Ask question

Ask question

All categories

nataly862011 [7]

3 years ago

11

A car with a mass of 860 kg sits at the top of a 32 meter high hill. Describe the transformations between potential and kinetic

energy that occur when the car rolls to the bottom of the hill and continues rolling.

1 answer:

I am Lyosha [343]3 years ago

5 0

Answer:

Kinetic energy at the bottom equals potential energy at the top hence 269971.2 J

Explanation:

Assuming no friction and neglecting air resistance, from the law of conservation of energy, the potential energy that the car posses at 32 m is all converted to kinetic energy hence

PE= mgh

$KE= 0.5mv^{2}$

Here m denotes the mass of the car, g is acceleration due to gravity, h is the height and v is the speed of the car

The potential energy will be PE= 860*9.81*32=269971.2 J

You might be interested in

Air flows through a nozzle at a steady rate. At the inlet the density is 2.21 kg/m3 and the velocity is 20 m/s. At the exit, the

Vitek1552 [10]

To solve the problem, it is necessary to apply the concepts related to the change of mass flow for both entry and exit.

The general formula is defined by

$\dot{m}=\rho A V$

Where,

$\dot{m} =$ mass flow rate

$\rho =$ Density

V = Velocity

Our values are divided by inlet(1) and outlet(2) by

$\rho_1 = 2.21kg/m^3$

$V_1 = 20m/s$

$A_1 = 60*10^{-4}m^2$

$\rho_2 = 0.762kg/m^3$

$V_2 = 160m/s$

PART A) Applying the flow equation we have to

$\dot{m} = \rho_1 A_1 V_1$

$\dot{m} = (2.21)(60*10^{-4})(20)$

$\dot{m} = 0.2652kg/s$

PART B) For the exit area we need to arrange the equation in function of Area, that is

$A_2 = \frac{\dot{m}}{\rho_2 V_2}$

$A_2 = \frac{0.2652}{(0.762)(160)}$

$A_2 = 2.175*10^{-3}m^2$

Therefore the Area at the end is $21.75cm^2$

3 0

3 years ago

Which statements describe the image produced by a concave lens?check all that apply.

kotegsom [21]

<em>Answer:</em>

<em>The answers are: </em>

<em>A-which is the image is always right side up.</em>

<em>E-the image is virtual</em>

<em></em>

<em>Explanation: MY EXPLANATION IS YOU ARE WELCOME BIG DOG 100..</em>

<em></em>

<h2 />

7 0

4 years ago

Read 2 more answers

Given that R1 is 13 Ω, R2 is 10 Ω, and R3 is 4 Ω, what is the current coming out of the 9-V battery? What is the power dissipati

Kisachek [45]

So sorry I need point good luck

8 0

3 years ago

During a testing process, a worker in a factory mounts a bicycle wheel on a stationary stand and applies a tangential resistive

Katyanochek1 [597]

Answer:

The force is $F = 1041.7N$

Explanation:

The moment of Inertia I is mathematically evaluated as

$I = MR_A^2$

Substituting $1.9kg$ for M(Mass of the wheel) and $\frac{66cm}{2} * \frac{1m}{100cm} = 0.33m$ for $R_A$ (Radius of wheel)

$I = 1.9 * 0.33^2$

$= 0.207kgm^2$

The torque on the wheel due to net force is mathematically represented as

$\tau = FR_B - F_rR_A$

Substituting 135 N for $F_r$ (Force acting on sprocket), $\frac{8.7cm}{2} * \frac{1m}{100cm} = 0.0435m$ for $R_B$ (radius of the chain) and F is the force acting on the sprocket due to the chain which is unknown for now

$\tau = F (0.0435) - 135 (0.33)$

This same torque due to the net force is the also the torque that is required to rotate the wheel to have an angular acceleration of $\alpha = 3.70 rad/s^2$ and this torque can also be represented mathematically as

$\tau = \alpha I$

Now equating the two equation for torque

$F (0.0435) - 135 (0.33) = \alpha I$

Making F the subject

$F = \frac{\alpha I + (135*0.33) }{0.0435}$

Substituting values

$F = \frac{(3.70 * 0.207) + (135*0.33)}{0.0435}$

$= 1041.7N$

4 0

3 years ago

How much does 1 liter of water weigh

Zinaida [17]

Weight = (mass) x (gravity)

On Earth ...

Weight = (1 kg) x (9.8 m/s^2)

Weight = 9.8 Newtons

8 0

3 years ago