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Luba_88 [7]
3 years ago
15

The vector position of an object is given by r what is the torque acting on the object about the origin when a force f = (−12.5i

hat) n acts on it? express your answer using unit vector notation.

Physics
1 answer:
attashe74 [19]3 years ago
7 0
Let the vector position of the object in the (x-y) plane be 
\vec{r} = x \hat{i} + y \hat{j}

The applied force is
\vec{f} = -12.5 \hat{i}


By definition, the applied torque is
\vec{T} = \vec{r} \times \vec{f} = (x\hat{i} + y\hat{j}) \times (-12.5y \hat{i}) = 12.5\hat{k}

Answer: 12.5y \, \hat{k}

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A pump uses a piston of 15 cm diameter that moves at 2.0 cm/s as it pushes a fluid through a pipe. what is the speed of the flui
Nitella [24]
The important point here is that volumetric flow rate in the pump and the pipe is the same.

Q = AV, where Q = Volumetric flow  rate, A = Cross sectional area, V = velocity

Q (pump) = (π*15^2)/4*2 = 353.43 cm^3/s
Q (pipe) = (π*(3/10)^2)/4*V = 0.071V

Q (pump) = Q (pipe)
0.071V = 353.43 => V = 5000 cm/s

Therefore, the flow  of water in the pipe is 5000 cm/s.
4 0
3 years ago
WILL GIVE BRAINLIEST TO THE CORRECT ANSWER
LenKa [72]

Answer:

See the answers below.

Explanation:

We can solve both problems using Newton's second law, which tells us that the sum of forces on a body is equal to the product of mass by acceleration.

∑F =m*a

where:

F = force [N] (units of newtons)

m = mass = 1000 [kg]

a = acceleration = 3 [m/s²]

F = 1000*3\\F=3000[N]

And the weight of any body can be calculated by means of the mass product by gravitational acceleration.

W=m*g\\W=1000*9.81\\W=9810 [N]

4 0
3 years ago
(a) Calculate the absolute pressure at the bottom of a freshwater lake at a point whose depth is 30.0 m. Assume the density of t
Law Incorporation [45]

Answer:

(a) The absolute pressure at the bottom of the freshwater lake is 395.3 kPa

(b) The force exerted by the water on the window is 36101.5 N

Explanation:

(a)

The absolute pressure is given by the formula

P = P_{o} + \rho gh

Where P is the absolute pressure

P_{o} is the atmospheric pressure

\rho is the density

g is the acceleration due to gravity (Take g = 9.8 m/s^{2} )

h is the height

From the question

h = 30.0 m

\rho = 1.00 × 10³ kg/m³ = 1000 kg/m³

P_{o} = 101.3 kPa = 101300 Pa

Using the formula

P = P_{o} + \rho gh

P = 101300 + (1000×9.8×30.0)

P = 101300 + 294000

P =395300 Pa

P = 395.3 kPa

Hence, the absolute pressure at the bottom of the freshwater lake is 395.3 kPa

(b)

For the force exerted

From

P = F/A

Where P is the pressure

F is the force

and A is the area

Then, F = P × A

Here, The area will be area of the window of the underwater vehicle.

Diameter of the circular window = 34.1 cm = 0.341 m

From Area = πD²/4

Then, A = π×(0.341)²/4 = 0.0913269 m²

Now,

From F = P × A

F = 395300 × 0.0913269

F = 36101.5 N

Hence, the force exerted by the water on the window is 36101.5 N

5 0
3 years ago
9- Under what circumstances would a vector have components that are equal in
valkas [14]

Explanation:

c. if the vector is oriented at 0° from the X -axis.

6 0
2 years ago
How do you solve for a net force
AVprozaik [17]

Answer:

Learning the formula.multiply mass accelebrations.the force(F)required to move an object of mass(M) with an acceleration (a) is given by the formula F = m x a.so, force = mass multiplied by accelebration.

8 0
3 years ago
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