Answer:
noun. a person who is connected with another or others by blood or marriage. something having, or standing in, some relation or connection to something else. something dependent upon external conditions for its specific nature, size, etc. (opposed to absolute).
Density is calculated by dividing the mass of an object by its volume. The Sun has a mass of 1.99×1030 kg and a radius of 6.96×108 m. What is the average density of the Sun?
Answer:
Density is calculated by dividing the mass of an object by its volume. The Sun has a mass of 1.99×1030 kg and a radius of 6.96×108 m. What is the average density of the Sun?
Jumping on a trampoline cause you to fly up in the air. What type of newton’s law is it ?
Answer:
The Third law
Explanation:
For every action there is an equal and opposite reaction.
Answer:
First Law
Explanation:
An object at rest (not moving) will stay at rest unless an unbalanced force acts on it.
An object in motion will stay in motion (in a straight line and at a constant speed) unless an unbalanced force acts on it.
You jump down on a trampoline and fly up in the air as a result.
Is it true or false that the displacement always equals the product of the average velocity and the time interval?
Answer:
True.
Explanation:
Applying the definition of average velocity, we know that we can always write the following expression:[tex]v_{avg} = \frac{\Delta x}{\Delta t}[/tex] (1)
By definition, Δx is just the displacement, and Δt is the time interval.So, just rearranging terms in (1), we get:[tex]\Delta x} = v_{avg}* {\Delta t}[/tex]
In the absence of a gravitational field, you could determine the mass of an object (of unknown composition) by:
A) applying a known force and measuring it's acceleration.
B) measuring the volume.
C) weighing it.
Answer:
A) By applying a known force, and measuring it's acceleration.
Explanation:
This is actually something that astronauts do in space as a mathmatical exercise when calculating the mass of an object since F = m × a.
Once the force, and acceleration are applied, the only unknown is the mass which can be solved by dividing force over acceleration. This is because inertial mass is equal to gravitational mass.
Sometimes we will want to write vectors in terms of a coordinate grid. To show a vector points
horizontally (along the x-axis), place an x after the magnitude of the vector. To show a vector point
vertically (along the y-axis), place a y after the magnitude.
4) Using the notation above,
i. How would you write d1?
ii. How would you write d2?
iii. How would you write dtotal?
d1=(0,5)
d2=(5,5)
Answer:
III) [tex]d_{1}+ d_{2}=d_{t}[/tex]
Explanation:
I) coordinate (0,5) is the head for [tex]d_{1}[/tex] I will put the tail coordinate as (0,0) but it could be any other number in the x just not in the 5 with the the y being any other value.
II) coordinate (5,5) is the head for [tex]d_{2}[/tex] the tail needs to be in the head of [tex]d_{1}[/tex] being (0,5)
III) coordinates for [tex]d_{t}[/tex] is connecting the tail from [tex]d_{1}[/tex] and the head of [tex]d_{2}[/tex] making it (0,0)[tex](tail)[/tex] and (0,5)[tex](head)[/tex] and is written as [tex]d_{1}+ d_{2}=d_{t}[/tex]
(i) using coordinate grid notation to represent d₁, d₁ = 5y
(ii) using coordinate grid notation to represent d₂, d₂ = 5x + 5y
(ii) The sum of d₁ and d₂ is written as 5x + 10y
In order to show the horizontal direction of a vector, we will place x after the magnitude of the vector.
Also, to show the vertical direction of a vector, we will place a y after the magnitude of the vector.
(i) Using coordinate grid to represent d₁ = (0, 5)
[tex]d_1 = 0(x) + 5(y)\\\\d_1 = 5y[/tex]
(ii) Using coordinate grid to represent d₂ = (5, 5)
[tex]d_2 = 5x + 5y[/tex]
(iii) The total vector is written as;
[tex]d_1 + d_2 = 5y + (5x + 5y)\\\\d_1 + d_2 = 5y + 5x + 5y\\\\d_1 + d_2 = 5x + 10y[/tex]
Learn more here: https://brainly.com/question/17212749
help me get the answer in Physical Science.
Answer:
lithium
Explanation:
I took physical science 2 years ago and passed with an A
While making some observations at the top of the 66 m tall Astronomy tower, Ron
accidently knocks a 0.5 kg stone over the edge. How long will a student at the bottom
have to get out of the way before being hit?
Analysing the question:
Since the stone was dropped, there was no initial velocity applied on it and hence it's initial velocity of the stone is 0 m/s
We are given:
height of the tower (h) = 66 m
mass of the stone (m) = 0.5 kg
initial velocity of the stone (u) = 0 m/s
time taken by the stone to reach the ground (t) = t seconds
acceleration due to gravity = 10 m/s²
** Neglecting air resistance**
Finding the time taken by the stone to reach the ground:
from the second equation of motion
h = ut + 1/2at²
replacing the variables
66 = (0)(t) + 1/2 (10)(t)²
66 = 5t²
t² = 13.2
t = 3.6 seconds
I initially wanted to subtract the height of the student from the height of the tower since the time i calculated is the time taken by the stone to reach the ground and that means that the stone has already hit the student before 3.6 seconds
but since we were NOT given the height of a student, the person who posed this question wants the time taken by the stone to reach the ground and that is what we solved