Answer:
Number of Significant Figures: 2
The Significant Figures are 3 6
Explanation:
= 3.60 × 102
(scientific notation)
= 3.60e2
(scientific e notation)
= 360 × 100
(engineering notation)
(one)
= 360
(real number)
Answer:
a) F = 4.9 10⁴ N, b) F₁ = 122.5 N
Explanation:
To solve this problem we use that the pressure is transmitted throughout the entire fluid, being the same for the same height
1) pressure is defined by the relation
P = F / A
to lift the weight of the truck the force of the piston must be equal to the weight of the truck
∑F = 0
F-W = 0
F = W = mg
F = 5000 9.8
F = 4.9 10⁴ N
the area of the pisto is
A = pi r²
A = pi d² / 4
A = pi 1 ^ 2/4
A = 0.7854 m²
pressure is
P = 4.9 104 / 0.7854
P = 3.85 104 Pa
2) Let's find a point with the same height on the two pistons, the pressure is the same
where subscript 1 is for the small piston and subscript 2 is for the large piston
F₁ =
the force applied must be equal to the weight of the truck
F₁ =
F₁ = (0.05 / 1) ² 5000 9.8
F₁ = 122.5 N
Answer: a=-2.4525 m/s^2
d=s=190.3 m
Explanation:The only force that is stopping the car and causing deceleration is the frictional force Fr
Fr = 25% of weight
W=mg
W=1750*9.81
W=17167.5
Hence
Frictional force is negative as it acts in opposite direction
According to newton second law of motion
F=ma
hence
given
u= 110 km/h
u=110*1000/3600
u=30.55 m/s
to get t we know that final velocity v=0
Answer:
d = 421.83 m
Explanation:
It is given that,
Height, h = 396.9 m
Horizontal speed, v = 46.87 m/s
We need to find the distance traveled by the ball horizontally. Let t is the time taken by the ball. Using second equation of motion for vertical direction. So,
Now d is the distance covered by the cannonball. So,
Hence, this is the required solution.