Answer: An aspect of the event of various types of balls bouncing off the same floor, being matter is that all the balls consist of matter. They all occupy space and have a form of energy when moved by a force, such as a person. And for energy, like I just said, when they bounce they create energy as they bounce up and down, so if the ball were to hit some other object, it would have an impact on the still object.
The combination of the material properties of a ball (surface textures, actual materials, amount of air, hardness/ softness, and so on) affects the height of its bounce.
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Explanation:
For #5 It's helpful to draw a free body diagram so you know which way the forces are acting on the block.
the weight mg is acting downwards, and you need to find the vertical and horizontal components of mg using sin and cosine. so do 15x9.8xsin40 which is the force. Assuming no friction, this is the only force acting on the block, as the forces on the vertical plane cancel out i.e the normal force and weight of the block.
after, just do F=ma And since you know F and m, solve for a.
CORRECT ANSWER:
a- Cell-surface receptors bind polar signaling molecules; intracellular receptors bind nonpolar signaling molecules.
STEP-BY-STEP EXPLANATION:
The complete question from book is
According to Figure 9.6, what is a key difference between cell signaling by a cell-surface receptor and cell signaling by an intracellular receptor?
a- Cell-surface receptors bind polar signaling molecules; intracellular receptors bind nonpolar signaling molecules.
b- Signaling molecules that bind to cell-surface receptors lead to cellular responses restricted to the cytoplasm; signaling molecules that bind to intracellular receptors lead to cellular responses restricted to the nucleus.
c- Cell-surface receptors bind to specific signaling molecules; intracellular receptors bind any signaling molecule.
d- Cell-surface receptors typically bind to signaling molecules that are smaller than those bound by intracellular receptors.
e- None of the other answer options is correct.
Answer:
a.) magnitude __49.7__ unit(s)
b.) direction __123.6°_ counterclockwise from the +x axis
Explanation:
Let Vector is v
x-component of Vector v = x = -27.5 units (minus sign indicate that x-component is along the minus x-axis )
y-component of Vector v = y = 41.4 units
Magnitude of v = ?
Direction of v = ?
To find the magnitude of the vector
v =
v = 
v = 49.7 units
To find direction
θ = tan⁻¹(y/x)
θ = tan⁻¹(41.4/-27.5)
θ = -56.4°
This Angle is in the clockwise direction with respect to -x axis.
We need to find Angle counterclockwise from the +x axis.
So,
θ = 180° - 56.4°
θ = 123.6°
The given vector is in 2nd quadrant