<em>The answer is </em>Ninth <em>and </em>Tenth <em>grade so the answer would be</em> B
<em>I hope this helps you </em>
Answer:
Energy due to air resistance = 31.8 Joules
Explanation:
According to the law of conservation of energy, energy can neither be created nor destroyed but can be transformed from one form to another
Kinetic Energy + Energy due to air resistance = Potential energy..........(1)
If there is no energy loss due to air resistance, potential energy = kinetic energy
mass, m = 1.5 kg
height, h = 4.0 m
speed, v = 6 m/s
Kinetic energy = 0.5 mv²
Kinetic energy = 0.5 * 1.5 * 6²
Kinetic energy = 27 Joules
Potential Energy = mgh
Potential energy = 1.5 * 9.8 * 4
Potential energy = 58.8 Joules
From equation (1)
27 + Energy due to air resistance = 58.8
Energy due to air resistance = 58.8 - 27
Energy due to air resistance = 31.8 Joules
Answer:
F = 147,78*10⁻⁹ [N]
Explanation:
By symmetry the Fy components of the forces acting on charge in point x = 0,7 m canceled each other, and the total force will be twice Fx ( Fx is x axis component of one of the forces .
The angle β ( angle between the line running through one of the charges in y axis and the charge in x axis) is
tan β = 0,5/0,7
tan β = 0,7142 then β = arctan 0,7142 ⇒ β = 35 ⁰
cos β = 0,81
d = √ (0,5)² + (0,7)² d1stance between charges
d = √0,25 + 0,49
d = √0,74 m
d = 0,86 m
Now Foce between two charges is:
F = K* q₁*q₂/ d² (1)
Where K = 9*10⁹ N*m²/C²
q₁ = 2,5* 10⁻⁹C
q₂ = 3,0*10⁻⁹C
d² = 0,74 m²
Plugging these values in (1)
F = 9*10⁹* 2,5* 10⁻⁹*3,0*10⁻⁹ / 0,74 [N*m²/C²]*C*C/m²
F = 91,21 * 10⁻⁹ [N]
And Fx = F*cos β
Fx = 91,21 * 10⁻⁹ *0,81
Fx =73,89*10⁻⁹ [N]
Then total force acting on charge located at x = 0,7 m is:
F = 2* Fx
F = 2*73,89*10⁻⁹ [N]
F = 147,78*10⁻⁹ [N]
Answer:
2 m/s
Explanation:
From the conservation of momentum, the initial momentum of the system must be equal to the final momentum of the system.
Let the 10.00 kg mass be 
and the 12.0 kg mass be 
. When they collide and stick, they have a combined mass of 
.
Momentum is given by 
. Set up the following equation:
, where 
is the desired final velocity of the masses.
Call the right direction positive. To indicate the 12.0 kg object is travelling left, its velocity should be substitute as -8.00 m/s.
Solving yields:
