Answer:
No, there won't be a collision.
Explanation:
We will use the constant acceleration formulas to calculate,
v = u + a*t
0 = 25 + (-0.1)*t
t = 250 seconds (the time taken for the passenger train to stop)
v^2 = u^2 + 2*a*s
0 = (25)^2 + 2*(-0.1)*s
s = 3125 m (distance traveled by passenger train to stop)
If the distance traveled by freight train in 250 seconds is less than (3125-200=2925 m) than the collision will occur
Speed*time = distance
Distance = (15)*(250)
Distance = 3750 m
As the distance is way more, there won’t be a collision
The mass of the hoop is the only force which is computed by:F net = 2.8kg*9.81m/s^2 = 27.468 N
the slow masses that must be quicker are the pulley, ring, and the rolling sphere.
The mass correspondent of M the pulley is computed by torque τ = F*R = I*α = I*a/R F = M*a = I*a/R^2 --> M = I/R^2 = 21/2*m*R^2/R^2 = 1/2*m
The mass equal of the rolling sphere is computed by: the sphere revolves around the contact point with the table. So using the proposition of parallel axes, the moment of inertia of the sphere is I = 2/5*mR^2 for spin about the midpoint of mass + mR^2 for the distance of the axis of rotation from the center of mass of the sphere. I = 7/5*mR^2 M = 7/5*m
the acceleration is then a = F/m = 27.468/(2.8 + 1/2*2 + 7/5*4) = 27.468/9.4 = 2.922 m/s^2
Answer and Explanation:
The computation of the shortest wavelength in the series is shown below:-
Where
represents wavelength
R represents Rydberg's constant
represents Final energy states
and 
represents initial energy states
Now Substitute is
now we will put the values into the above formula
Now we will rewrite the answer in the term of
So, the whole Paschen series is in the part of the spectrum.
Answer:
The first minimum would be observed at 41.57°
Explanation:
v = 340m/s = speed of sound
f = 610Hz
d = 0.840m
λ = ?
Mλ = wsinθ
m = mth order minima
λ = wavelength incident on the single slit
θ = angular position of the mth minima
But, λ = v / f
λ = 340 / 610 = 0.557m
θ = sin⁻(mλ/d)
θ = sin⁻ [(1 * 0.557) / 0.840]
θ = sin⁻ 0.6635
θ = 41.57°
The first minimum would be observed at 41.57°
Answer:
v = 2.45 m/s
Explanation:
first we find the time taken during this motion by considering the vertical motion only and applying second equation of motion:
h = Vi t + (1/2)gt²
where,
h = height of cliff = 15 m
Vi = Initial Vertical Velocity = 0 m/s
t = time taken = ?
g = 9.8 m/s²
Therefore,
15 m = (0 m/s) t + (1/2)(9.8 m/s²)t²
t² = (15 m)/(4.9 m/s²)
t = √3.06 s²
t = 1.75 s
Now, we consider the horizontal motion. Since, we neglect air friction effects. Therefore, the horizontal motion has uniform velocity. Therefore,
s = vt
where,
s = horizontal distance covered = 4.3 m
v = original horizontal velocity = ?
Therefore,
4.3 m = v(1.75 s)
v = 4.3 m/1.75 s
<u>v = 2.45 m/s</u>
