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2 years ago

A small glider is coasting horizontally when suddenly a very heavy piece of cargo falls out of the bottom of the plane.

Physics

1 answer:

myrzilka [38]2 years ago

5 0

Answer:

a. The plane speeds up but the cargo does not change speed.

Explanation:

Just to make it clear, the question is as follows from what I understand.

A small glider is coasting horizontally when suddenly a very heavy piece of cargo falls out of the bottom of the plane. You can neglect air resistance.

Just after the cargo has fallen out:

a. The plane speeds up but the cargo does not change speed.

b. The cargo slows down but the plane does not change speed.

c. Neither the cargo nor the plane change speed.

d. The plane speeds up and the cargo slows down.

e. Both the cargo and the plane speed up.

And we are requested to choose the right answer under the given conditions. We know the glider has no motor, then it must be in free fall movement, then it is experiencing some force that pulls it to the from due to the gravity effect on it, and a force in general is calculated by

F=m*a, m:= mass of the object, a:= acceleration.

Here we are only considering the horizontal effect of the forces, then since the mass is reduced the acceleration must increase to compensate and maintain the equilibrium of the forces, then the glider being lighter can travel faster due to the acceleration. On the other hand by the time the cargo left the glider there was no acceleration and the speed it had at the moment he left the plane continues, then the cargo does not change its speed, then horizontally speaking the answer would be a. The plane speeds up but the cargo does not change speed.

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A 26 foot ladder is lowered down a vertical wall at a rate of 3 feet per minute. The base of the ladder is sliding away from the

lakkis [162]

Answer:

(i) 7.2 feet per minute.

(ii) No, the rate would be different.

(iii) The rate would be always positive.

(iv) the resultant change would be constant.

(v) 0 feet per min

Explanation:

Let the length of ladder is l, x be the height of the top of the ladder from the ground and y be the length of the bottom of the ladder from the wall,

By making the diagram of this situation,

Applying Pythagoras theorem,

$l^2 = x^2 + y^2-----(1)$

Differentiating with respect to t ( time ),

$0=2x\frac{dx}{dt} + 2y\frac{dy}{dt}$ ( l = 26 feet = constant )

$\implies 2y\frac{dy}{dt} = -2x\frac{dx}{dt}$

$\implies \frac{dy}{dt}=-\frac{x}{y}\frac{dx}{dt}$

We have,

$y = 10, \frac{dx}{dt}= -3\text{ feet per min}$

$\frac{dy}{dt}=\frac{3x}{10}-----(X)$

(i) From equation (1),

$26^2 = x^2 + 10^2$

$676=x^2 + 100$

$576 = x^2$

$\implies x = 24\text{ feet}$

From equation (X),

$\frac{dy}{dt}=\frac{3\times 24}{10}=7.2\text{ feet per min}$

(ii) From equation (X),

$\frac{dy}{dt}\propto x$

Thus, for different value of x the value of $\frac{dy}{dt}$ would be different.

(iii) Since, distance = Positive number,

So, the value of y will always a positive number.

Thus, from equation (X),

The rate would always be a positive.

(iv) The length of the ladder is constant, so, the resultant change would be constant.

i.e. x = increases ⇒ y = decreases

y = decreases ⇒ y = increases

(v) if ladder hit the ground x = 0,

So, from equation (X),

$\frac{dy}{dt}=0\text{ feet per min}$

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2 years ago

|| Climbing ropes stretch when they catch a falling climber, thus increasing the time it takes the climber to come to rest and r

Otrada [13]

To solve this problem it is necessary to apply the concepts related to Newton's second law and the kinematic equations of movement description.

Newton's second law is defined as

$F = ma$

Where,

m = mass

a = acceleration

From this equation we can figure the acceleration out, then

$a = \frac{F}{m}$

$a = \frac{11*10^3}{80}$

$a = 137.5m/s$

From the cinematic equations of motion we know that

$v_f^2-v_i^2 = 2ax$

Where,

$v_f =$ Final velocity

$v_i =$ Initial velocity

a = acceleration

x = displacement

There is not Final velocity and the acceleration is equal to the gravity, then

$v_f^2-v_i^2 = 2ax$

$0-v_i^2 = 2(-g)x$

$v_i =\sqrt{2gx}$

$v_i = \sqrt{2*9.8*4.8}$

$v_i = 9.69m/s$

From the equation of motion where acceleration is equal to the velocity in function of time we have

$a = \frac{v_i}{t}$

$t = \frac{v_i}{a}$

$t =\frac{9.69}{137.5}$

$t = 0.0705s$

Therefore the time required is 0.0705s

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2 years ago

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hoa [83]

Answer: C.

Explanation:

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2 years ago

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Answer:

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2 years ago

A spaceship of frontal area 10 m2 moves through a large dust cloud with a speed of 1 x 106 m/s. The mass density of the dust is

Step2247 [10]

Answer:

The decelerating force is $3\times 10^{- 11}\ N$

Solution:

As per the question:

Frontal Area, A = $10\ m^{2}$

Speed of the spaceship, v = $1\times 10^{6}\ m/s$

Mass density of dust, $\rho_{d} = 3\times 10^{- 18}\ kg/m^{3}$

Now, to calculate the average decelerating force exerted by the particle:

$Mass,\ m = \rho_{d}V$ (1)

Volume, $V = A\times v\times t$

Thus substituting the value of volume, V in eqn (1):

$m = \rho_{d}(Avt)$

where

A = Area

v = velocity

t = time

$m = \rho_{d}(A\times v\times t)$ (2)

$Momentum,\ p = \rho_{d}(Avt)v = \rho_{d}Av^{2}t$

From Newton's second law of motion:

$F = \frac{dp}{dt}$

Thus differentiating w.r.t time 't':

$F_{avg} = \frac{d}{dt}(\rho_{d}Av^{2}t) = \rho_{d}Av^{2}$

where

$F_{avg}$ = average decelerating force of the particle

Now, substituting suitable values in the above eqn:

$F_{avg} = 3\times 10^{- 18}\times 10\times 1\times 10^{6} = 3\times 10^{- 11}\ N$

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2 years ago