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sergeinik [125]
3 years ago
10

Which statement is true about the ideal mechanical advantage of a third-class lever?

Physics
2 answers:
Flura [38]3 years ago
8 0
WhT exactly is this for can you give more information
Nadya [2.5K]3 years ago
8 0

Answer:

It is always less than one.

Explanation:

Because the load covers more distance than the force when they move in the same direction.

♥️Mark as brainliest♥️

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Un móvil con velocidad inicial de 19,8km/h adquiere una aceleración constante de 2,4m/s^2. Determine la velocidad y el espacio r
nata0808 [166]

Responder:

Velocidad = 41.5m / s

Espacio recorrida = 352.5 metros

Explicación:

Dado lo siguiente:

Velocidad inicial (u) = 19.8 km / h

Aceleración (a) = 2.4m / s ^ 2

Tiempo de viaje (t) = 15 s

A.) velocidad después de 15 s

Velocidad inicial = (19.8 × 1000) m / 3600s Velocidad inicial = 19800m / 3600 = 5.5m / s

Usando la ecuación: v = u + at, donde v es la velocidad

v = 5.5 + 2.4 (15)

v = 5.5 + 36

v = 41.5m / s

Espacio recorrida:

v ^ 2 = u ^ 2 + 2aS; donde S es la distancia recorrida

41.5 ^ 2 = 5.5 ^ 2 + 2 × (2.4) × S

1722.25 = 30.25 + 4.8S

1722.25 - 30.25 = 4.8S

1692 = 4.8S S = 1692 / 4.8 S = 352.5m

8 0
3 years ago
The inner cylinder of a long, cylindrical capacitor has radius r and linear charge density +λ. It is surrounded by a coaxial cyl
Ulleksa [173]

Hi there!

a)

We can begin by using the equation for energy density.

U = \frac{1}{2}\epsilon_0 E^2

U = Energy (J)

ε₀ = permittivity of free space

E = electric field (V/m)

First, derive the equation for the electric field using Gauss's Law:
\Phi _E = \oint E \cdot dA = \frac{Q_{encl}}{\epsilon_0}

Creating a Gaussian surface being the lateral surface area of a cylinder:
A = 2\pi rL\\\\E \cdot 2\pi rL = \frac{Q_{encl}}{\epsilon_0}\\\\Q = \lambda L\\\\E \cdot 2\pi rL = \frac{\lambda L}{\epsilon_0}\\\\E = \frac{\lambda }{2\pi r \epsilon_0}

Now, we can calculate the energy density using the equation:
U = \frac{1}{2} \epsilon_0 E^2

Plug in the expression for the electric field and solve.

U = \frac{1}{2}\epsilon_0 (\frac{\lambda}{2\pi r \epsilon_0})^2\\\\U = \frac{\lambda^2}{8\pi^2r^2\epsilon_0}

b)

Now, we can integrate over the volume with respect to the radius.

Recall:
V = \pi r^2L \\\\dV = 2\pi rLdr

Now, we can take the integral of the above expression. Let:
r_i = inner cylinder radius

r_o = outer cylindrical shell inner radius

Total energy-field energy:

U = \int\limits^{r_o}_{r_i} {U_D} \, dV =   \int\limits^{r_o}_{r_i} {2\pi rL *U_D} \, dr

Plug in the equation for the electric field energy density and solve.

U =   \int\limits^{r_o}_{r_i} {2\pi rL *\frac{\lambda^2}{8\pi^2r^2\epsilon_0}} \, dr\\\\U = \int\limits^{r_o}_{r_i} { L *\frac{\lambda^2}{4\pi r\epsilon_0}} \, dr\\

Bring constants in front and integrate. Recall the following integration rule:
\int {\frac{1}{x}} \, dx  = ln(x) + C

Now, we can solve!

U = \frac{\lambda^2 L}{4\pi \epsilon_0}\int\limits^{r_o}_{r_i} { \frac{1}{r}} \, dr\\\\\\U = \frac{\lambda^2 L}{4\pi \epsilon_0} ln(r)\left \| {{r_o} \atop {r_i}} \right. \\\\U = \frac{\lambda^2 L}{4\pi \epsilon_0} (ln(r_o) - ln(r_i))\\\\U = \frac{\lambda^2 L}{4\pi \epsilon_0} ln(\frac{r_o}{r_i})

To find the total electric field energy per unit length, we can simply divide by the length, 'L'.

\frac{U}{L} = \frac{\lambda^2 L}{4\pi \epsilon_0} ln(\frac{r_o}{r_i})\frac{1}{L} \\\\\frac{U}{L} = \boxed{\frac{\lambda^2 }{4\pi \epsilon_0} ln(\frac{r_o}{r_i})}

And here's our equation!

3 0
2 years ago
What is the density of a box measuring 100 grams and 10 ml?
Alisiya [41]

Answer:

In order to convert density to grams, you have to put the mass on one side of the equation, and the density and the volume on the other. Therefore, d * v = m. Multiply the density by the volume. Using the example in step 1, you would multiply 2 g/mL by 4mL.

Explanation:

ok

6 0
3 years ago
A 3.0 kg pendulum swings from point A of height ya = 0.04 m to point B of height yb = 0.12 m, as seen in the diagram below.
lakkis [162]

i dont know soory pls give brainlyist

5 0
3 years ago
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Lifting a rope A mountain climber is about to haul up a 50-m length of hanging rope. How much work will it take if the rope weig
timama [110]

Answer:

780 J

Explanation:

W=\int _{\:0}^{50}0.624xdx

4 0
2 years ago
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