In the Netherlands, there is an annual ice-skating race called the "Tour of the Eleven Towns." The total distar.rce of the cours

To tackle this question we will apply the kinematic equations that describe the motion of a projectile, where both maximum height and distance traveled are defined. This scenario demonstrates a lion achieving a height (H) of 3m and covering a horizontal distance (R) of 10m. The equations governing this kind of motion are expressed as follows:

By dividing the two equations, we determine:

Plugging in the values for H and R yields:

After substituting \theta into the relevant equation, we find:

In conclusion, the mountain lion's launch speed upon takeoff is approximately 9.98m/s at an angle of 50.2°.

Depth = 5.0 × 10^2 m
Density of seawater = 1.025 x 10^3
Pd = Po + pgh
Standard atmospheric pressure is Patm = 1.01325 x 10^5 Pa
Since the pressure inside the hull is normal, we can disregard Po.
Thus, Pd = pgh = 1.025 x 10^3 x 9.8 x 5.0 x 10^2 = 50.225 x 10^5
Now calculating Pd / Patm gives us 50.225 x 10^5 / 1.01325 x 10^5 = 49.56
This indicates the pressure is 49.56 times greater.

According to Newton's second law, Force equals the rate of change of momentum over time. Momentum change is equal to Force times time. So, F=ma can be rearranged to a=F/m, a more recognizable formulation of Newton's second law
Using a relevant kinematic equation for mass m: V=u+at; where initial speed u=0; thus, acceleration a=F/m gives V=(F/m)xt, which translates to t=mV/F. For mass 2m, applying the same formula: V=u+at; u=0; a=F/2m indicates V=(F/2m)xt, leading to t=2mV/F (possibly double the initial time)
I might have erred somewhere along the line, but the fundamental concept seems valid... using another kinematic equation for m: s=ut + (1/2)at²; with s=d; and initial speed u=0; a=F/m; t=1; results in d=(1/2)(F/m) = F/2m. Similarly, for 2m: s=ut + (1/2)at²; s=d; u=0; a=F/2m; and t=1 gives d=(1/2)(F/2m)=F/4m (half the distance perhaps???? WHAT???!)

Answer:

20.353125 V

Explanation:

m = Mass of proton =

q = Charge of proton =

= Velocity of proton at point A = 50 km/s

= Velocity of proton at point B = 80 km/s

The relationship derived from energy conservation is as follows:

The determined potential difference is 20.353125 V

Answer:

the time it takes after impact for the puck is 2.18 seconds

Explanation:

initially given information

mass = 30 g = 0.03 kg

diameter = 100 mm = 0.1 m

thickness = 0.1 mm = 1 × m

dynamic viscosity = 1.75 × Ns/m²

temperature of air = 15°C

to determine

time needed for the puck to reduce its speed by 10%

solution

we note that velocity changes from 0 to v

assuming initial velocity = v

therefore final velocity = 0.9v

implying a change in velocity is du = v

and clearance dy = h

shear stress acting on the surface is expressed as

= µ

therefore

= µ ............1

substituting the values

= 1.75 × ×

= 0.175 v

the area between the air and puck is given by

Area =

area =

area = 7.85 × m²

thus, the force on the puck can be represented as

Force = × area

force = 0.175 v × 7.85 ×

force = 1.374 × v

now applying Newton's second law

force = mass × acceleration

- force =

- 1.374 × v =

solving for t =

the time needed after impact for the puck is 2.18 seconds