Answer:
= 1.9792 × 10^10
Significant Figures= 5
Explanation:
Look at the attachment below
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2.10-kg box is moving to the right with speed 8.50 m>s on a horizontal, frictionless surface. At t = 0 a horizontal force is
applied to the box. The force is directed to the left and has magnitude F1t2 = 16.00 N>s22t2. (a) What distance does the box move from its position at t = 0 before its speed is reduced to zero? (b) If the force continues to be applied, what is the speed of the box at t = 3.00 s?
2 answers:
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1)
Answer:
Part 1)
H = 30.6 m
Part 2)
t = 2.5 s
Part 3)
t = 2.5 s
Part 4)
Explanation:
Part 1)
initial speed of the ball upwards
so maximum height of the ball is given by
Part 2)
As we know that final speed will be zero at maximum height
so we will have
Part 3)
Since the time of ascent of ball is same as time of decent of the ball
so here ball will same time to hit the ground back
so here it is given as
t = 2.5 s
Part 4)
since the acceleration due to earth will be same during its return path as well as the time of the motion is also same
so here its final speed will be same as that of initial speed
so we have
2)
Answer:
a = 9.76 m/s/s
Explanation:
As we know that the object is released from rest
so the displacement of the object in vertical direction is given as
3)
Answer:
v = 29.7 m/s
Explanation:
acceleration of the rocket is given as
time taken by the rocket
t = 0.33 min
final speed of the rocket is given as
4)
Answer:
Part 1)
y = 25.95 m
Part 2)
d = 6.72 m
Explanation:
Part 1)
As it took t = 2.3 s to hit the water surface
so here we will have
Part 2)
Distance traveled by it in horizontal direction is given as
Refer to the diagram shown below.
m = the mass of the object
x = the distance of the object from the equilibrium position at time t.
v = the velocity of the object at time t
a = the acceleration of the object at time t
A = the amplitude ( the maximum distance) of the mass from the equilibrium
position
The oscillatory motion of the object (without damping) is given by
x(t) = A sin(ωt)
where
ω = the circular frequency of the motion
T = the period of the motion so that ω = (2π)/T
The velocity and acceleration are respectively
v(t) = ωA cos(ωt)
a(t) = -ω²A sin(ωt)
In the equilibrium position,
x is zero;
v is maximum;
a is zero.
At the farthest distance (A) from the equilibrium position,
x is maximum;
v is zero;
a is zero.
In the graphs shown, it is assumed (for illustrative purposes) that
A = 1 and T = 1.
m = the mass of the object
x = the distance of the object from the equilibrium position at time t.
v = the velocity of the object at time t
a = the acceleration of the object at time t
A = the amplitude ( the maximum distance) of the mass from the equilibrium
position
The oscillatory motion of the object (without damping) is given by
x(t) = A sin(ωt)
where
ω = the circular frequency of the motion
T = the period of the motion so that ω = (2π)/T
The velocity and acceleration are respectively
v(t) = ωA cos(ωt)
a(t) = -ω²A sin(ωt)
In the equilibrium position,
x is zero;
v is maximum;
a is zero.
At the farthest distance (A) from the equilibrium position,
x is maximum;
v is zero;
a is zero.
In the graphs shown, it is assumed (for illustrative purposes) that
A = 1 and T = 1.