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3 years ago

A tank of water has a length 10.0 m, width 5.00 m, and depth 2.50 m. What the absolute pressure at the bottom of the tank?

Physics

1 answer:

Damm [24]3 years ago

6 0

Answer:

1.28 x 10^5 Pa

Explanation:

The absolute pressure at the bottom of the tank of water is given by:

$p= p_0 + \rho g h$

where

$p_0 = 1.03 \cdot 10^5 Pa$ is the atmospheric pressure

$\rho = 1000 kg/m^3$ is the water density

g = 9.8 m/s^2 is the acceleration of gravity

h = 2.50 m is the heigth of the column of water

Substituting into the formula, we find

$p=1.03\cdot 10^5 +(1000)(9.8)(2.50)=1.28\cdot 10^5 Pa$

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1. What is the potential energy of a 5.0-kg

babymother [125]

Answer:

C. 98 J

Explanation:

The appropriate formula is ...

PE = mgh . . . . . m is mass; below, m is meters

PE = (5 kg)(9.8 m/s^2)(2 m) = 98 kg·m^2/s^2

PE = 98 J

5 0

3 years ago

A carbon resistor is to be used as a thermometer. On a winter day when the temperature is 4.00°C, the resistance of the carbon r

dusya [7]

Answer:

28 degree C

Explanation:

We are given that

$T_1=4.00^{\circ}$

$R_1=217.7 \Ohm$

$R_2=215.1\Ohm$

$\alpha=-5.00\times 10^{-4}C^{-1}$

We have to find the temperature on a spring day when resistance is 215.1 ohm.

We know that

$\alpha(T_2-T_1)=\frac{R_2}{R_2}-1$

Using the formula

$-5.00\times 10^{-4}(T_2-4)=\frac{215.1}{217.7}-1$

$-5\times 10^{-4}(T_2-4)=0.988-1=-0.012$

$T_2-4=\frac{0.012}{5\times 10^{-4}}=24$

$T_2=24+4=28^{\circ}C$

Hence, the temperature on a spring day 28 degree C.

7 0

3 years ago

A _____ is used in car headlights.

uysha [10]

E concave mirror because it reflects the light

7 0

3 years ago

Read 2 more answers

Determine experimentally which rotational axis yields the maximum rotational inertia (i.e., moment of inertia) and which yields

Law Incorporation [45]

Answer:

the maximum is I₁ axis of rotation at the end

the minimum moment is I₂ axis of rotation at the center of mass

Explanation:

For this exercise we use the definition moment of inertia

I = ∫ r² dm

for bodies of high symmetry it is tabulated; In this case we can approximate a broomstick to a thin rod, the moment of inertia with respect to a perpendicular axis when varying are

at one end

I₁ = ⅓ mL²

in in center

I₂ = $\frac{1}{12}$ m L²

There is another possible axis of rotation around the axis of the broom, in this case we have a solid cylinder

I₃ = $\frac{1}{2}$ m r²

remember that the diameter of the broom is much smaller than its length, therefore this moment of inertia is very small

when examining the different moments of inertia:

the maximum is I₁ axis of rotation at the end

the minimum moment is I₂ axis of rotation at the center of mass

3 0

2 years ago

An antelope moving with constant acceleration covers the distance 70.0 m between two points in time 6.60 s. Its speed as it pass

Veronika [31]

Answer:

A) The speed at the first point is 6.91 m/s.

B) The acceleration is 1.12 m/s²

Explanation:

The equations of velocity and position of the antelope will be as follows:

x = x0 + v0 · t + 1/2 · a · t²

v = v0 + a · t

Where:

x = position of the antelope at time t

x0 = initial position

v0 = initial velocity

t = time

a = acceleration

v = velocity at time t

A) If we place the origin of the frame of reference at the first point, we can say that at t = 6.60 the position of the antelope is 70.0 m and its velocity is 14.3 m/s. In this way, we will have 2 equations with 2 unknowns, the initial velocity (the velocity at the first point) and the acceleration.

Let´s start finding the speed at the first point:

v = v0 + a · t (solving for "a")

(v - v0)/t = a

Replacing a = (v - v0)/t in the equation of the position:

x = x0 + v0 · t + 1/2 · (v - v0)/t · t² (x0 = 0)

x = v0 · t + 1/2 · v · t - 1/2 · v0 · t

x - 1/2 · v · t = 1/2 · v0 · t

2/t · (x - 1/2 · v · t) = v0

2/6.60 s · (70.0 m - 1/2 · 14.3 m/s · 6.60s) = v0

v0 = 6.91 m/s

The speed at the first point is 6.91 m/s.

B) Using the equation of velocity

a = (v - v0)/t

a = (14,3 m/s - 6.91 m/s) / 6.60 s

a = 1.12 m/s²

The acceleration is 1.12 m/s²

4 0

3 years ago