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N76 [4]
3 years ago
13

A cylindrical can is partially filled with water. The radius of the cylindrical can is 8 inches, and its height is 9 inches. If

the volume of the water is decreasing at a rate of 527 cubic inches per minute, what is the rate, in inches per minute, at which the height of the water is changing when the height of the water is 5 inches?
Physics
1 answer:
VashaNatasha [74]3 years ago
7 0

The volume of the cylindrical can is given by:

V = πr²h

V = volume, r = base radius, h = height

Differentiate both sides of the equation with respect to time t. The radius r doesn't change over time, so we treat it as a constant:

dV/dt = πr²(dh/dt)

Given values:

dV/dt = -527in³/min

r = 8in

Plug in and solve for dh/dt:

-527 = π(8)²(dh/dt)

dh/dt = -2.62in/min

The height of the water is decreasing at a rate of 2.62in/min

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GREYUIT [131]

Answer:

h = 1.8 m

Explanation:

The initial velocity of the glove, u =- 6 m/s

We need to find the maximum height of the glove. Let it is equal to h. Using equation of kinematics. At the maximum height v = 0

v^2-u^2=2ah, h is the maximum height and a = -g

0^2-(6)^2=2\times (-10)\times h\\\\h=\dfrac{36}{20}\\\\h=1.8\ m

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4 0
3 years ago
What could be the possible answer to the question ?<br><br>thankyou ~​
Ganezh [65]

The value of the force, F₀, at equilibrium is equal to the horizontal

component of the tension in string 2.

Response:

  • The value of F₀ so that string 1 remains vertical is approximately <u>0.377·M·g</u>

<h3>How can the equilibrium of forces be used to find the value of F₀?</h3>

Given:

The weight of the rod = The sum of the vertical forces in the strings

Therefore;

M·g = T₂·cos(37°) + T₁

The weight of the rod is at the middle.

Taking moment about point (2) gives;

M·g × L = T₁ × 2·L

Therefore;

T_1 = \mathbf{\dfrac{M \cdot g}{2}}

Which gives;

M \cdot g = \mathbf{T_2 \cdot cos(37 ^{\circ})+ \dfrac{M \cdot g}{2}}

T_2 = \dfrac{M \cdot g - \dfrac{M \cdot g}{2}}{cos(37 ^{\circ})}  = \mathbf{\dfrac{M \cdot g}{2 \cdot cos(37 ^{\circ})}}}

F₀ = T₂·sin(37°)

Which gives;

F_0 = \dfrac{M \cdot g \cdot sin(37 ^{\circ})}{2 \cdot cos(37 ^{\circ})}} = \dfrac{M \cdot g \cdot tan(37 ^{\circ})}{2}  \approx  \mathbf{0.377  \cdot M \cdot g}

  • F₀ ≈ <u>0.377·M·g</u>

<u />

Learn more about equilibrium of forces here:

brainly.com/question/6995192

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A substance made up of only one type of element
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A skier descends a mountain at an angle of 35.0º to the horizontal. If the mountain is 235 m long, what are the horizontal and v
neonofarm [45]
Total displacement along the length of mountain is given as
L = 235 m
angle of mountain with horizontal = 35 degree
now we will have horizontal displacement as
x = L cos35
x = 235 cos35 = 192.5 m
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