Answer:
A 10 N force pointing up
Explanation:
If the net acceleration of the object is horizontal pointing to the right, that means that all vertical forces must have canceled out, and the only ones "unbalanced" are the horizontal ones (10 N to the right minus 5 N to the left giving a net force of 5 N to the right).
Since they mentioned only one vertical force pointing down (10 N), there must be another one of same magnitude but pointing in opposite direction (up).
Then there must also be a 10 N force pointing up acting on the object.
Answer: The correct answer is "metalloid".
Explanation:
Metal is the material which can conduct electricity as it contains free electrons. It is good conductor of electricity. For example, copper, silver.
Metal can be drawn into wires. This property of material is called ductile.
Metal can be beaten into sheets. This property of material is called malleable.
Non metal is the material which cannot conduct electricity as it does not contain free electrons. They are poor conductor of electricity. For example, oxygen.
Metalloid: It has properties of both metals and non metal. It is electrical conductor. For example, semiconductor- silicon and germanium. But they cannot be pulled into wires.
Therefore, a material you are testing conducts electricity but cannot be pulled into wires. It is most likely a metalloid.
Hello,
<u>Solution for A:</u>
Force = 3.00N
Mass = 0.50 Kgs
Time = 1.50 Seconds
According to newton's second law of motion;
Force = Mass times Acceleration(a)
3.00 = 0.50 * a
a = 3.00/0.50 = 6.00 m/s^2
We know that acceleration = Velocity / time
So Velocity = time * acceleration = 1.50 * 6 = 9.00 m/s^2
<u>Solution for B:</u>
The net force = 4.00N - 3.00N = 1.00N to the left
Force = 1.00N
Mass = 0.50Kg
Time = 3.00 Seconds
Again; F = MA (Where F is force, M is mass and A is acceleration)
1.00N = 0.5 * A
A = 1/0.5 = 2 m/s^2
Velocity = Acceleration * Time = 2 * 3 = 6 m/s
Incomplete question.The Complete question is here
A flat uniform circular disk (radius = 2.00 m, mass = 1.00 ✕ 102 kg) is initially stationary. The disk is free to rotate in the horizontal plane about a friction less axis perpendicular to the center of the disk. A 40.0-kg person, standing 1.25 m from the axis, begins to run on the disk in a circular path and has a tangential speed of 2.00 m/s relative to the ground.
a.) Find the resulting angular speed of the disk (in rad/s) and describe the direction of the rotation.
b.) Determine the time it takes for a spot marking the starting point to pass again beneath the runner's feet.
Answer:
(a)ω = 1 rad/s
(b)t = 2.41 s
Explanation:
(a) initial angular momentum = final angular momentum
0 = L for disk + L............... for runner
0 = Iω² - mv²r ...................they're opposite in direction
0 = (MR²/2)(ω²) - mv²r
................where is ω is angular speed which is required in part (a) of question
0 = [(1.00×10²kg)(2.00 m)² / 2](ω²) - (40.0 kg)(2.00 m/s)²(1.25 m)
0=200ω²-200
200=200ω²
ω = 1 rad/s
b.)
lets assume the "starting point" is a point marked on the disk.
The person's angular speed is
v/r = (2.00 m/s) / (1.25 m) = 1.6 rad/s
As the person and the disk are moving in opposite directions, the person will run part of a revolution and the turning disk would complete the whole revolution.
(angle) + (angle disk turns) = 2π
(1.6 rad/s)(t) + ωt = 2π
t[1.6 rad/s + 1 rad/s] = 2π
t = 2.41 s
