Ask question

Ask question

All categories

3 years ago

6

With fuel prices for combustible engine automobiles increasing, researchers and manufacturers have given more attention to the c

oncept of an ultralight car. Using carbon composites, lighter steels, and plastics, a fuel-efficient car can be manufactured at 540 kg. How much less does an ultralight car weigh compared to a 1450-kg Honda Accord (2007)

1 answer:

Alinara [238K]3 years ago

4 0

Answer:

Then the difference of weight between the two cars are:

Δw = 14210 - 5292 = 8918 N

Explanation:

An object's weigh due to the gravitational attraction force of the earth is:

w = mg

Where: m is the object's mass

g is the gravitational acceleration in the surface earth

g = 9.8 m/s2

The the ultralight car's weight is:

$w_{uc} = (540)(9.8)$

$w_{uc} = 5292 N$

And the Honda Accord's weight is:

$w_{HA} = (1450)(9.8)$

$w_{HA} = 14210 N$

Then the difference of weight between the two cars are:

Δw = 14210 - 5292 = 8918 N

You might be interested in

A 1.20-m cylindrical rod of diameter 0.570 cm is connected to a power supply that maintains a constant potential difference of 1

nasty-shy [4]

(a) $1.72\cdot 10^{-5} \Omega m$

The resistance of the rod is given by:

$R=\rho \frac{L}{A}$ (1)

where

$\rho$ is the material resistivity

L = 1.20 m is the length of the rod

A is the cross-sectional area

The radius of the rod is half the diameter: $r=0.570 cm/2=0.285 cm=2.85\cdot 10^{-3} m$ , so the cross-sectional area is

$A=\pi r^2=\pi (2.85\cdot 10^{-3} m)^2=2.55\cdot 10^{-5} m^2$

The resistance at 20°C can be found by using Ohm's law. In fact, we know:

- The voltage at this temperature is V = 15.0 V

- The current at this temperature is I = 18.6 A

So, the resistance is

$R=\frac{V}{I}=\frac{15.0 V}{18.6 A}=0.81 \Omega$

And now we can re-arrange the eq.(1) to solve for the resistivity:

$\rho=\frac{RA}{L}=\frac{(0.81 \Omega)(2.55\cdot 10^{-5} m^2)}{1.20 m}=1.72\cdot 10^{-5} \Omega m$

(b) $8.57\cdot 10^{-4} /{\circ}C$

First of all, let's find the new resistance of the wire at 92.0°C. In this case, the current is

I = 17.5 A

So the resistance is

$R=\frac{V}{I}=\frac{15.0 V}{17.5 A}=0.86 \Omega$

The equation that gives the change in resistance as a function of the temperature is

$R(T)=R_0 (1+\alpha(T-T_0))$

where

$R(T)=0.86 \Omega$ is the resistance at the new temperature (92.0°C)

$R_0=0.81 \Omega$ is the resistance at the original temperature (20.0°C)

$\alpha$ is the temperature coefficient of resistivity

$T=92^{\circ}C$

$T_0 = 20^{\circ}$

Solving the formula for $\alpha$ , we find

$\alpha=\frac{\frac{R(T)}{R_0}-1}{T-T_0}=\frac{\frac{0.86 \Omega}{0.81 \Omega}-1}{92C-20C}=8.57\cdot 10^{-4} /{\circ}C$

5 0

3 years ago

Identify which type of source is being described.

frez [133]

Answer:

Primary, secondary

Explanation:

3 0

3 years ago

Read 2 more answers

A piece of copper wire with thin insulation, 200 m long and 1.00 mm in diameter, is wound onto a plastic tube to form a long sol

bearhunter [10]

Answer:

N= 3

Explanation:

For this exercise we must use Faraday's law

E = - dФ / dt

Ф = B . A = B Acos θ

tje bold indicate vectors. As it indicates that the variation of the field is linear, we can approximate the derivatives

E = - A cos θ (B - B₀) / t

The angle enters the magnetic field and the normal to the area is zero

cos 0 = 1

A = π r²

In the length of the wire there are N turns each with a length L₀ = 2π r

L = N (2π r)

r = L / 2π N

we substitute

A = L² / (4π N²)

The magnetic field produced by a solenoid is

B = μ₀ N/L I

for which

B₀ = μ₀ N/L I

The final field is zero, because the current is zero

B = 0

We substitute

E = - (L² / 4π N²) (0 - μ₀ N/L I) / t

E = μ₀ L I / (4π N t)

N = μ₀ L I / (4π t E)

The electromotive force is E = 0.80 mV = 0.8 10⁻³ V

let's calculate

N = 4π 10⁻⁷ 200 1.60 / (4π 0.120 0.8 10⁻³)]

N = 320 10⁻⁷ / 9.6 10⁻⁶

N = 33.3 10⁻¹

N= 3

7 0

3 years ago

A 160.-kilogram space vehicle is traveling along a

mario62 [17]

If the object is moving in a straight line at a constant speed, then that's
the definition of zero acceleration. It can only happen when the sum of
all forces (the 'net' force) on the object is zero.

And it doesn't matter what the object's mass is. That argument is true
for specks of dust, battleships, rocks, stars, rock-stars, planets, and
everything in between.

4 0

3 years ago

Which word in the sentence is a gerund?

OleMash [197]

A gerund usually ends wit -ing, so we must find the word that includes that.

Loading includes ing at the end.

Final answer: b. loading

8 0

3 years ago

Read 2 more answers