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2 years ago

A loaded tractor-trailer has a mass of 36,000 kg, what is it’s weight?

Physics

1 answer:

Cerrena [4.2K]2 years ago

8 0

That depends on WHERE the rig is, because

weight = (mass) x (acceleration of gravity where the object is) .

-- If the truck is on Mars, then

Weight = (36,000 kg) x (3.71² m/s²) = 133,560N.

-- If the truck is on the moon, then

Weight = (36,000 kg) x (1.62 m/s²) = 58,320N.

-- If the truck is on Earth, then

Weight = (36,000 kg) x (9.81 m/s²) = 353,160N.

It all depends.

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For the airfoil and conditions in Problem 2.2, calculate the lift-to-drag ratio. Comment on its magnitude.

raketka [301]

Answer:

L/D= 112

Explanation:

Aerodynamics can be defined as the branch of dynamics which deals with the motion of air, their properties and the interaction between the air and solid bodies.

Aerodynamics law explains how an airplane is able to fly. There are four forces of flight, and they are; lift, weight, thrust and drag. The amount of lift generated by a wing divided by the aerodynamic drag is known as the lift to drag ratio.

Lift increases proportionally to the square of the speed.

The solutions to the question is the file attached to this explanation.

Lift,L= qC(l). S---------------------------(1).

and,

Drag,D = qC(d).S ----------------------(2).

Hence, Lift to drag ratio,L/D= C(l)/C(d).

Therefore, we have to compute various angle of attack.(check attached file)...

Then, (L/D) will then be equal to 112.

8 0

3 years ago

A car has a mass of 1.00 × 103 kilograms, and it has an acceleration of 4.5 meters/second2. What is the net force on the car?

aksik [14]

ANSWER

C. $F=4.5 \times10^3$ . newtons

EXPLANATION

According to Newton's second law,

$F_{net}=ma$ , where

$m=1.00\times 10^3kg$ is the mass measured in kilograms.

and

$a=4.5ms^{2}$ is the acceleration in metres per second square.

We substitute these values to obtain,

$F=1.00\times10^3 \times 4.5$ .

We rearrange to get,

$F=1.00\times4.5 \times10^3$ .

We multiply out the first two numbers and leave our answer in standard form to get,

$F=4.5 \times10^3 N$ .

The correct answer is C

3 0

3 years ago

Which force is there between an ice skate and the ice

andrew-mc [135]

The answer is friction

7 0

2 years ago

Read 2 more answers

One cycle of the power dissipated by a resistor ( R = 800 Ω R=800 Ω) is given by P ( t ) = 60 W , 0 ≤ t < 5.0 s P(t)=60 W, 0≤

OLga [1]

Answer:

42.5W

Explanation:

To solve this problem we must go back to the calculations of a weighted average based on the time elapsed thus,

$Power_{avg} = \frac{P_1(t_1)+P_2(t_2)}{t_1+t_2}$

We need to calculate the average power dissipated by the 800\Omega resistor.

Our values are given by:

$P(t)=60 W, 0\leq t$

$P(t)=25 W, 5.0\leq t$

Aplying the values to the equation we have:

$Power_{avg} = \frac{P_1(t_1)+P_2(t_2)}{t_1+t_2}$

$Power_{avg} = \frac{60(5-0)+25(10-5)}{(5-0)+(10-5)}$

$Power_{avg} = 42.5W$

5 0

2 years ago

An inclined plane makes an angle of 30.0º with the horizontal.

Wittaler [7]

The force required to push the box upward is 145.3N and the force to pus the box downward is -109.3N

Data given;

mass = 15kg
angle = 30 degree
acceleration = 1.2 m/s^2
acceleration due to gravity = 9.8 m/s^2

<h3>Force against gravity</h3>

To move the plane upward, the box will move against gravity.

$F = F - mgcos X\\
ma = F - mg cosX$

Let's solve for F

$ma = F - mgcosX\\
15*1.2 = F - 15*9.8 * cos 30\\
18 = F - 127.30\\
F = 18 + 127.30\\
F = 145.3N$

<h3>Force towards gravity</h3>

When the force pushes the box down the inclined plane, it moves towards gravity.

$F = F + mgcosX\\
ma = F + mgcosX\\
15*1.2 = F + 15*9.8*cos30\\
F = -109.3N$

The force required to push the box upward is 145.3N and the force required to push the box downward is -109.3N

Learn more on force across an inclined plane here;

brainly.com/question/11888124

4 0

2 years ago