The bicyclist accelerates with magnitude <em>a</em> such that
25.0 m = 1/2 <em>a</em> (4.90 s)²
Solve for <em>a</em> :
<em>a</em> = (25.0 m) / (1/2 (4.90 s)²) ≈ 2.08 m/s²
Then her final speed is <em>v</em> such that
<em>v</em> ² - 0² = 2<em>a</em> (25.0 m)
Solve for <em>v</em> :
<em>v</em> = √(2 (2.08 m/s²) / (25.0 m)) ≈ 10.2 m/s
Convert to mph. If you know that 1 m ≈ 3.28 ft, then
(10.2 m/s) • (3.28 ft/m) • (1/5280 mi/ft) • (3600 s/h) ≈ 22.8 mi/h
Here, Initial momentum = mu = 6*2 = 12 Kg m/s
Final momentum = mv = 6*4 = 24 Kg m/s
In short, Your Answer would be Option C
Hope this helps!
Answer:
b. Friction decreased when he went from pavement to ice and then increased two more times.
Explanation:
Frictional force depends on the normal force of the surface and a friction coefficient.
Since we're talking about the same car, the value of 
will remain constant whereas μ will represent the change in the frictional coefficient of the surface. Now we consider the different surfaces, cars will slide in an icy road which means that the frictional coefficient is smaller than the pavement.
After Joshua returns to the pavement road, the resulting frictional force increases and will do so one more time when he reaches the gravel road. Gravel roads have greater frictional coefficients than pavement roads which means the frictional force will increase a second time.
Let us take east and north as the positive x and y-axes should the motion be plotted in a cartesian plane. Thus, the x value is 45 miles and the y value is 20. The tangent of an angle is equal to the ratio of y to x.
tanθ = y / x
Substituting,
tanθ = 20/45 = 0.44
The value of θ is 23.96°.
