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2 years ago

8

A bike, a truck, and a train—all without passengers, motors, or engines—roll down the same hill. Put the vehicles in order from

the least amount of motion energy to the most.

1 answer:

Bezzdna [24]2 years ago

3 0

Answer:

Train Bike Truck

Explanation:

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Assuming the same current is running through two separate coils, why is it easier to thrust a magnet into a wire coil with one l

s2008m [1.1K]

The magnetic field strength in a coil is directly proportional to the number of turns, or loops, in the coil.
Therefore, when there are four loops instead of one, the magnetic field strength has increased four times, making it harder to push the magnet in.

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2 years ago

Read 2 more answers

An object is dropped from a height of 25 meters. At what velocity will it hit the ground? A. 7.0 meters/second B. 11 meters/seco

Basile [38]

$Final^{2}=Initial^{2}+2ad \\ x^{2}=0^{2}+2(9.8)(25) \\ x^{2}=490 \\ x=22.13 \\ C$

4 0

3 years ago

Delivery trucks that operate by making use of energy stored in a rotating flywheel have been used in Europe. The trucks are char

Marta_Voda [28]

Answer:

KE = 2.535 x 10⁷ Joules

Explanation:

given,

angular speed of the fly wheel = 940 rad/s

mass of the cylinder = 630 Kg

radius = 1.35 m

KE of flywheel = ?

moment of inertia of the cylinder

$I = \dfrac{1}{2}mr^2$

= $\dfrac{1}{2}\times 630\times 1.35^2$

= 574 kg m²

kinetic energy of the fly wheel

$KE = \dfrac{1}{2}I\omega^2$

$KE =\dfrac{1}{2}\times 574\times 940^2$

KE = 2.535 x 10⁷ Joules

the kinetic energy of the flywheel is equal to KE = 2.535 x 10⁷ Joules

5 0

2 years ago

Conductors have ___<br> resistance.

vazorg [7]

Answer:

little/no

Explanation:

Conductors are materials, which conduct electricity and/or heat. That means, that their resistance to such energy is so little, that an electric current is able to pass through.

5 0

3 years ago

One of the most important properties of materials in many applications is strength. Two of the qualitative measures of the stren

katrin2010 [14]

To solve this exercise it is necessary to take into account the concepts related to Tensile Strength and Shear Strenght.

In Materials Mechanics, generally the bodies under certain loads are subject to both Tensile and shear strenghts.

By definition we know that the tensile strength is defined as

$\sigma = \frac{F}{A}$

Where,

$\sigma =$ Tensile strength

F = Tensile Force

A = Cross-sectional Area

In the other hand we have that the shear strength is defined as

$\sigma_y = \frac{F_y}{A}$

where,

$\sigma_y =$ Shear strength

$F_y =$ Shear Force

$A_0 =$ Parallel Area

PART A) Replacing with our values in the equation of tensile strenght, then

$311*10^6 = \frac{F}{(15*10^{-6})(30*10^{-2})}$

Resolving for F,

$F= 1399.5N$

PART B) We need here to apply the shear strength equation, then

$\sigma_y = \frac{F_y}{A}$

$210*10^6 = \frac{F_y}{15*10^{-6}30*10^{-2}}$

$F_y = 945N$

In such a way that the material is more resistant to tensile strength than shear force.

6 0

3 years ago