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nexus9112 [7]
3 years ago
9

Anyone can help me with this

Physics
2 answers:
Fantom [35]3 years ago
8 0

Answer:

this is in the middle of the wheel and helps it rotate it is an axle

Flura [38]3 years ago
8 0
The center of a wheel is called an AXLE
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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 of Newton’s laws of accounts for the following statement?
Neko [114]
The answer is B


second law
4 0
3 years ago
An object falls from a high building and hits the ground in 8.0 seconds. Ignoring air resistance, what is the distance that it f
d1i1m1o1n [39]

Answer:

310 meters

Explanation:

Given:

v₀ = 0 m/s

t = 8.0 s

a = -9.8 m/s²

Find: Δy

Δy = v₀ t + ½ at²

Δy = (0 m/s) (8.0 s) + ½ (-9.8 m/s²) (8.0 s)²

Δy = -313.6

Rounded to two significant figures, the object fell 310 meters.

4 0
3 years ago
the weight of 80 kg of mass on mercury is 296 N and almost identical to the weight of the same mass on Mars but mercury has much
Dmitrij [34]
Where Gravity rely's on only mass and distance and nothing else, so the weight on the planets will vary like you have stated. However Mars is smaller than Mercury, so the weight on Mars will be less, and the weight on Mercury will be more. Think this way.

More Mass = More Gravity = More Weight

Less Mass = Less Gravity = Less Weight
3 0
3 years ago
A bike and rider together have a mass of 60 kg. If the bike and rider have an acceleration of 2.0 m/s^2, what is the force on th
Mice21 [21]

The net force on the bike and the rider is 120 N

Explanation:

We can solve this problem by applying Newton's second law of motion, which states that:

F = ma

where

F is the net force exerted on an object

m is the mass of the object

a is its acceleration

For the bike and the rider in this problem, we have

m = 60 kg is their combined mass

a=2.0 m/s^2 is their acceleration

Therefore, the net force on them is

F=(60)(2.0)=120 N

Learn more about Newton's second law:

brainly.com/question/3820012

#LearnwithBrainly

8 0
4 years ago
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