Answer:
Momentum P is 840000kgm/s or 8.4 × 10^6
Explanation:
Data :
Mass = 21000 kg
Velocity = 400 m/s
So momentum is given as
P = mv
P = 21000×400
P = 8400000 kgm/s
P = 8.4 × 10^6
m = mass of the box
N = normal force on the box
f = kinetic frictional force on the box
a = acceleration of the box
μ = coefficient of kinetic friction
perpendicular to incline , force equation is given as
N = mg Cos30 eq-1
kinetic frictional force is given as
f = μ N
using eq-1
f = μ mg Cos30
parallel to incline , force equation is given as
mg Sin30 - f = ma
mg Sin30 - μ mg Cos30 = ma
"m" cancel out
a = g Sin30 - μ g Cos30
inserting the values
1.20 = (9.8) Sin30 - (9.8) Cos30 μ
μ = 0.44
Answer:
The speed at the bottom of the driveway is3.67m/s.
Explanation:
Height,h= 5sin20°= 1.71m
Potential energy PE=mgh= 2000×9.8×1.71
PE= 33516J
KE= PE- Fk ×d
0.5mv^2= 33516 - (4000×5)
0.5×2000v^2= 33516 - 20000
1000v^2= 13516
v^2= 13516/1000
v =sqrt 13.516
v =3.67m/s
Answer:
750 J
Explanation:
We have a student that pushes a 50N block across the floor for a distance of 15m. The question is asking how much work was done to move the block.
To solve this, we must know that we are looking for a certain thing called joules. And to get the answer, we must follow the formula of W = FS
F being the force and S being the distance.
W = FS
W = (50)(15)
W = 750
Therefore, 750 joules is our answer.
Answer:
Explanation:
(a) It is given that Joseph jogs on a straight road of 300m in a time interval of 2 minutes and 30 seconds, which is equal to 150seconds. Therefore, when Joseph jogs from point A to point B, he covers a distance of 300m in time of 150seconds. Hence, his average speed is 300m/150s=2ms^−1. Since it is a straight road and he jogs in a single direction in this case, his displacement is equal to 300m. Since it is a straight road and he jogs in a single direction in this case, his displacement is equal to 300m.
Hence, his average velocity is 300m/150s=2ms^−1
(b) Then it is given that he turns back and points B and jogs on the same road but in the opposite direction for a time interval for 1 minute and covers a distance of 100m.If we consider the whole motion of Joseph, i.e. from point A to point C, then he covers a total distance of 300m+100m=400m. And he covers this total distance in a time interval of 2.5min+1min=3.5min=210s.
Therefore, his average speed for this journey is 400m210s=1.9ms−1.
For the same journey is displacement is equal to the distance between the points A and C,i.e. 300m−100m=200m.
Hence, his average velocity for this case is 200m/210s=0.95ms^−1