1) Displacement is to velocity as ____________ is to acceleration. A) direction B) speed C) time D) velocity

Question 9

Anastasy [175]

nAnswer:

uyui

Explanation:

4 0

3 years ago

A 52 kg and a 95 kg skydiver jump from an airplane at an altitude of 4750 m, both falling in the pike position. Assume all value

Scilla [17]

Answer: 52 kg skydiver: 9.09 m/s and 522.55 s

95 kg skydiver: 12.3 m/s and 386.2 s

Explanation: Drag Force is an opposite force when an object is moving in a fluid.

For skydivers, when falling through the air, the forces acting on it are gravitational and drag forces. At a certain point, drag force equals gravitational force, which is constant on any part of the planet, producing a net force that is zero. Since there is no net force, there is no acceleration and, consequently, velocity is constant. When that happens, the person reached the Terminal Velocity.

Drag Force and Velocity are proportional to the squared speed. So, terminal velocity is given by:

$F_{G}=F_{D}$

$mg=\frac{1}{2}C \rho Av_{T}^{2}$

$v_{T}=\sqrt{\frac{2mg}{\rho CA} }$

where

m is mass in kg

g is acceleration due to gravitational force in m/s²

ρ is density of the fluid in kg/m³

C is drag coefficient

A is area of the object in the fluid in m²

Calculating:

The 52kg skydiver has terminal velocity of:

$v_{T}=\sqrt{\frac{2(52)(9.8)}{(1.21)(0.7)(0.14)} }$

$v_{T}=$ 9.09

The 95kg skydiver's terminal velocity is

$v_{T}=\sqrt{\frac{2(95)(9.8)}{(1.21)(0.7)(0.14)} }$

$v_{T}=$ 12.3

The 52 kg and 95kg skydivers' terminal velocity are 9.09m/s and 12.3m/s, respectively.

The time each one will reach the floor will be:

52 kg at 9.09 m/s:

$t=\frac{4750}{9.09}$

t = 522.5

95 kg at 12.3 m/s:

$t=\frac{4750}{12.3}$

t = 386.2

The 52 kg and 95kg skydivers' time to reach the floor are 522.5 s and 386.2 s, respectively.

3 0

3 years ago

Give an example of a situation in which you would describe an

Viktor [21]

It would be A because it would make sense

5 0

3 years ago

We can model a pine tree in the forest as having a compact canopy at the top of a relatively bare trunk. Wind blowing on the top

frez [133]

We need to consider for this exercise the concept Drag Force and Torque. The equation of Drag force is

$F_D = c_D A \frac{\rho V^2}{2}$

Where,

F_D = Drag Force

$c_D$ = Drag coefficient

A = Area

$\rho$ = Density

V = Velocity

Our values are given by,

$c_D = 0.5$ (That is proper of a cone-shape)

$A = 9m^2$

$\rho = 1.2Kg/m^3$

$V = 6.5m/s$

Part A ) Replacing our values,

$F_D = 0.5*9*\frac{1.2*6.5^2}{2}$

$F_D = 114.075N$

Part B ) To find the torque we apply the equation as follow,

$\tau = F*d$

$\tau = (114.075N)(7)$

$\tau = 798.525N.m$

3 0

3 years ago

"how do the fundamental laws of physics make manifest that space has 3 dimensions?"

bazaltina [42]

If you are asking for a proof on having at least 3 dimensions in space, you can find the physical proof anywhere in your daily life activities. Just the fact that solids have volumes is a proof already that we live in a three-dimensional space. We can move forwards, backwards, sidewards and in all other directions possible.

When you go right into detail, the fundamental laws governing these proofs are very technical. They have differential equations to show as proof. It is too detailed to discuss here. The important things is that, these fundamental laws are what explains the science in our basic activities and natural phenomena:

*Gravitation and planetary motion
* Translation, rotation, magnetic field, forces
* Integrals of equations:

8 0

3 years ago