Answer:
3600joules
Explanation:
formula :W=FS
W=work done (J)
F=force (N)
S=displacement moved in the direction of force (m)
200N×18m
=3600J
Answer:
6.54 × 10⁻⁵ Pa-s
Explanation:
Since the shear force, F = μAu/y where μ = viscosity of fluid between plates, A = area of plates, u = velocity of fluid = 0.6 m/s and y = separation of plates = 0.02 mm = 2 × 10⁻⁵ m
Since F = μAu/y
F/A = μu/y where F/A = force per unit area
Since we are given force per unit area, F/A = 1.962 N per unit area = 1.962 N/m²
So, μ = F/A ÷ u/y
substituting the values of the variables into the equation, we have
μ = F/A ÷ u/y
μ = 1.962 N/m² ÷ 0.6 m/s/2 × 10⁻⁵ m
μ = 1.962 N/m² ÷ 0.3 × 10⁵ /s
μ = 6.54 × 10⁻⁵ Ns/m²
μ = 6.54 × 10⁻⁵ Pa-s
De-celleration because speed is lower as time goes on
Answer:
-19.9m/s²
Explanation:
Given that
vi = 14cm/s
xi = 2.94 cm
xf=-5.00 cm
t= 1.85s
So
Using xf-xi= vi t + 1/2at²
Which is = -5-2.94= 14*1.85+1/2a1.85²
= -7.94= 25.9+1.7a
-33.9= 1.7a
a= - 19.9m/s²
A) force = mass • acceleration
Convert: 1.870 KN -> 1870 N
1870N = mass • 1.83 m/s^2
———- —————-
1.83 m/s^2. 1.83 m/s^2
-> approx. 1,021.9 grams :D
b) Vi = Vf - at
0 m/s = Vf - 1.83 m/s^2 • 16s
0m/s = Vf - 29.28
+29.28 + 29.28
Vf = 29.28 m/s :D
c) x = Vi + 1/2 • at^2
A to B
T= 15s
Vi = 16m / s
A = 1.83m/s^2
x= 16 + 1/2 • 1.83(15)^2)
x = approx. 221. 9 meters
C to D
T = 9s
Vi= 30m/s
x = 30 + 1/2 1.83(9)^2
x= approx. 104.1 meters
Therefore, when the car is moving at a CONSTANT RATE, it’s traveling farther than when slowing down cause it travel 221.9 meters while slowing down, it’s going only 104.1 meters.
