Answer:
d = 39.7 km
Explanation:
initial position of the boat is 45 km away at an angle of 15 degree East of North
so we will have
after some time the final position of the boat is found at 30 km at 15 Degree North of East
so we have
now the displacement of the boat is given as
so the magnitude is given as
Answer:
N/kg
Explanation:
newton/kilogram is the answer.please mark me brainliest
According to the big bang theory, the universe was once a very C. hot and hostile place, and it expanded rapidly. This rapid expansion was the very reason it eventually cooled down and eventually got to its present state where all the planets and galaxies and what not were created.
Answer:
Time - taken = 2.5 s
deceleration= -8 m/s²
Solution:
Given:
speed, v = 8 m/s
distance, d = 20m
To Find:
deacceleration = ?
As we know speed is defined as
v = d/t
plugging in the values
t = 20/ 8
t = 2.5s
Now from deceleration formula
a = - v/ t
a = - 20/ 2.5
a = - 8 m/s²
Thus, the time taken and acceleration is 2.5 s and -8 m/s²
respectively.
Learn more about deceleration here:
brainly.com/question/13354629
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Complete question is;
a. Two equal sized and shaped spheres are dropped from a tall building. Sphere 1 is hollow and has a mass of 1.0 kg. Sphere 2 is filled with lead and has a mass of 9.0 kg. If the terminal speed of Sphere 1 is 6.0 m/s, the terminal speed of Sphere 2 will be?
b. The cross sectional area of Sphere 2 is increased to 3 times the cross sectional area of Sphere 1. The masses remain 1.0 kg and 9.0 kg, The terminal speed (in m/s) of Sphere 2 will now be
Answer:
A) V_t = 18 m/s
B) V_t = 10.39 m/s
Explanation:
Formula for terminal speed is given by;
V_t = √(2mg/(DρA))
Where;
m is mass
g is acceleration due to gravity
D is drag coefficient
ρ is density
A is Area of object
A) Now, for sphere 1,we have;
m = 1 kg
V_t = 6 m/s
g = 9.81 m/s²
Now, making D the subject, we have;
D = 2mg/((V_t)²ρA))
D = (2 × 1 × 9.81)/(6² × ρA)
D = 0.545/(ρA)
For sphere 2, we have mass = 9 kg
Thus;
V_t = √[2 × 9 × 9.81/(0.545/(ρA) × ρA))]
V_t = 18 m/s
B) We are told that The cross sectional area of Sphere 2 is increased to 3 times the cross sectional area of Sphere 1.
Thus;
Area of sphere 2 = 3A
Thus;
V_t = √[2 × 9 × 9.81/(0.545/(ρA) × ρ × 3A))]
V_t = 10.39 m/s
