 0

3 years ago

7. If the magnitude of the gravitational force of

JulijaS [17]

Newton's Third Law states that for every action there is an opposite and equal reaction:

If the gravitational force of the Earth on the Moon is F then the gravitational force of the Moon on the Earth is also F

3 0

2 years ago

A baseball of mass m = 0.31 kg is spun vertically on a massless string of length L = 0.51m. The string can only support a tensio

natulia [17]

Given data:

* The mass of the baseball is 0.31 kg.

* The length of the string is 0.51 m.

* The maximum tension in the string is 7.5 N.

Solution:

The centripetal force acting on the ball at the top of the loop is,

$\begin{gathered} T+mg=\frac{mv^2}{L}_{} \\ v^2=\frac{L(T+mg)}{m} \\ v=\sqrt[]{\frac{L(T+mg)}{m}} \end{gathered}$

For the maximum velocity of the ball at the top of the vertical circular motion,

$v_{\max }=\sqrt[]{\frac{L(T_{\max }+mg)}{m}}$

where g is the acceleration due to gravity,

Substituting the known values,

$\begin{gathered} v_{\max }=\sqrt[]{\frac{0.51(7.5_{}+0.31\times9.8)}{0.31}} \\ v_{\max }=\sqrt[]{\frac{0.51(10.538)}{0.31}} \\ v_{\max }=\sqrt[]{17.34} \\ v_{\max }=4.16\text{ m/s} \end{gathered}$

Thus, the maximum speed of the ball at the top of the vertical circular motion is 4.16 meters per second.

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1 year ago

Which do you think changes the land more frozen water or flowing water?

Maksim231197 [3]

I believe flowing water changes the land because eventually frozen water has to melt to flowing water, right?

8 0

3 years ago

Read 2 more answers