Answer:
Mass of the cart = 146 kg
Explanation:
A cart is pulled by a force of 250 N at an angle of 35° above the horizontal.
The cart accelerates at 1.4 m/s² horizontally.
Horizontal force = Fcosθ = 250 cos35° = 204.79N
We have F = ma
Substituting
204.79 = m x 1.4
m = 146.28 kg = 146 kg
Mass of the cart = 146 kg
Answer:
Acceleration = 2.35 m/
Speed = 8.67 m/s
Explanation:
The coefficient of friction , u =0.3
The angle of incline = 30°
The two forces acting on block are weight and friction.
weight along the incline = mg cos60° = 
= 0.5 mg
Friction along incline = umg cos30° = mg 
Friction along incline = 0.26 mg
Net force acting on the weight = (0.5 - 0.26) mg = 0.24 mg
Acceleration = 
= 0.24 g = 2.35 m/
The height of incline = 8 m
Length of the inclined edge = 16 m
v= 8.67 m/s
If I had to go with any of those answers, It would be A maybe D, But im not too sure on how to decide between them. Because Einstein mentioned the sun in his theory which has a very large mass <span> 1.989 x 10 with a exponent of 30 to be exact. Hope this helped though.</span>
Answer:
<h2>C. maintenance </h2>
Explanation:
I personally believe one key disadvantages is the cost of maintaining the equipment unlike the gym where you have to subscribe for the month or the year and forget about anything, owning the gym equipment comes with the extra cost and responsibilities of maintenance to the owners.
Answer:
serie Ceq=0.678 10⁻⁶ F and the charge Q = 9.49 10⁻⁶ C
Explanation:
Let's calculate all capacity values
a) The equivalent capacitance of series capacitors
1 / Ceq = 1 / C1 + 1 / C2 + 1 / C3 + 1 / C4 + 1 / C5
1 / Ceq = 1 / 1.5 + 1 / 3.3 + 1 / 5.5 + 1 / 6.2 + 1 / 6.2
1 / Ceq = 1 / 1.5 + 1 / 3.3 + 1 / 5.5 + 2 / 6.2
1 / Ceq = 0.666 + 0.3030 +0.1818 +0.3225
1 / Ceq = 1,147
Ceq = 0.678 10⁻⁶ F
b) Let's calculate the total system load
Dv = Q / Ceq
Q = DV Ceq
Q = 14 0.678 10⁻⁶
Q = 9.49 10⁻⁶ C
In a series system the load is constant in all capacitors, therefore, the load in capacitor 5.5 is Q = 9.49 10⁻⁶ C
c) The potential difference
ΔV = Q / C5
ΔV = 9.49 10⁻⁶ / 5.5 10⁻⁶
ΔV = 1,725 V
d) The energy stores is
U = ½ C V²
U = ½ 0.678 10-6 14²
U = 66.4 10⁻⁶ J
e) Parallel system
Ceq = C1 + C2 + C3 + C4 + C5
Ceq = (1.5 +3.3 +5.5 +6.2 +6.2) 10⁻⁶
Ceq = 22.7 10⁻⁶ F
f) In the parallel system the voltage is maintained
Q5 = C5 V
Q5 = 5.5 10⁻⁶ 14
Q5 = 77 10⁻⁶ C
g) The voltage is constant V5 = 14 V
h) Energy stores
U = ½ C V²
U = ½ 22.7 10-6 14²
U = 2.2 10⁻³ J
