What is the velocity of a narwhal that swims 76 kilometers North in 3 hours
1 answer:
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Answer:
A) The acceleration is zero
<em>B) The total distance is 112 m</em>
Explanation:
<u>Velocity vs Time Graph</u>
It shows the behavior of the velocity as time increases. If the velocity increases, then the acceleration is positive, if the velocity decreases, the acceleration is negative, and if the velocity is constant, then the acceleration is zero.
The graph shows a horizontal line between points A and B. It means the velocity didn't change in that interval. Thus the acceleration in that zone is zero.
A. To calculate the acceleration, we use the formula:
Let's pick the extremes of the region AB: (0,8) and (12,8). The acceleration is:
This confirms the previous conclusion.
B. The distance covered by the body can be calculated as the area behind the graph. Since the velocity behaves differently after t=12 s, we'll split the total area into a rectangle and a triangle.
Area of rectangle= base*height=12 s * 8 m/s = 96 m
Area of triangle= base*height/2 = 4 s * 8 m/s /2= 16 m
The total distance is: 96 m + 16 m = 112 m
Answer: 7.38 km
Explanation: The attachment shows the illustration diagram for the question.
The range of the bomb's motion as obtained from the equations of motion,
H = u(y) t + 0.5g(t^2)
U(y) = initial vertical component of velocity = 0 m/s
That means t = √(2H/g)
The horizontal distance covered, R,
R = u(x) t = u(x) √(2H/g)
Where u(x) = the initial horizontal component of the bomb's velocity = 287 m/s, H = vertical height at which the bomb was thrown = 3.24 km = 3240 m, g = acceleration due to gravity = 9.8 m/s2
R = 287 √(2×3240/9.8) = 7380 m = 7.38 km
Answer:
0.266 m
Explanation:
Assuming the lump of patty is 3 Kg then applying the principal of conservation of linear momentum,
P= mv where p is momentum, m is mass and v is the speed of an object. In this case
where sunscripts p and b represent putty and block respectively, c is common velocity.
Substituting the given values then
3*8=v(15+3)
V=24/18=1.33 m/s
The resultant kinetic energy is transferred to spring hence we apply the law of conservation of energy
where k is spring constant and x is the compression of spring. Substituting the given values then