A box sliding on a horizontal floor surface starts out moving at 4.9 m/s and stops after 2 seconds. The surface and the box have a kinetic friction has friction coefficient of 0.25 (g=9.8 ms-2).
The speed of an object at the start of a measurement, or an initial state, is known as the initial speed. The ratio of distance travelled to travel time, also known as the average speed, is the sum of the initial and final speeds. The difference between initial and ending speeds is the speed change. A force that acts between moving surfaces is referred to as kinetic friction. A force acting in opposition to the direction of a moving body on the surface is felt.
Interval (t) equals 2 seconds
Kinetic friction coefficient is 0.25.
gravity-induced acceleration (g) = 9.8 m/s2.
Coming to a stop at a final velocity of 0 m/s.
Regular force equals mg
Kinetic friction force is given by = k.
N = μkmg
I = (delta P) (delta P)
Vo = (0.25)(9.8)(2) (2)
Vo = 4.9m/s
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Answer:
The Answer is C. the student changed too many variables.
Explanation:
For the quiz in K12
Answer:
0.12 mm ; 140.50 rad/m ; 628.32 rad/sec ; +
Explanation:
Given the wave equation of the form :
y(x, t) = ym sin(kx ± ωt)
Mas per unit length (u) = 5 g/cm = (5÷1000)kg / 0.01m) = 0.005kg/0.01m = 0.5kg/m
Tension, T = 10 N
Amplitude, A = 0.12 mm
Frequency, F = 100 Hz
Comparing with the general wave equation :
y = Asin(kx ± ωt)
A = amplitude = ym = 0.12 mm
2.) k = 2π / λ
Recall :
v = fλ
v = sqrt(T/u) = sqrt(10/0.5) = sqrt(20) = 4.472
λ = v/ f = 4.472 / 100 = 0.04472
Hence,
k = (2 * π) / 0.04472
k = 140.50 rad/m
3.) Angular frequency, ω
ω = 2πf = 2 * 3.14 * 100 = 628.32 rad/sec
4.) sign is +ve
Direction of wave propagation as given is in the negative x axis
21+10=31 because you can see that 21 and 10 are in metres while 12 is in seconds so 21+10=31 is the answer.
