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Inessa [10]
3 years ago
6

If the acceleration of the projective is: a = c s m/s 2 Where c is a constant that depends on the initial gas pressure behind th

e projectile. The initial position of the projectile is s=1.5m and the projectile is initially at rest. The projectile accelerates until it reaches the end of the barrel at s=3m. What is the value of the constant c such that the projectile leaves the barrel with a velocity of 200 m/s?
Physics
1 answer:
Sedbober [7]3 years ago
6 0

Answer:

c = 4,444.44

Explanation:

You have the following expression for the acceleration of the projectile:

a=cs   (1)

s: distance to the ground of the projectile

To find the value of the constant c you use the following formula:

v^2=v_o^2+2a \Delta s   (2)

vo: initial  velocity = 0 m/s

v: final speed = 200 m/s

Δs: distance traveled by the projectile = 3m - 1.5m = 1.5m

You replace the expression (1) into the expression (2):

v^2=2(cs)\Delta s

You do the constant c in the last equation, then you replace the values of v, s and Δs:

c=\frac{v^2}{2s\Delta s}=\frac{(200m/s)^2}{2(3m/s^2)(1.5m)}=4444.44

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the volume of a water tank is 5m×4m×2m. If the tank is half filled with water. calculate the pressure exerted at the bottom of t
Natali5045456 [20]

Answer:

10⁴ Pa

Explanation:

  • Volume = 5m * 4m * 2m

When the tank is half filled ,

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Also we know that ,

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\longrightarrow Density = Mass/Volume

\longrightarrow 10³ kg/m³ = m/20m³

\longrightarrow m = 2 * 10⁴ kg

So that ,

\longrightarrow Weight = mg

\longrightarrow F = 2*10⁴ × 10 N

\longrightarrow F = 2 * 10⁵ N

And ,

\longrightarrow Pressure = Force/Area

\longrightarrow P = 2 *10⁵/ 5 * 4 Pa

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3 years ago
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An offshore oil well is 2 kilometers off the coast. The refinery is 4 kilometers down the coast. Laying pipe in the ocean is twi
shusha [124]

Answer:

Rectangular path

Solution:

As per the question:

Length, a = 4 km

Height, h = 2 km

In order to minimize the cost let us denote the side of the square bottom be 'a'

Thus the area of the bottom of the square, A = a^{2}

Let the height of the bin be 'h'

Therefore the total area, A_{t} = 4ah

The cost is:

C = 2sh

Volume of the box, V = a^{2}h = 4^{2}\times 2 = 128            (1)

Total cost, C_{t} = 2a^{2} + 2ah            (2)

From eqn (1):

h = \frac{128}{a^{2}}

Using the above value in eqn (1):

C(a) = 2a^{2} + 2a\frac{128}{a^{2}} = 2a^{2} + \frac{256}{a}

C(a) = 2a^{2} + \frac{256}{a}

Differentiating the above eqn w.r.t 'a':

C'(a) = 4a - \frac{256}{a^{2}} = \frac{4a^{3} - 256}{a^{2}}

For the required solution equating the above eqn to zero:

\frac{4a^{3} - 256}{a^{2}} = 0

\frac{4a^{3} - 256}{a^{2}} = 0

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The path in order to minimize the cost must be a rectangle.

8 0
3 years ago
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