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Nimfa-mama [501]
3 years ago
9

How many legs does an arachnid have??

Physics
1 answer:
Virty [35]3 years ago
6 0
An arachnid has eight legs.
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What important event takes place during the interphase of the cell cycle?
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The dna is multiplied obviously
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3 years ago
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A particle moves through an xyz coordinate system while a force acts on it. When the particle has the position vector r with arr
Paha777 [63]

Answer:

The question is incomplete, below is the complete question "A particle moves through an xyz coordinate system while a force acts on it. When the particle has the position vector r with arrow = (2.00 m)i hat − (3.00 m)j + (2.00 m)k, the force is F with arrow = Fxi hat + (7.00 N)j − (5.00 N)k and the corresponding torque about the origin is vector tau = (4 N · m)i hat + (10 N · m)j + (11N · m)k.

Determine Fx."

F_{x}=-1N.m

Explanation:

We asked to determine the "x" component of the applied force. To do this, we need to write out the expression for the torque in the in vector representation.

torque=cross product of force and position . mathematically this can be express as

T=r*F

Where

F=F_{x}i+(7N)j-(5N)k  and the position vector

r=(2m)i-(3m)j+(2m)k

using the determinant method to expand the cross product in order to determine the torque we have

\left[\begin{array}{ccc}i&j&k\\2&-3&2\\ F_{x} &7&-5\end{array}\right]\\\\

by expanding we arrive at

T=(18-14)i-(-12-2F_{x})j+(12+3F_{x})k\\T=4i-(-12-2F_{x})j+(12+3F_{x})k\\\\

since we have determine the vector value of the toque, we now compare with the torque value given in the question

(4Nm)i+(10Nm)j+(11Nm)k=4i-(-12-2F_{x})j+(12+3F_{x})k\\

if we directly compare the j coordinate we have

10=-(-12-2F_{x})\\10=12+2F_{x}\\ 10-12=2F_{x}\\ F_{x}=-1N.m

8 0
3 years ago
Which action describes an experimental investigation?
EastWind [94]
I think the answer is c
3 0
3 years ago
A pear hangs in a tree at a height of 1.8 m. The pear has a mass of 0.2 kg. The pear falls out of the tree and lands on the grou
TiliK225 [7]

a) PE=mgh=0.2*9.8*1.2=2.352 J

b) KE=PE=2.352 J

c) v=\sqrt{\frac{2KE}{m}}=4.85 m/s

6 0
4 years ago
A toroidal solenoid has an inner radius of 12.0 cm and an outer radius of 15.0 cm . It carries a current of 1.50 A . Part A How
AleksAgata [21]

Answer:

The number of turns, N = 1750

Explanation:

It is given that,

The inner radius of a toroid, r = 12 cm

Outer radius, r' = 15 cm

The magnetic field at points within the coils 14 cm from its center is, B=3.75\ mT=3.75\times 10^{-3}\ T

R = 14 cm = 0.14 m

Current, I = 1.5 A

The formula for the magnetic field at some distance from its center is given by :

B=\dfrac{\mu_o NI}{2\pi R}

N=\dfrac{2\pi R B}{\mu_o I}

N=\dfrac{2\pi \times 0.14\times 3.75\times 10^{-3}}{4\pi \times 10^{-7}\times 1.5}

N = 1750

So, the number of turns must have in a toroidal solenoid is 1750. Hence, this is the required solution.

3 0
3 years ago
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