Help with velocity math pls help asap

According to the theory of plate tectonics, the lithosphere is separated into sections that are called tectonic plates. These pl

loris [4]

Answer:

Transform :j

Explanation:

4 0

3 years ago

Mr. Rudman drives his race car for 4 hrs at 150<br> miles/hr. How far will he travel?

Lilit [14]

150*4=600
So the answer is 600

7 0

2 years ago

In the sport of parasailing, a person is attached to a rope being pulled by a boat while hanging from a parachute-like sail. A r

IrinaK [193]

Answer:

570 N

Explanation:

Draw a free body diagram on the rider. There are three forces: tension force 15° below the horizontal, drag force 30° above the horizontal, and weight downwards.

The rider is moving at constant speed, so acceleration is 0.

Sum of the forces in the x direction:

∑F = ma

F cos 30° - T cos 15° = 0

F = T cos 15° / cos 30°

Sum of the forces in the y direction:

∑F = ma

F sin 30° - W - T sin 15° = 0

W = F sin 30° - T sin 15°

Substituting:

W = (T cos 15° / cos 30°) sin 30° - T sin 15°

W = T cos 15° tan 30° - T sin 15°

W = T (cos 15° tan 30° - sin 15°)

Given T = 1900 N:

W = 1900 (cos 15° tan 30° - sin 15°)

W = 570 N

The rider weighs 570 N (which is about the same as 130 lb).

6 0

2 years ago

A particle moves in a straight line with the velocity function v ( t ) = sin ( w t ) cos 3 ( w t ) . find its position function

Sunny_sXe [5.5K]

Integrating the velocity equation, we will see that the position equation is:

$$f(t)=\frac{\cos ^3(\omega t)-1}{3}$

<h3>How to get the position equation of the particle?</h3>

Let the velocity of the particle is:

$$v(t)=\sin (\omega t) * \cos ^2(\omega t)$

To get the position equation we just need to integrate the above equation:

$$f(t)=\int \sin (\omega t) * \cos ^2(\omega t) d t$

$$\mathrm{u}=\cos (\omega \mathrm{t})$

Then:

$$d u=-\sin (\omega t) d t$

$\Rightarrow d t=-d u / \sin (\omega t)$

Replacing that in our integral we get:

$\int \sin (\omega t) * \cos ^2(\omega t) d t$

$-\int \frac{\sin (\omega t) * u^2 d u}{\sin (\omega t)}-\int u^2 d t=-\frac{u^3}{3}+c$

Where C is a constant of integration.

Now we remember that $u=\cos (\omega t)$

Then we have:

$$f(t)=\frac{\cos ^3(\omega t)}{3}+C$

To find the value of C, we use the fact that f(0) = 0.

$$f(t)=\frac{\cos ^3(\omega * 0)}{3}+C=\frac{1}{3}+C=0$

C = -1 / 3

Then the position function is:

$$f(t)=\frac{\cos ^3(\omega t)-1}{3}$

Integrating the velocity equation, we will see that the position equation is:

$$f(t)=\frac{\cos ^3(\omega t)-1}{3}$

To learn more about motion equations, refer to:

brainly.com/question/19365526

#SPJ4

4 0

1 year ago

A rod of length L and mass M has a nonuniform mass distribution. The linear mass density (mass per length) is λ=cx2, where x is

grigory [225]

Answer:

Explanation:

λ=c x²

c = λ / x²

λ is mass / length

so its dimensional formula is ML⁻¹

x is length so its dimensional formula is L

c = λ / x²

= ML⁻¹ / L²

= ML⁻³

B )

We shall find out the mass of the rod with the help of given expression of mass per unit length and equate it with given mass that is M

The mass in the rod is symmetrically distributed on both side of middle point.

we consider a small strip of rod of length dx at x distance away from middle point

its mass dm = λdx = cx² dx

By integrating it from -L to +L we can calculate mass of whole rod , that is

M = ∫cx² dx

= [c x³ / 3] from -L/2 to +L/2

= c/3 [ L³/8 + L³/8]

M = c L³/12

c = 12 M L⁻³

C ) Moment of inertia of rod

∫dmx²

= ∫λdxx²

= ∫cx²dxx²

= ∫cx⁴dx

= c x⁵ / 5 from - L/2 to L/2

= c / 5 ( L⁵/ 32 +L⁵/ 32)

= (2c / 160)L⁵

= (c / 80) L⁵

= (12 M L⁻³/80)L⁵

= 3/20 ML²

=

4 0

2 years ago