Answer: 288.8 m
Explanation:
We have the following data:
is the time it takes to the child to reach the bottom of the slope
is the initial velocity (the child started from rest)
is the angle of the slope
is the length of the slope
Now, the Force exerted on the sled along the ramp is:
(1)
Where
is the mass of the sled and
its acceleration
In addition, if we draw a free body diagram of this sled, the force along the ramp will be:
(2)
Where
is the acceleration due gravity
Then:
(3)
Finding
:
(4)
(5)
(6)
Now, we will use the following kinematic equations to find
:
(7)
(8)
Where
is the final velocity
Finding
from (7):
(9)
(10)
Substituting (10) in (8):
(11)
Finding
:

Answer:
The induced emf in the coil is 0.522 volts.
Explanation:
Given that,
Radius of the circular loop, r = 9.65 cm
It is placed with its plane perpendicular to a uniform 1.14 T magnetic field.
The radius of the loop starts to shrink at an instantaneous rate of 75.6 cm/s , 
Due to the shrinking of radius of the loop, an emf induced in it. It is given by :

So, the induced emf in the coil is 0.522 volts.
Imagine an object is moving in one dimension on a number line, and for this we'll say that the numbers on the line are a metre apart. If the object moves from 2 m to 7 m, the change in position is 7-2=+5 metres. But if the object moves back from 7 m to 2 m, the change in position is 2-7=-5 metres. since

, and time is always positive, velocity will be positive in one direction and negative in the other direction.
Answer:
Explanation:
The mass of the block is 0.5kg
m = 0.5kg.
The spring constant is 50N/m
k =50N/m.
When the spring is stretch to 0.3m
e=0.3m
The spring oscillates from -0.3 to 0.3m
Therefore, amplitude is A=0.3m
Magnitude of acceleration and the direction of the force
The angular frequency (ω) is given as
ω = √(k/m)
ω = √(50/0.5)
ω = √100
ω = 10rad/s
The acceleration of a SHM is given as
a = -ω²A
a = -10²×0.3
a = -30m/s²
Since we need the magnitude of the acceleration,
Then, a = 30m/s²
To know the direction of net force let apply newtons second law
ΣFnet = ma
Fnet = 0.5 × -30
Fnet = -15N
Fnet = -15•i N
The net force is directed to the negative direction of the x -axis