Answer:
The accurate statements are
2. The train is not an inertial frame of reference.
5. The train could be moving at a constant velocity in a circular path.
8. The train must be undergoing acceleration.
Explanation:
As we observe that the string forms an angle with the horizontal
we can formulate the force equation relevant to the given ball
similarly in the Y direction
Thus we conclude
Answer:
v₀ = 3.8 m/s
Explanation:
According to Newton's second law relating to the box:
∑F = m*a Formula (1)
∑F: the net force in Newton (N)
m: mass expressed in kilograms (kg)
a: acceleration measured in meters per second squared (m/s²)
Information known:
m = 2.1 kg, the mass of the box
d = 5.4m, the length of the roof
θ = 20° is the angle between the roof and the horizontal
μk = 0.51, the coefficient of kinetic friction between the box and the roof
g = 9.8 m/s², gravitational acceleration
Forces influencing the box:
The x-axis is oriented parallel to the box's movement on the roof, and the y-axis is oriented perpendicularly.
W: Weight of the box: directed vertically
N: Normal force: perpendicular to the roof's angle
fk: Frictional force: parallel to the direction along the roof
Calculating the weight of the box:
W = m*g = (2.1 kg)*(9.8 m/s²)= 20.58 N
x-y components of weight:
Wx= Wsin θ=(20.58)*sin(20)°=7.039 N
Wy= Wcos θ=(20.58)*cos(20)°= 19.34 N
Finding the Normal force:
∑Fy = m*ay ay = 0
N-Wy = 0
N=Wy = 19.34 N
Calculating the Friction force:
fk=μk*N= 0.51* 19.34 N = 9.86 N
We substitute into Formula (1) to determine the box's acceleration:
∑Fx = m*ax ax=a: acceleration of the box
Wx-fk = (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
Considering the box's Kinematics:
Since the box undergoes uniformly accelerated motion, we use the following to find the final speed of the box:
vf² = v₀² + 2*a*d Formula (2)
Where:
d refers to displacement = 5.4 m
v₀ is the initial speed
vf represents the final speed = 0
a is the box's acceleration = -1.34 m/s²
Plugging in the values into Formula (2):
0² = v₀² + 2*(-1.34)*(5.4)
2*(1.34)*(5.4) = v₀²
v₀ = 3.8 m/s
Response:
Once it has crossed, the locomotive requires 17.6 seconds to achieve a speed of 32 m/s.
Details:
The locomotive's acceleration is 1.6
The duration taken to pass the crossing is 2.4 seconds.
We can apply the motion equation, v = u + at, where v represents final velocity, u indicates initial velocity, a denotes acceleration, and t signifies time.
When the speed reaches 32 m/s, we have v = 32 m/s, u = 0 m/s, and a= 1.6 .
32 = 0 + 1.6 * t
t = 20 seconds.
Therefore, the locomotive attains a speed of 32 m/s after 20 seconds, and it passes the crossing in 2.4 seconds.
Thus, after clearing the crossing, it takes an additional 17.6 seconds to reach the speed of 32 m/s.
Answer:
Explanation:
Let us denote the launch angle as .
,
, and
are complementary angles which means their ranges are identical.
Time of flight is indicated as .
For