Which set of items listed can be arranged to transform electrical energy into mechanical energy?

An incline plane has Ф = 40.0° and μ k = 0.15. Starting from rest, how long will it take a 4.0 kg block to reach a speed of 12 m

lesya [120]

The block on the action of two forces, the Force of Friction and the Tangential Weight. Using the Newton's Secound Law, we have:

$P_{t}-Fat=ma \\ mgsen\O-mgucos\O=ma \\ a=g(sen\O-ucos\O)$

Using the Velocity Hourly Equation, we get:

$V=V_{o}+a\Delta t \\ \Delta t= \frac{V}{a}$

Uniting the equations:

$\Delta t= \frac{V}{g(sen\O-ucos\O)}$

Entering the unknowns:

$\Delta t= \frac{V}{g(sen\O-ucos\O)} \\ \Delta t= \frac{12}{10(sen40^o-0.15cos^o)} \\ \Delta t= \frac{12}{10(0,64-0.15x0.77)} \\ \Delta t= \frac{12}{5.27} \\ \boxed {\Delta t=2.28s}$

Obs: Approximate results

If you notice any mistake in my english, please let me know, because i am not native.

5 0

3 years ago

What is the final velocity of a drag racer that has constant acceleration and finishes a

Svetradugi [14.3K]

Answer:

120 mph

Explanation:

Given:

Δx = 0.25 mi

v₀ = 0 mi/s

t = 15 s

Find: v

Δx = ½ (v + v₀) t

0.25 mi = ½ (v + 0 mi/s) (15 s)

v = 0.0333 mi/s

v = 120 mi/h

8 0

3 years ago

Can someone help meeeeee... show how to solve it plzzzzzzzz

liubo4ka [24]

<h2>Right answer: 64 units</h2><h2></h2>

According to the law of universal gravitation, which is a classical physical law that describes the gravitational interaction between different bodies with mass:

$F=G\frac{m_{1}m_{2}}{r^2}$

Where:

$F$ is the module of the force exerted between both bodies

$G$ is the universal gravitation constant.

$m_{1}$ and $m_{2}$ are the masses of both bodies.

$r$ is the distance between both bodies

In this case we have a gravitation force $F_{1}=16units$ , given by the formula written at the beginning. Let’s rename the distance $r$ as $d$ :

$F_{1}=G\frac{m_{1}m_{2}}{d^2}$ (1)

And we are asked to find the gravitation force $F_{2}$ with a given distance of $\frac{d}{2}$ :

$F_{2}=G\frac{m_{1}m_{2}}{({\frac{d}{2})}^{2}}$

$F_{2}=G\frac{m_{1}m_{2}}{{\frac{d^{2}}{4}}}$ (2)

The gravity constant is the same for both equations, and we are assuming both masses are constants, as well. So, let’s isolate $G m_{1}m_{2}$ in both equations: