Argon, neon, and xenon are examples of

Physics

1 answer:

Luda [366]3 years ago

3 0

1. your in highschool you should know this but its A

You might be interested in

Two golf balls are hit from the same point on a flat field. Both are hit at an angle of 55 degree above the horizontal. Ball 2 h

juin [17]

Answer:

d_2 = 4d_1

Explanation:

The range or horizontal distance covered by a projectile projected with a velocity U at an angel of θ to the horizontal is given by

R = U²sin2θ/g

Let the range or horizontal distance of ball 1 with initial velocity U projected at an angle θ = 55° be

d_1 = U²sin2θ/g

Let the range or horizontal distance of ball 2 with initial velocity V = 2U projected at an angle θ = 55° be

d_2 = V²sin2θ/g

= (2U)²sin2θ/g

= 4U²sin2θ/g

= 4d_1 (since d_1 = U²sin2θ/g)

So, the ball 2 lands a distance d_2 = 4d_1 from the initial point.

4 0

3 years ago

A light platform is suspended from the ceiling by a spring. A student with a mass of 90 kg climbs onto the platform. When it sto

Ilya [14]

Refer to the diagram shown.

When the student climbs onto the platform, the spring stretches by 0.82 m to reach the equilibrium position.
The mass of the student is m = 90 kg, so his weight is
mg = (90 kg)*(9.8 m/s²) = 882 N

By definition, the spring constant is
k = (882 N)/(0.82 m) = 1075.6 N/m

When the spring is stretched by x from the equilibrium position, the restoring force is
F = - k*x.

If damping is ignored, the equation of motion is
F = m * acceleration
or
$m \frac{d^{2}x}{dt^{2}} = -kx \\ \frac{d^{2}x}{dt^{2}} + \frac{k}{m} x = 0$

Define ω² = k/m = 11.751 => ω = 3.457.
Then the solution of the ODE is
x(t) = c₁ cos(ωt) + c₂ sin(ωt)

x'(t) = -c₁ω sin(ωwt) + c₂ω cos(ωt)
When t=0, x' =0, therefore c₂ = 0

The solution is of the form
x(t) = c₁ cos(ωt)
When t = 0, x = 0.32 m. Therefore c₁ = 0.32

The motion is
x(t) = 0.32 cos(3.457t)
The single amplitude is 0.32 m, and the double amplitude is 0.64 m.

Answer:
0.32 m (single amplitude), or
0.64 m (double amplitude)