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2 years ago

6

A car is moving in the positive direction along a straight highway and accelerates at a constant rate while going from point A t

o point B. If the acceleration is positive, increasing the speed of the car, when does the position where the average speed equals the instantaneous speed occur during the time interval from A to B? Assume the time interval is T.

1 answer:

rusak2 [61]2 years ago

8 0

Answer:

The time where the avergae speed equals the instaneous speed is T/2

Explanation:

The velocity of the car is:

v(t) = v0 + at

Where v0 is the initial speed and a is the constant acceleration.

Let's find the average speed. This is given integrating the velocity from 0 to T and dividing by T:

$v_{ave} = \frac{1}{T}\int\limits^T_0 {v(t)} \, dt$

v_ave = v0+a(T/2)

We can esaily note that when <u><em>t=T/2</em></u><u><em> </em></u>

v(T/2)=v_ave

Now we want to know where the car should be, the osition of the car is:

$x(t) = x_A + v_0 t + \frac{1}{2}at^2$

Where x_A is the position of point A. Therefore, the car will be at:

<u><em>x(T/2) = x_A + v_0 (T/2) + (1/8)aT^2</em></u>

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A large, cylindrical water tank with diameter 3.60 m is on a platform 2.00 m above the ground. The vertical tank is open to the

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To solve this problem it is necessary to apply the concepts related to the geometry of a cylindrical tank and its respective definition.

The volume of a tank is given by

$V = \frac{\pi d^2}{4}h$

Where

d = Diameter

h = Height

Considering that there are two stages, let's define the initial and final volume as,

$V_0 = \frac{\pi d^2}{4}H$

$V_f = \frac{\pi d^2}{4}h$

We know as well by definition that

$1gal = 3.785*10^{-3}m^3$

Then we have for the statement that

$V_f = V_0 -1gal$

$V_f = V_0 - 3.785*10^{-3}$

Replacing the previous data

$\frac{\pi d^2}{4}h = \frac{\pi d^2}{4}H- 3.785*10^{-3}$

$\frac{\pi (3.6)^2}{4}h = \frac{\pi (3.6)^2}{4}(2)- 3.785*10^{-3}$

Solving to get h,

$h = 1.99963m$

Therefore the change is

$\Delta h = H-h$

$\Delta h = 2- 1.99963$

$\Delta h = 3.7*10^{-4}m=0.37mm$

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3 years ago

One block rests upon a horizontal surface. A second identical block rests upon the first one. The coefficient of static friction

Sophie [7]

Answer:

F = 156.3 N

Explanation:

Let's start with the top block, apply Newton's second law

F - fr = 0

F = fr

fr = 52.1 N

Now we can work with the bottom block

In this case we have two friction forces, one between the two blocks and the other between the block and the surface. In the exercise, indicate that the two friction coefficients are equal

we apply Newton's second law

Y axis

N - W₁ -W₂ = 0

N = W₁ + W₂

as the two blocks are identical

N = 2W

X axis

F - fr₁ - fr₂ = 0

F = fr₁ + fr₂

indicates that the lower block is moving below block 1, therefore the upper friction force is

fr₁ = 52.1 N

fr₁ = μ N

a

s the normal in the lower block of twice the friction force is

fr₂ = μ 2N

fr₂ = 2 μ N

fr₂ = 2 fr₁

we substitute

F = fr₁ + 2 fr₁

F = 3 fr₁

F = 3 52.1

F = 156.3 N

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2 years ago

11) A man is on a 1/4 on a bridge. A train is coming the same direction he is going. The man can run across the bridge in the sa

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15mph

If the man turns and runs toward point A, he will cover

3/8 of the length of the bridge in the time that it takes

the train to reach A.

If the man runs forward toward point B, what part of the bridge

will he cover before the train reaches A? Well, he will cover

3/8 of the bridge, only heading forward toward B. This will put

him 3/8 + 3/8 = 6/8 = 3/4 of the way across the bridge by the

time the train reaches A.

since we know that the man and the train will meet at B, this

means that in the time it takes the man to run the remaining

1/4 of the bridge, the train will cover the entire length of

the bridge.

If it takes the man the same time to cover 1/4 of the bridge

that it takes the train to cover the whole bridge, then the train

must be going four times as fast as the man. Another way of saying

this is that the man runs at 1/4 the speed of the train.

Since the train's speed is known to be 60 mph, this means that

the man runs at (1/4) 60 = 15 mph.

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2 years ago

A small box slides down a ramp on a friction with surface. If the total energy of the system is conserved, which computational m

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3 years ago

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A car is traveling up a hill that is inclined at an angle θ above the horizontal. Determine the ratio of the magnitude of the no

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Explanation:

The weight of the car is equal to, $W_c=m\times g$ ...........(1)

Where

m is the mass of car

g is the acceleration due to gravity

The normal or vertical component of the force is, $F_N=mg\ cos\theta$

or

$F_N=mg\ cos(13)$ .............(2)

The horizontal component of the force is, $F_H=mg\ sin\theta$

Taking ratio of equation (1) and (2) as :

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$\dfrac{F_N}{W_c}=cos(13)$

$\dfrac{F_N}{W_c}=0.97$

or

$\dfrac{F_N}{W_c}=\dfrac{97}{100}$

Hence, this is the required solution.

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2 years ago