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3 years ago

What minimum speed must the rocket have just before impact in order to save the explorer's life?

1 answer:

34kurt3 years ago

8 0

<span>Answer: Assuming that I understand the geometry correctly, the combine package-rocket will move off the cliff with only a horizontal velocity component. The package will then fall under gravity traversing the height of the cliff (h) in a time T given by h = 0.5*g*T^2 However, the speed of the package-rocket system must be sufficient to cross the river in that time v2 = L/T Conservation of momentum says that m1*v1 = (m1 + m2)*v2 where m1 is the mass of the rocket, v1 is the speed of the rocket, m2 is the mass of the package, and v2 is the speed of the package-rocket system. Expressing v2 in terms of v1 v2 = m1*v1/(m1 + m2) and then expressing the time in terms of v1 T = (m1 + m2)*L/(m1*v1) substituting T in the first expression h = 0.5*g*(m1 + m2)^2*L^2/(m1*v1)^2 solving for v1, the speed before impact is given by v1 = sqrt(0.5*g/h)*(m1 + m2)*L/m1</span>

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Which of the following is most likely to be an observation made by a physiologist

Naya [18.7K]

Do you have the answer choices ?

3 0

3 years ago

By the time she managed/had managed to open the door, the postman already went/had already gone.

leva [86]

Answer:

<h3>Answer Sentence: </h3><h2><em><u>By the time she had managed to open the door, the postman had already gone</u></em>.</h2>

8 0

2 years ago

A tow truck exerts a net horizontal force of 1050 N on a 760-kg car. What is the acceleration of the car during this time?

vovikov84 [41]

We can solve the problem by using Newton's second law of motion:
$F=ma$
where
F is the net force applied to the object
m is the object's mass
a is the acceleration of the object

In this problem, the force applied to the car is F=1050 N, while the mass of the car is m=760 kg. Therefore, we can rearrange the equation and put these numbers in, in order to find the acceleration of the car:
$a= \frac{F}{m}= \frac{1050 N}{760 kg}=1.4 m/s^2$

The equation also tells us that the acceleration and the force have same directions: therefore, since the force exerted on the car is horizontal, the correct answer is
<span>B) 1.4 m/s2 horizontally.</span>

5 0

3 years ago

A heat engine accepts 200,000 Btu of heat from a source at 1500 R and rejects 100,000 Btu of heat to a sink at 600 R. Calculate

diamong [38]

To solve the problem it is necessary to apply the concepts related to the conservation of energy through the heat transferred and the work done, as well as through the calculation of entropy due to heat and temperatra.

By definition we know that the change in entropy is given by

$\Delta S = \frac{Q}{T}$

Where,

Q = Heat transfer

T = Temperature

On the other hand we know that by conserving energy the work done in a system is equal to the change in heat transferred, that is

$W = Q_{source}-Q_{sink}$

According to the data given we have to,

$Q_{source} = 200000Btu$

$T_{source} = 1500R$

$Q_{sink} = 100000Btu$

$T_{sink} = 600R$

PART A) The total change in entropy, would be given by the changes that exist in the source and sink, that is

$\Delta S_{sink} = \frac{Q_{sink}}{T_{sink}}$

$\Delta S_{sink} = \frac{100000}{600}$

$\Delta S_{sink} = 166.67Btu/R$

On the other hand,

$\Delta S_{source} = \frac{Q_{source}}{T_{source}}$

$\Delta S_{source} = \frac{-200000}{1500}$

$\Delta S_{source} = -133.33Btu/R$

The total change of entropy would be,

$S = \Delta S_{source}+\Delta S_{sink}$

$S = -133.33+166.67$

$S = 33.34Btu/R$

Since $S\neq 0$ the heat engine is not reversible.

PART B)

Work done by heat engine is given by

$W=Q_{source}-Q_{sink}$

$W = 200000-100000$

$W = 100000 Btu$

Therefore the work in the system is 100000Btu

4 0

3 years ago

The power lines are at a high potential relative to the ground, so there is an electric field between the power lines and the gr

Amanda [17]

Answer:

The tube should be held vertically and perpendicular to the ground.

Explanation:

Answer: The tube should be held vertically and perpendicular to the ground. The reason is as follows:

Reasoning:

The power lines are parallel to the ground hence, their electric field will be perpendicular to the ground and equipotential surface will be cylindrical.

Hence, if you will put fluorescent tube parallel to the ground then both the ends of the tube will lie on the same equipotential surface and the potential difference will be zero.

So, to maximize the potential the ends of the tube must be on different equipotential surfaces. The surface which is near to the power line has high potential value and the surface which is farther from the line has lower potential value.

hence, to maximize the potential difference, the tube must be placed perpendicular to the ground.

6 0

2 years ago