The answer is electron
Explanation: An atom consists of a small but massive nucleus that include protons (positive charge) and neutrons (neutral charge). around the nucleus is a cloud of rapidly moving electrons (negative charge)
Answer:
Shawn's kinetic energy is 61.45 J.
Explanation:
Given;
Mass of the system is, 
Displacement of the bike is, 
Time taken is, 
Let the constant velocity be 
.
For constant velocity, magnitude of velocity is given as distance by time.
Therefore, 
Now, kinetic energy of a body is given as:
Here, 
Therefore, Shawn's kinetic energy is 61.45 J.
Answer:
The time after which the two stones meet is tₓ = 4 s
Explanation:
Given data,
The height of the building, h = 200 m
The velocity of the stone thrown from foot of the building, U = 50 m/s
Using the II equation of motion
S = ut + ½ gt²
Let tₓ be the time where the two stones meet and x be the distance covered from the top of the building
The equation for the stone dropped from top of the building becomes
x = 0 + ½ gtₓ²
The equation for the stone thrown from the base becomes
S - x = U tₓ - ½ gtₓ² (∵ the motion of the stone is in opposite direction)
Adding these two equations,
x + (S - x) = U tₓ
S = U tₓ
200 = 50 tₓ
∴ tₓ = 4 s
Hence, the time after which the two stones meet is tₓ = 4 s
Answer:
(a) 1.21 m/s
(b) 2303.33 J, 152.27 J
Explanation:
m1 = 95 kg, u1 = - 3.750 m/s, m2 = 113 kg, u2 = 5.38 m/s
(a) Let their velocity after striking is v.
By use of conservation of momentum
Momentum before collision = momentum after collision
m1 x u1 + m2 x u2 = (m1 + m2) x v
- 95 x 3.75 + 113 x 5.38 = (95 + 113) x v
v = ( - 356.25 + 607.94) / 208 = 1.21 m /s
(b) Kinetic energy before collision = 1/2 m1 x u1^2 + 1/2 m2 x u2^2
= 0.5 ( 95 x 3.750 x 3.750 + 113 x 5.38 x 5.38)
= 0.5 (1335.94 + 3270.7) = 2303.33 J
Kinetic energy after collision = 1/2 (m1 + m2) v^2
= 0.5 (95 + 113) x 1.21 x 1.21 = 152.27 J
Answer:
r = 2161.9 m
Explanation:
Aerodynamic lift(L) is perpendicular to the wing, which is tilted 40 degrees to the horizontal.
Since the plane is moving in a horizontal circle, the vertical component of the lift must cancel the weight W of the airplane, but the horizontal component is the centripetal force that keeps it in a circle.
L is perpendicular to wing at angle θ with respect to horizontal
Thus,
Vertical component of lift is:
L cosθ = W = mg
Thus, m = L cosθ / g - - - - (eq1)
Horizontal component of lift is:
L sinθ = centripetal force = mv² / r - - - - (eq2)
Combining equations 1 and 2,we have;
L sinθ = (L cosθ / g)(v² / r)
L cancels out on both sides to give;
tanθ = v²/ rg
r = v² / (g tanθ)
We are given;
velocity; v = 480 km/hr = 480 x 10/36 = 133.33 m/s
r = 133.33²/[(9.8) tan(40)] = 2161.9 m
