Answer:
A change that produces one or more new substances is called
<u>a chemical change</u><u>.</u>
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Answer:
The normal force the ground exerts on the block, F = -300 N
Explanation:
Given data,
The block pulled up with a tension force, T = 100 N
The weight of the block, W = 300 N
The weight of the block is due to the force of attraction of gravitation.
The surface exerts a force that is equal and opposite to the force acting on the block due to gravitation.
The weight of the block,
W = mg
300 N
The normal force the ground exerts on the block,
F = - mg
= - 300 N
Hence, the normal force the ground exerts on the block, F = -300 N
Answer:
Please find the answer in the explanation
Explanation:
Given that a scientist conducting a field investigation records measurements of very low pressure and high relative humidity at the top of a mountain.
Since a weather map indicates that a warm front is approaching the mountain, according to the conventional current, the warm front is approaching because the weather must have been in higher relative humidity in cool air.
The warm front is approaching to replace it so that the cool air can conventionally replace the warmth air too.
The condition the scientists will most likely observe at the top of the mountain will be high relative humidity.
The eggs is a example of a chemical change bc your cooking it to something different
Answer:
vi = 4.77 ft/s
Explanation:
Given:
- The radius of the surface R = 1.45 ft
- The Angle at which the the sphere leaves
- Initial velocity vi
- Final velocity vf
Find:
Determine the sphere's initial speed.
Solution:
- Newton's second law of motion in centripetal direction is given as:
m*g*cos(θ) - N = m*v^2 / R
Where, m: mass of sphere
g: Gravitational Acceleration
θ: Angle with the vertical
N: Normal contact force.
- The sphere leaves surface at θ = 34°. The Normal contact is N = 0. Then we have:
m*g*cos(θ) - 0 = m*vf^2 / R
g*cos(θ) = vf^2 / R
vf^2 = R*g*cos(θ)
vf^2 = 1.45*32.2*cos(34)
vf^2 = 38.708 ft/s
- Using conservation of energy for initial release point and point where sphere leaves cylinder:
ΔK.E = ΔP.E
0.5*m* ( vf^2 - vi^2 ) = m*g*(R - R*cos(θ))
( vf^2 - vi^2 ) = 2*g*R*( 1 - cos(θ))
vi^2 = vf^2 - 2*g*R*( 1 - cos(θ))
vi^2 = 38.708 - 2*32.2*1.45*(1-cos(34))
vi^2 = 22.744
vi = 4.77 ft/s
