Athlete’s foot is also called...

Assume that the position vector of A is r=i+j+k . Determine the moment about the origin O if the force F=(1)i+(0)j+(5)k . The mo

ddd [48]

Answer:

M₀ = 5i - 4j - k

Explanation:

Using the cross product method, the moment vector(M₀) of a force (F) is about a given point is equal to cross product of the vector A from the point (r) to anywhere on the line of action of the force itself. i.e

M₀ = r x F

From the question,

r = i + j + k

F = 1i + 0j + 5k

Therefore,

M₀ = (i + j + k) x (1i + 0j + 5k)

M₀ = $\left[\begin{array}{ccc}i&j&k\\1&1&1\\1&0&5\end{array}\right]$

M₀ = i(5 - 0) -j(5 - 1) + k(0 - 1)

M₀ = i(5) - j(4) + k(-1)

M₀ = 5i - 4j - k

Therefore, the moment about the origin O of the force F is

M₀ = 5i - 4j - k

3 0

2 years ago

At each corner of a square of side l there are point charges of magnitude Q, 2Q, 3Q, and 4Q.What is the magnitude and direction

bogdanovich [222]

Answer:

magnitude of force on charge 2Q = $\frac{KQ^{2} }{I^{2} }$

Direction of force on charge = 61 ⁰

Explanation:

The magnitude on the force on the charge can be evaluated by finding the net force acting on the charge 2Q i.e x-component of the net force and the y-component of the net force

║F║ = $\sqrt{f_{x}^{2} + f_{y}^{2} }$ = after considering the forces coming from Q, 3Q and 4Q AND APPLYING COULOMBS LAW

magnitude of force acting on 2Q = $\frac{KQ^{2} }{I^{2} }$

The direction of the force on charge 2Q is calculated as

tan ∅ = $\frac{f_{y} }{f_{x} }$ = 1.8284

therefore ∅ = $tan^{-1}$ 1.8284

= 61⁰

3 0

3 years ago

Like the filters falling through the air, a car on the freeway represents an object with a high Reynolds number traveling throug

Goshia [24]

Answer:

ΔF=125.22 %

Explanation:

We know that drag force on the car given as

$F_D=\dfrac{1}{2}\rho C_DA v^2$

$C_D$ =Drag coefficient

A=Projected area

v=Velocity

ρ=Density

All other quantity are constant so we can say that drag force and velocity can be given as

$\dfrac{F_D_1}{F_D_2}=\dfrac{v_1^2}{v_2^2}$

Now by putting the values

$\dfrac{F_D_1}{F_D_2}=\dfrac{v_1^2}{v_2^2}$

$\dfrac{F_D_1}{F_D_2}=\dfrac{50^2}{75^2}$

$\dfrac{F_D_1}{F_D_2}=0.444$

Percentage Change in the drag force

$\Delta F=\dfrac{F_D_2-F_D_1}{F_D_1}\times 100$

$\Delta F=\dfrac{F_D_2-0.444F_D_2}{0.444F_D_2}\times 100$

$\Delta F=\dfrac{1-0.444}{0.444}\times 100$

ΔF=125.22 %

Therefore force will increase by 125.22 %.

3 0

3 years ago

Sally takes two bar magnets and randomly tapes one end of each bar magnet. she labels the magnets A and B. She brings the taped

vlabodo [156]

Since the two taped poles of the magnets labeled A and B attracts one to each other, we know that the two taped poles are oppsosite.

So, you can predict with total certainty that when she brings the taped end of the third magnet (magnet C) near each of the first two magntes, in one case they will attract each other and in the other case they will repele mutually.

You are certain of that because, since the taped poles of the first two magnets are opposite, the pole of the third magnet has to be equal to one of the two first taped poles and opposite to the other of the two firest taped poles.

8 0

3 years ago

A 77.0 kg woman slides down a 42.6 m long waterslide inclined at 42.3. at the bottom, she is moving 20.3 m/s.how much work did f

zhuklara [117]

Answer:

5.791244495 KNm

Explanation:

The height h is given by, $h=42.6sin42.3^{o}$

Potential energy, PE is given by

PE=mgh where m is mass of the woman, g is acceleration due to gravity whose value is taken as $9.81 m/s^{2}$ and h is already given hence substituting 77 Kg for m we obtain

$PE=77*9.81*42.6sin42.3^{o}=21656.7095 Nm$

PE=21.6567095 KNm

We also know that Kinetic energy is given by $0.5mv^{2}$ where v is the velocity and substituting v for 20.3 we obtain

$KE=0.5*77*20.3^{2}=15865.465 Nm$

KE=15.865465 KNm

Friction work is the difference between PE and KE hence

Friction work=21.6567095 KNm-15865.465 Nm=5.791244495 KNm

8 0

3 years ago