When is the gravitation force equal to the value of universal<br>gravitational constant?

Find the position vector of a particle that has the given acceleration and the specified initial velocity and position. a(t) = 1

kondor19780726 [428]

Answer:

Explanation:

Given

Acceleration $a(t)=14t\hat{i]+\sin (t)\hat{j}+\cos (2t)\hat{k}$ [/tex]

and $v(0)=\hat{i}$

$r(0)=\hat{j}$

we know $a=\frac{\mathrm{d} v}{\mathrm{d} t}$

$\int dv=\int adt$

$v(t)=\int (14t\hat{i}+\sin (t)\hat{j}+\cos (2t)\hat{k})dt$

$v(t)=7t^2\hat{i}-\cos t\hat{j}+\frac{\sin (2t)\hat{k}}{2}+c$

at $t=0$

$v(0)=0-1\cdot \hat{j}+0+c$

$c=\hat{i}+\hat{j}$

$v(t)=(7t^2+1)\hat{i}+(1-\cos t)\hat{j}+\frac{\sin (2t)\hat{k}}{2}$

and $\frac{\mathrm{d} r}{\mathrm{d} t}=v(t)$

$\int dr=\int vdt$

$r(t)=\int ((7t^2+1)\hat{i}+(1-\cos t)\hat{j}+\frac{\sin (2t)\hat{k}}{2})dt$

$r(t)=(\frac{7}{2}t^3+t)\hat{i}+(t-\sin (t))\hat{j}+\frac{1}{2}\times (-\frac{1}{2}\cos 2t)\hat{k}+c_2$

at $t=0$

$r(0)=\hat{j}$

$r(t)=(\frac{7}{3}t^3+t)\hat{i}+(1+t-\sin t)\hat{j}+\frac{1}{4}(1-\cos 2t)\hat{k}$

4 0

3 years ago

Boyle’s Law: When____ is held constant, the pressure and volume of a gas are___ proportional.

Aleksandr-060686 [28]

Answer:

When temperature is held constant, the pressure and volume of a gas are not proportional.

Explanation:

That is Boyle's Law

7 0

3 years ago

A small sphere with mass m is attached to a massless rod of length L that is pivoted at the top, forming a simple pendulum. The

USPshnik [31]

Answer:

a) see attached, a = g sin θ

b)

c) v = √(2gL (1-cos θ))

Explanation:

In the attached we can see the forces on the sphere, which are the attention of the bar that is perpendicular to the movement and the weight of the sphere that is vertical at all times. To solve this problem, a reference system is created with one axis parallel to the bar and the other perpendicular to the rod, the weight of decomposing in this reference system and the linear acceleration is given by

Wₓ = m a

W sin θ = m a

a = g sin θ

b) The diagram is the same, the only thing that changes is the angle that is less

θ' = 9/2 θ

c) At this point the weight and the force of the bar are in the same line of action, so that at linear acceleration it is zero, even when the pendulum has velocity v, so it follows its path.

The easiest way to find linear speed is to use conservation of energy

Highest point

Em₀ = mg h = mg L (1-cos tea)

Lowest point

Emf = K = ½ m v²

Em₀ = Emf

g L (1-cos θ) = v² / 2

v = √(2gL (1-cos θ))

4 0

3 years ago

A truck with 28-in.-diameter wheels is traveling at 50 mi/h. Find the angular speed of the wheels in rad/min, *hint convert mile

solong [7]

Answer:

Angular speed ω=3771.4 rad/min

Revolution=5921 rpm

Explanation:

Given data

$d=28in\\r=d/2=28/2=14in\\v=50mi/hr$

To find

Angular speed ω

Revolution per minute N

Solution

First we need to convert the speed of truck to inches per mile

as

1 mile=63360 inches

1 hour=60 minutes

so

$v=(50*\frac{63360}{60} )\\v=52800in/min$

Now to solve for angular speed ω by substituting the speed v and radius r in below equation

$w=\frac{v}{r}\\ w=\frac{52800in/min}{14in}\\ w=3771.4rad/min$

To solve for N(revolutions per minute) by substituting the angular speed ω in the following equation

$N=\frac{w}{2\pi }\\ N=\frac{3771.4rad/min}{2\pi }\\ N=5921RPM$

3 0

3 years ago

A car accelerates uniformly from +10.0 m/s to +40.0 m/s over a distance of 125 m. How long did it take to go that distance? Show

melamori03 [73]

3 seconds my dude i think

5 0

4 years ago

When is the gravitation force equal to the value of universalgravitational constant?​

When is the gravitation force equal to the value of universalgravitational constant?