Solving this using the time, we know that range = horizontal velocity x time of flight
since
there are no horizontal forces acting on the ball, there are no
horizontal accelerations and the initial horizontal velocity of 36 cos
28 will be constant throughout. If we use the correct time of flight given the launch parameters, we have
range = 36 cos 28 x 3.44 s = 109.3 m
Answer:
a). V = 3.13*10⁶ m/s
b). T = 1.19*10^-7s
c). K.E = 2.04*10⁵
d). V = 1.02*10⁵V
Explanation:
q = +2e
M = 4.0u
r = 5.94cm = 0.0594m
B = 1.10T
1u = 1.67 * 10^-27kg
M = 4.0 * 1.67*10^-27 = 6.68*10^-27kg
a). Centripetal force = magnetic force
Mv / r = qB
V = qBr / m
V = [(2 * 1.60*10^-19) * 1.10 * 0.0594] / 6.68*10^-27
V = 2.09088 * 10^-20 / 6.68 * 10^-27
V = 3.13*10⁶ m/s
b). Period of revolution.
T = 2Πr / v
T = (2*π*0.0594) / 3.13*10⁶
T = 1.19*10⁻⁷s
c). kinetic energy = ½mv²
K.E = ½ * 6.68*10^-27 * (3.13*10⁶)²
K.E = 3.27*10^-14J
1ev = 1.60*10^-19J
xeV = 3.27*10^-14J
X = 2.04*10⁵eV
K.E = 2.04*10⁵eV
d). K.E = qV
V = K / q
V = 2.04*10⁵ / (2eV).....2e-
V = 1.02*10⁵V
Answer: d = 33 cm or 0.33 m
Explanation: In physics, Work is the amount of energy transferred to an object to make it move. It can be expressed by:
W = F.d.cosθ
F is the force applied to the object, d is the displacement and θ is the angle formed between the force and the displacement.
For the ice block, the angle is 0, i.e., force and distance are at the same direction, so:
W = F.d.cos(0)
W = F.d
To determine d:
d = 
d = 
d = 0.33 m
The distance d the block ice moved is 33 cm.
James E. Hansen studied climate change
Answer:
The value of acceleration that accomplishes this is 8.61 ft/s² .
Explanation:
Given;
maximum distance to be traveled by the car when the brake is applied, d = 450 ft
initial velocity of the car, u = 60 mph = (1.467 x 60) = 88.02 ft/s
final velocity of the car when it stops, v = 0
Apply the following kinematic equation to solve for the deceleration of the car.
v² = u² + 2as
0 = 88.02² + (2 x 450)a
-900a = 7747.5204
a = -7747.5204 / 900
a = -8.61 ft/s²
|a| = 8.61 ft/s²
Therefore, the value of acceleration that accomplishes this is 8.61 ft/s² .