Norma kicks a soccer ball with an initial velocity of 10.0 meters per second at an angle of 30.0°. If the ball moves through the
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Answer:
v₀ =3.8 m/s
Explanation:
Newton's second law of the box:
∑F = m*a Formula (1)
∑F : algebraic sum of the forces in Newton (N)
m : mass in kilograms (kg)
a : acceleration in meters over second square (m/s²)
Known data
m=2.1 kg mass of the box
d= 5.4m length of the roof
θ = 20° angle θ of the roof with respect to the horizontal direction
μk= 0.51 : coefficient of kinetic friction between the box and the roof
g = 9.8 m/s² : acceleration due to gravity
Forces acting on the box
We define the x-axis in the direction parallel to the movement of the box on the roof and the y-axis in the direction perpendicular to it.
W: Weight of the box : In vertical direction
N : Normal force : perpendicular to the direction the roof
fk : Friction force: parallel to the direction to the roof
Calculated of the weight of the box
W= m*g = (2.1 kg)*(9.8 m/s²)= 20.58 N
x-y weight components
Wx= Wsin θ= (20.58)*sin(20)° =7.039 N
Wy= Wcos θ =(20.58)*cos(20)°= 19.34 N
Calculated of the Normal force
∑Fy = m*ay ay = 0
N-Wy= 0
N=Wy =19.34 N
Calculated of the Friction force:
fk=μk*N= 0.51* 19.34 N = 9.86 N
We apply the formula (1) to calculated acceleration of the block:
∑Fx = m*ax , ax= a : acceleration of the block
Wx-f = ( 2.1)*a
7.039 - 9.86 = ( 2.1)*a
-2.821 = ( 2.1)*a
a=(-2.821) /( 2.1)
a= -1.34 m/s²
Kinematics of the box
Because the box moves with uniformly accelerated movement we apply the following formula to calculate the final speed of the block :
vf²=v₀²+2*a*d Formula (2)
Where:
d:displacement = 5.4 m
v₀: initial speed
vf: final speed = 0
a : acceleration of the box = -1.34 m/s²
We replace data in the formula (2)
0²=v₀²+2*(-1.34)*(5.4)
2*(1.34)*(5.4)= v₀²
v₀ = 3.8 m/s
I attached the missing picture.
We can figure this one out using the law of conservation of energy.
At point A the car would have potential energy and kinetic energy.
Then, while the car is traveling down the track it loses some of its initial energy due to friction:
So, we know that the car is approaching the point B with the following amount of energy:
The law of conservation of energy tells us that this energy must the same as the energy at point B.
The energy at point B is the sum of car's kinetic and potential energy:
As said before this energy must be the same as the energy of a car approaching the loop:
Now we solve the equation for
:
We can figure this one out using the law of conservation of energy.
At point A the car would have potential energy and kinetic energy.
Then, while the car is traveling down the track it loses some of its initial energy due to friction:
So, we know that the car is approaching the point B with the following amount of energy:
The law of conservation of energy tells us that this energy must the same as the energy at point B.
The energy at point B is the sum of car's kinetic and potential energy:
As said before this energy must be the same as the energy of a car approaching the loop:
Now we solve the equation for
Answer:
Charge enter a 0.100 mm length of the axon is
Explanation:
Electric field E at a point due to a point charge is given by
where is the constant =
is the magnitude of point charge and
is the distance from the point charge
Charges entering one meter of axon is
Charges entering 0.100 mm of axon is
substituting the value of in above equation, we get charge enter a 0.100 mm length of the axon is