Energy is the capacity to do what ? motion or work

describe the relationship between the direction of the velocity vector and the direction of the acceleration for a body moving i

Xelga [282]

Answer:

a = v²/r

Explanation:

The acceleration of a body moving in a circular path is known as the centripetal acceleration. This is the acceleration of a body that keeps the body within the circular path. It is written in terms of the linear velocity v and the radius of the circle of rotation as shown;

a = v²/r where

v is the linear velocity

r is the radius

a is the centripetal acceleration

7 0

2 years ago

What is the acceleration of a 10kg mass pushed by a 5N force?

GenaCL600 [577]

Answer:

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

$a = \frac{f}{m} \\$

From the question we have

$a = \frac{5}{10} = \frac{1}{2} \\ = 0.5$

We have the final answer as

Hope this helps you

5 0

2 years ago

The function x = (1.2 m) cos[(3πrad/s)t + π/5 rad] gives the simple harmonic motion of a body. At t = 9.7 s, what are the (a) di

nlexa [21]

Answer and Explanation:

Let:

$x(t)=Acos(\omega t+ \phi)$

The equation representing a simple harmonic motion, where:

$x=Displacement\hspace{3}from\hspace{3}the\hspace{3}equilibrium\hspace{3}point\\A=Amplitude \hspace{3}of\hspace{3} motion\\\omega= Angular \hspace{3}frequency\\\phi=Initial\hspace{3} phase\\t=time$

As you may know the derivative of the position is the velocity and the derivative of the velocity is the acceleration. So we can get the velocity and the acceleration by deriving the position:

$v(t)=\frac{dx(t)}{dt} =- \omega A sin(\omega t + \phi)\\\\a(t)=\frac{dv(t)}{dt} =- \omega^2 A cos(\omega t + \phi)$

Also, you may know these fundamental formulas:

$f=\frac{\omega}{2 \pi} \\\\T=\frac{2 \pi}{\omega}$

Now, using the previous information and the data provided by the problem, let's solve the questions:

(a)

$x(9.7)=1.2 cos((3 \pi *(9.7))+\frac{\pi}{5} ) \approx -0.70534m$

(b)

$v(9.7)=-(3\pi) (1.2) sin((3\pi *(9.7))+\frac{\pi}{5} ) \approx 9.1498 m/s$

(c)

$a(9.7)=-(3 \pi)^2(1.2)cos((3\pi*(9.7))+\frac{\pi}{5} )\approx -62.653m/s^2$

(d)

We can extract the phase of the motion, the angular frequency and the amplitude from the equation provided by the problem:

$\phi = \frac{\pi}{5}$

(e)

$f=\frac{\omega}{2 \pi} =\frac{3\pi}{2 \pi} =\frac{3}{2} =1.5 Hz$

(f)

$T=\frac{2 \pi}{\omega} =\frac{2 \pi}{3 \pi} =\frac{2}{3} \approx 0.667s$

8 0

2 years ago

Read 2 more answers

A proton traveling due west in a region that contains only a magnetic field experiences a vertically upward force (away from the

Zanzabum

Answer:

D

Explanation:

6 0

2 years ago

Read 2 more answers

sara and tory are out fishing on the lake on a hot summer day when they both decide to go for a swim. sara dives off the front o

crimeas [40]

Here it is an application of Newton's III law

as we know by Newton's III law that every action has equal and opposite reaction

So here as we know that two boys jumps off the boat with different forces

from front side of the boat the boy jumps off with force 45 N which means as per Newton's III law if boy has a force of 45 N in forward direction then he must apply a reaction force on the boat in reverse direction of same magnitude

So boat must have an opposite force on front end with magnitude 45 N

Now similar way we can say

from back side of the boat the boy jumps off with force 60 N which means as per Newton's III law if boy has a force of 60 N in backward direction then he must apply a reaction force on the boat in reverse direction of same magnitude

So boat must have an opposite force on front end with magnitude 60 N

So here net force due to both jump on the boat is given by

$F_{net} = F_1 - F_2$

$F_{net} = 60 - 45$

$F_{net} = 15 N$

so boat will have net force F = 15 N in forward direction due to both jumps

3 0

2 years ago