The answer in Meters is going to to 1265.341
Answer:
a) b = -5
b) slope = 3/2
Explanation:
a) The equation of a line is given as y = mx + b, where m is the slope of the line and b is the intercept on the y axis.
Given that y = 3x + b and it passes through the point (2, 1). Hence when x = 2, y = 1. Therefore, substituting for x and y:
1 = 3(2) + b
1 = 6 + b
b = 1 - 6
b = -5
b) The equation of a line passing through two points (
) and
is given by:

The equation of the line passing through the two points (0,3) and (4,9) is:

Comparing y = (3/2)x + 3 with y = mx + b, the slope (m) is 3/2
Answer:
the wagon should be used as frame of reference if an observer said the child was not moving.
Explanation:
The state of motion of a body depends upon the frame of reference. It is the set of co-ordinates according to which the motion is analyzed. If a child is riding in a wagon, then he will be considered in motion to a person standing outside the wagon. Hence, if we take a frame of reference outside the wagon then the child must be in motion with respect to the observer. On the other hand if the observer is inside the wagon, then the child must be in rest with respect to the observer. Hence, if we take the wagon to be the frame of reference, then the child will be at rest with respect to the observer.
<u>Therefore, the wagon should be used as frame of reference if an observer said the child was not moving.</u>
Answer:
V = 331.59m/s
Explanation:
First we need to calculate the time taken for the shell fire to hit the ground using the equation of motion.
S = ut + 1/2at²
Given height of the cliff S = 80m
initial velocity u = 0m/s²
a = g = 9.81m/s²
Substitute
80 = 0+1/2(9.81)t²
80 = 4.905t²
t² = 80/4.905
t² = 16.31
t = √16.31
t = 4.04s
Next is to get the vertical velocity
Vy = u + gt
Vy = 0+(9.81)(4.04)
Vy = 39.6324
Also calculate the horizontal velocity
Vx = 1330/4.04
Vx = 329.21m/s
Find the magnitude of the velocity to calculate speed of the shell as it hits the ground.
V² = Vx²+Vy²
V² = 329.21²+39.63²
V² = 329.21²+39.63²
V² = 108,379.2241+1,570.5369
V² = 109,949.761
V = √ 109,949.761
V = 331.59m/s
Hence the speed of the shell as it hits the ground is 331.59m/s
Answer:
a = 2.22 [m/s^2]
Explanation:
First we have to convert from kilometers per hour to meters per second
![40 [\frac{km}{h}]*[\frac{1h}{3600s}]*[\frac{1000m}{1km}] = 11.11 [m/s]](https://tex.z-dn.net/?f=40%20%5B%5Cfrac%7Bkm%7D%7Bh%7D%5D%2A%5B%5Cfrac%7B1h%7D%7B3600s%7D%5D%2A%5B%5Cfrac%7B1000m%7D%7B1km%7D%5D%20%3D%2011.11%20%5Bm%2Fs%5D)
We have to use the following kinematics equation:

where:
Vf = final velocity = 11.11 [m/s]
Vi = initial velocity = 0
a = acceleration [m/s^2]
t = time = 5 [s]
The initial speed is taken as zero, as the car starts from zero.
11.11 = 0 + (a*5)
a = 2.22 [m/s^2]