Answer:
option B. Limestone
<em>h</em><em>o</em><em>p</em><em>e</em><em>f</em><em>u</em><em>l</em><em>l</em><em>y</em><em>,</em>
<em>Z</em><em>a</em><em>r</em><em>a</em><em>♡</em>
Answer:
Force = -91.7 Newton
Explanation:
Given the following data;
Mass = 47 kg
Time = 4.1 seconds
Initial velocity = 8 m/s
Since the object comes to a stop, its final velocity would be equal to zero.
To find the force required to bring it to stop;
First of all, we would determine the acceleration of the object;
Mathematically, acceleration is given by the equation;

Substituting into the equation;
Acceleration, a = -1.95 m/s²
Next, we would determine the force required to bring the object to stop;
Force = -91.65 ≈ 91.7 Newton
Answer:
the coefficient of Kinetic friction between the tires and road is 0.38
Option A) .38 is the correct answer
Explanation:
Given that;
final velocity v = 0
initial velocity u = 15m/s
time taken t = 4 s
acceleration a = ?
from the equation of motion
v = u + at
we substitute
0 = 15 + a × 4
acceleration a = -15/4 = - 3.75 m/s²
the negative sign tells us that its a deacceleration so the sign can be ignored.
Deacceleration due to friction a = μ × g
we substitute
3.75 = μ × 9.8
μ = 3.75 / 9.8 = 0.3826 ≈ 0.38
Therefore the coefficient of Kinetic friction between the tires and road is 0.38
Option A) .38 is the correct answer
Answer:
0.21%
Explanation:
We are given;
Mass; m = 100 kg
Diameter; d = 2.2 mm = 2.2 × 10^(-3) m
Young's modulus; E = 12.5 x 10^(10) N/m².
Formula for area is;
A = πd²/4
A = (π/4) x (2.2 x 10^(-3))²
A = 3.8 x 10^(-6) m²
Force; F = mg
g is acceleration due to gravity and has a constant value of 9.8 m/s²
F = 100 × 9.8
F = 980 N
Formula for young's modulus is;
E = Stress/strain
Formula for stress = F/A
Formula for strain = ΔL/L
Thus;
E = (F/A)/(ΔL/L)
Making ΔL/L the subject, we have;
ΔL/L = (F/A)/E
Plugging in the relevant values;
ΔL/L = 980/(3.8 x 10^(-6) × 12.5 × 10^(10))
ΔL/L = 0.0021
Then percentage increase in length of a wire = 0.0021 × 100% = 0.21%
Horizontal speed = 24.0 m/s
height of the cliff = 51.0 m
For the initial vertical speed will are considering the vertical component. Therefore,
Since the student fires the canonical ball at the maximum height of 51 m, the initial vertical velocity will be zero. This means

let's find how long the ball remained in the air.
![\begin{gathered} 0=51-\frac{1}{2}(9.8)t^2 \\ 4.9t^2=51 \\ t^2=\frac{51}{4.9} \\ t^2=10.4081632653 \\ t=\sqrt[]{10.4081632653} \\ t=3.22 \\ t=3.22\text{ s} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%200%3D51-%5Cfrac%7B1%7D%7B2%7D%289.8%29t%5E2%20%5C%5C%204.9t%5E2%3D51%20%5C%5C%20t%5E2%3D%5Cfrac%7B51%7D%7B4.9%7D%20%5C%5C%20t%5E2%3D10.4081632653%20%5C%5C%20t%3D%5Csqrt%5B%5D%7B10.4081632653%7D%20%5C%5C%20t%3D3.22%20%5C%5C%20t%3D3.22%5Ctext%7B%20s%7D%20%5Cend%7Bgathered%7D)
Finally, let's find the how far from the base of the building the ball landed(horizontal distance)