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3 years ago

A second-class lever will always multiply distance. a. true b. false

1 answer:

4 0

Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions here.

I think the statement "<span>A second-class lever will always multiply distance" is true.
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What's the difference between a direct relationship and a positive one?

borishaifa [10]

A positive or direct relationship is one in which the two variables (we will generally call them x and y) move together, that is, they either increase or decrease together. In a negative or indirect relationship, the two variables move in opposite directions, that is, as one increases, the other descremases

5 0

2 years ago

Two dump trucks each have a mass of 1,500 kg. The distance of the dump truck

xxTIMURxx [149]

Answer:

6.00 x 10⁻⁸N

Explanation:

Given parameters:

Mass of each dump trucks = 1500kg

Distance between them = 50m

Unknown:

New gravitational force between them = ?

Solution:

From Newton's law of universal gravitation,

F = $\frac{G m1 m2}{r^{2} }$

F is the gravitational force

G is the universal gravitation constant

m is the mass

r is the distance

F = $\frac{6.67 x 10^{-11} x 1500 x 1500}{50^{2} }$ = 6.00 x 10⁻⁸N

4 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

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4 0

1 year ago

wo fixed charges, A and B are located at x axis. A is at x = 0 m, B is at x = 4 m. QA = +4.0 μC and QB = -5.0 μC. Calculate the

lys-0071 [83]

Answer:

10250 N/C leftwards

Explanation:

QA = 4 micro Coulomb

QB = - 5 micro Coulomb

AP = 6 m

BP = 2 m

A is origin, B is at 4 m and P is at 6 m .

The electric field due to charge QA at P is EA rightwards

$E_{A}=\frac{KQ_{A}}{AP^{2}}=\frac{9\times10^{9}\times4\times10^{-6}}{6^{2}}=1000 N/C (rightwards)$

The electric field due to charge QB at P is EB leftwards

$E_{B}=\frac{KQ_{B}}{BP^{2}}=\frac{9\times10^{9}\times5\times10^{-6}}{2^{2}}=11250 N/C (leftwards)$

The resultant electric field at P due the charges is given by

E = EB - EA

E = 11250 - 1000 = 10250 N/C leftwards

5 0

3 years ago

A proton in a particle accelerator is traveling at a speed of 0.99c

Gwar [14]

A proton in a particle accelerator is traveling at a speed of 0.99c has a speed magnitude of 2.97 x 10⁸ m/s.

<h3>What is speed of proton?</h3>

The speed of a proton is the rate at which a proton is moving through a given space.

The given speed of the proton is 0.99c

where;

c is speed of light

<h3>What is speed of light?</h3>

The speed of light in vacuum, commonly denoted c, is a universal physical constant that is important in many areas of physics.

The value of speed of light in a vacuum is given as 3 x 10⁸ m/s.

The speed of the proton is calculated as follows;

v = 0.99 x 3 x 10⁸ m/s.

v = 2.97 x 10⁸ m/s.

Thus, a proton in a particle accelerator is traveling at a speed of 0.99c has a speed magnitude of 2.97 x 10⁸ m/s.

Learn more about speed of proton here: brainly.com/question/14663642

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8 0

1 year ago