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zhuklara [117]
3 years ago
8

Vin Diesel jumps out of a plane. Gravity pulls on him with 184N of force and his parachute pushes him up with 82 N of force.

Physics
1 answer:
Lelu [443]3 years ago
4 0

Answer:

Fr = 102[N]

Explanation:

We must perform a free body diagram to understand the forces acting on Vin Diesel. In the attached image we see the forces related to his weight and the air resistance force.

We must analyze the forces in the Y axis, we see that we have the force of the air opposing the fall of Vin Diesel, in this way this force is negative since it opposes the movement, the fall.

∑Fy = 0

F_{weight}-F_{air}=F_{r}\\F_{r}= 184-82\\F_{r}=102[N]

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The bright yellow light emitted by a sodium vapor lamp consists of two emission lines at 589.0 and 589.6 nm. What are the freque
stiv31 [10]

Answer:

Explanation:

Given

wavelength of emissions are

\lambda _1=589 nm

\lambda _2=589.6 nm

Energy is given by

E=\frac{hc}{\lambda }

where h=Planck's constant

x=velocity of Light

\lambda=wavelength of emission

E_1=\frac{6.626\times 10^{-34}\times 3\times 10^8}{589\times 10^{-9}}

E_1=3.374\times 10^{-19} J

E_1 in kJ/mol

E_1=203.2 kJ/mol

frequency corresponding to this emission

\nu =\frac{c}{\lambda }

\nu _1=\frac{3\times 10^8}{589\times 10^{-9}}

\nu _1=5.09\times 10^{14} Hz

Energy corresponding to \lambda _2

E_2=\frac{6.626\times 10^{-34}\times 3\times 10^8}{589.6\times 10^{-9}}

E_2=3.371\times 10^{-19} J

E_2=203.02 kJ/mol

frequency corresponding to this emission

\nu =\frac{c}{\lambda }

\nu _1=\frac{3\times 10^8}{589.6\times 10^{-9}}

\nu _1=5.088\times 10^{14} Hz

6 0
3 years ago
A laser beam of wavelength λ=632.8 nm shines at normal incidence on the reflective side of a compact disc. The tracks of tiny pi
Juliette [100K]

To solve this problem we will apply the concepts given by the principles of superposition, specifically those described by Bragg's law in constructive interference.

Mathematically this relationship is given as

dsin\theta = n\lambda

Where,

d = Distance between slits

\lambda = Wavelength

n = Any integer which represent the number of repetition of the spectrum

\theta = sin^{-1} (\frac{n\lambda}{d})

Calculating the value for n, we have

n = 1

\theta_1 = sin^{-1} (\frac{\lambda}{d})\\\theta_1 = sin^{-1} (\frac{632.8*10^{-9}}{1.6*10^{-6}})\\\theta_1 = 23.3\°

n=2

\theta_2 = sin^{-1} (\frac{2\lambda}{d})\\\theta_2 = sin^{-1} (2\frac{632.8*10^{-9}}{1.6*10^{-6}})\\\theta_2 = 52.28\°

n =3

\theta_2 = sin^{-1} (\frac{2\lambda}{d})\\\theta_2 = sin^{-1} (3\frac{632.8*10^{-9}}{1.6*10^{-6}})\\\theta_2 = \text{not possible}

Therefore the intensity of light be maximum for angles 23.3° and 52.28°

7 0
4 years ago
A skater is spinning at 15.0 rad/s with a rotational inertia of 8.0 kg m? If she changes her rotational inertia to 5.0 kg mº, wh
mixas84 [53]

Answer:Idrk man

Explanation:

3 0
3 years ago
An olympic high diver has gravitational potential energy because of her height. as she dives, what becomes of her energy just be
sammy [17]

An Olympic high diver has gravitational potential energy because of her height. As she dives, kinetic energy becomes of her energy just before she hits the water.

Gravitational potential energy is the energy possessed or acquired by an object due to a change in its position when it is present in a gravitational field. In simple terms, it can be said that gravitational potential energy is an energy that is related to gravitational force or to gravity.

Kinetic energy is the energy of motion, observable as the movement of an object, particle, or set of particles.

When the high diver is standing stable and not moving , that diver has a gravitational potential energy because of the height . The moment she dives , before hitting the water , from being stationary she gained some momentum and come in motion , due to motion her gravitational potential energy will change to kinetic energy before hitting the ground.

To learn more about Gravitational potential energy  here

brainly.com/question/15978356

#SPJ4

6 0
1 year ago
A 250 kg flatcar 25 m long is moving with a speed of 3.0 m/s along horizontal frictionless rails. A 61 kg worker starts walking
Anna11 [10]

Answer:

x=31.09m

Explanation:

p1=p2

The momentum of flatcar and the momentum of the worker so

The velocity of the worker is:

m_{f}*v_{f}=m_{w}*v_{w}\\\\v_{f}=\frac{m_{f}*v_{f}}{m_{w}}\\v_{f}=\frac{61kg*3.0\frac{m}{s}}{250kg}\\v_{f}=0.732\frac{m}{s}

The total motion has a total velocity and is

Vt=v_{w}+v_{f}\\Vt=0.732\frac{m}{s}+3.0\frac{m}{s}\\Vt=3.732\frac{m}{s}

The time the worker take walking is

t=\frac{x}{v_{w}}\\t=\frac{25m}{3\frac{m}{s}}=8.33s

Now the total time and the total velocity determinate the motion of tha flatcar how far has moved

x=t*Vt\\x=8.33s*3.732\frac{m}{s} \\x=31.09m

5 0
3 years ago
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