Displacement magnitude and direction?

As a box slides down a ramp, friction does 23.0 joules of work. At the bottom of the ramp, the box has 3.8 joules of kinetic ene

tensa zangetsu [6.8K]

Answer:

The high of the ramp is 2.81[m]

Explanation:

This is a problem where it applies energy conservation, that is part of the potential energy as it descends the block is transformed into kinetic energy.

If the bottom of the ramp is taken as a potential energy reference point, this point will have a potential energy value equal to zero.

We can find the mass of the box using the kinetic energy and the speed of the box at the bottom of the ramp.

$E_{k}=0.5*m*v^{2}\\\\where:\\E_{k}=3.8[J]\\v = 2.8[m/s]\\m=\frac{E_{k}}{0.5*v^{2} } \\m=\frac{3.8}{0.5*2.8^{2} } \\m=0.969[kg]$

Now applying the energy conservation theorem which tells us that the initial kinetic energy plus the work done and the potential energy is equal to the final kinetic energy of the body, we propose the following equation.

$E_{p}+W_{f}=E_{k}\\where:\\E_{p}= potential energy [J]\\W_{f}=23[J]\\E_{k}=3.8[J]\\$

And therefore

$m*g*h + W_{f}=3.8\\ 0.969*9.81*h - 23= 3.8\\h = \frac{23+3.8}{0.969*9.81}\\ h = 2.81[m]$

8 0

3 years ago

What is/are the energy transformation(s) that take place when using a wind turbine to generate usable energy?

Cerrena [4.2K]

Answer:

the answer is C

Explanation:

C) friction - mechanical - electrical

4 0

3 years ago

A car of mass 1600 kg traveling at 27.0 m/s is at the foot of a hill that rises vertically 135 m after travelling a distance of

Zarrin [17]

Answer:

Neglecting any frictional losses, the average power delivered by the car's engine is 10565 W

Explanation:

The energy conservation law indicates that the energy must be the same at the bottom of the hill and at the top of the hill.

The energy at the bottom is only the Kinect energy (K_1) of the car in motion, but in the top, the energy is the sum of its Kinect energy (K_2), potential energy (P) and the work (W) done by the engine.

$K_1 = K_2 + P + W$

then, the work done by the engine is:

$W = K_1 - K_2 - P$

The formulas for the Kinetic and potential energy are:

$K=\frac{1}{2}mV^2\\P=mgh$

where, m is the mass of the car, V the velocity, g the gravity and h is the elevation of the hill.

Using the formulas:

$W=\frac{1}{2}mV_1^2-\frac{1}{2}mV_2^2-mgh$

Replacing the values:

$W=\frac{1}{2}(1600Kg)(27m/s)^2-\frac{1}{2}(1600Kg)(14m/s)^2-(1600Kg)(9.8m/s^2)(135m)\\W=-1690400 J$

The negative of this value indicates the direction of the work done, but for the problem, you only care about the magnitude, so the power is W=1690400 J. Now, the power is equal to work/time so you need to find the time the car took to get to the top of the hill.

The average speed of the car is (27+14)/2=20m/s, and t=d/v so the time is:

$t=\frac{3200m}{20m/s}=160s$

the power delivered by the car's engine was:

$power=\frac{work}{time}=\frac{1690400J}{160s}=10565W$

8 0

4 years ago

Two cylinders with the same mass density rhoC = 713 kg / m3 are floating in a container of water (with mass density rhoW = 1025

Tanzania [10]

Answer:

Explanation:

Given

density of cylinder is $\rho _c=713 kg/m^3$

Length of first cylinder is $L_1=20 cm$

radius $r_1=5 cm$

For cylinder 2 $L_2=10 cm$

$r_2=10 cm$

$h_1$ and $h_2$ are the height above water

E

as object is floating so its weight must be balanced with buoyant force

$\rho _c\frac{\pi }{4}d_1^2L_1g=\rho _w\frac{\pi }{4}d_1^2(L_1-h_1)g----1$

For 2nd cylinder

$\rho _c\frac{\pi }{4}d_2^2L_2g=\rho _w\frac{\pi }{4}d_2^2(L_2-h_2)g----2$

Dividing 1 and 2 we get

$\frac{L_1}{L_2}=\frac{L_1-h_1}{L_2-h_2}$

$\frac{20}{10}=\frac{20-h_1}{10-h_2}$

$2h_2=h_1$

$\\\Rightarrow\frac{h_2}{h_1}=\frac{1}{2}$

5 0

3 years ago

A 14 kg tennis ball moves at a velocity of 11 m/s. The ball is struck by a racket, causing it to rebound in the opposite directi

Charra [1.4K]

To solve this problem we will apply the concepts related to the momentum.

This is defined as the product between the change in velocity and the mass of the object, that is

$p = m\Delta v$

$p = m (v_f-v_i)$

Where,

m = mass

$v_f =$ Final velocity

$v_i =$ Initial velocity

Our values are given as,

m = 14kg

$v_i$ = 11m/s

$v_f= - 5m/s \rightarrow$ <em>the negative Symbol implies that the direction is opposite to the initial one and therefore there is also a change in the sense of magnitude</em>

$p = (14)(-5-11)$

$p = 14(-16)$

$p = -224 kg \cdot m/s$

The negative symbol indicates that the momentum has a direction opposite to that of the initial velocity. Or failing that, it has the same direction of the final speed

7 0

4 years ago