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poizon [28]
3 years ago
15

Two passenger cars, car A and car B, of the same weight are put to a 30 mph head-on crash test. In the test, cars are pulled for

ward by a cable at 30 mph speed into a flat concrete wall. After the crash test, the length of the cars became shorter because the frontal part of the cars got crumbled at the crash. If car A's length got shortened much more than car B's,
(A) the impact force on car A was much less than on car B during the collision.
(B) the impulse on the car A was much more than on car B during the collision.
(C) the impulse on the car A was much less than on car B during the collision.
(D) the impact force on car A was much more than on car B during the collision.
Physics
1 answer:
maksim [4K]3 years ago
7 0

Answer:

the statements the correct one is A

Explanation:

Let's analyze this exercise, vehicles have the same mass and speed, so we can use the momentum impulse ratio

          I = ∫ F dt = Δp

the Δp is the same for both cars since they have the same mass and the same speeds, so the momentum is the same in both vehicles

When they indicate that vehicle A was reduced more than vehicle B, this implies that the force acted for a longer time, to have the largest reduction in size, therefore the impact force was less in car A than in car B

Resisting the statements the correct one is A

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The amount of heat energy required to raise the temperature of a unit mass of a material one degree is
charle [14.2K]
The amount of heat energy required to raise the temperature of a unit mass of a material to one degree is called D. its heat capacity.

The relationship of the heat when applied to the object and the change in temperature of the object when heat is being applied is directly proportional to each other. This means that when heat is applied to the object, the temperature of the object increases and when heat is not applied to the object, the temperature of the object decreases.
5 0
3 years ago
We can model a pine tree in the forest as having a compact canopy at the top of a relatively bare trunk. Wind blowing on the top
notka56 [123]

Answer:

A,)FD= 114.1N

B)Torque=798.5Nm

Explanation:

We can model a pine tree in the forest as having a compact canopy at the top of a relatively bare trunk. Wind blowing on the top of the tree exerts a horizontal force, and thus a torque that can topple the tree if there is no opposing torque. Suppose a tree's canopy presents an area of 9.0 m^2 to the wind centered at a height of 7.0 m above the ground. (These are reasonable values for forest trees.)

If the wind blows at 6.5 m/s, what is the magnitude of the drag force of the wind on the canopy? Assume a drag coefficient of 0.50 and the density of air of 1.2 kg/m^3

B)What torque does this force exert on the tree, measured about the point where the trunk meets the ground?

A)The equation of Drag force equation can be expressed below,

FD =[ CD × A × ρ × (v^2/ 2)]

Where CD= Drag coefficient for cone-shape = 0.5

ρ = Density

Area of of the tree canopy = 9.0 m^2

density of air of = 1.2 kg/m^3

V= wind velocity= 6.5 m/s,

If we substitute those values to the equation, we have;

FD =[ CD × A × ρ × (v^2/ 2)]

F= [ 0.5 × 9.0 m^2 × 1.2 kg/m^3 ( 6.5 m/s/ 2)]

FD= 114.1N

B) the torque can be calculated using below formula below

Torque= (Force × distance)

= 114.1 × 7

= 798.5Nm

8 0
3 years ago
Consider a monochromatic electromagnetic plane wave propagating in the x direction. At a particular point in space, the magnitud
Scrat [10]

Answer:

a)   S = 2.35 10³   J/m²2 ,  

b)and the tape recorder must be in the positive Z-axis direction.

the answer is 5

c) the direction of the positive x axis

Explanation:

a) The Poynting vector or intensity of an electromagnetic wave is

          S = 1 /μ₀ E x B

if we use that the fields are in phase

          B = E / c

we substitute

         S = E² /μ₀ c

let's calculate

        s = 941 2 / (4π 10⁻⁷  3 10⁸)

        S = 2.35 10³   J/m²2

 

b) the two fields are perpendicular to each other and in the direction of propagation of the radiation

In this case, the electro field is in the y direction and the wave propagates in the ax direction, so the magnetic cap must be in the y-axis direction, and the tape recorder must be in the positive Z-axis direction.

the answer is 5

C) The poynting electrode has the direction of the electric field, by which or which should be in the direction of the positive x axis

5 0
3 years ago
Consider a system of a cliff diver and the Earth. The gravitational potential energy of the system decreases by 25,000 J as the
salantis [7]

Answer:

568.18 N

Explanation:

From the question,

The formula for gravitational potential is given as

Ep = mgh........................ Equation 1

Where Ep = Gravitational potential, m = mass of the diver,h = Height.

But,

W = mg.................... Equation 2

Where W = weight of the diver.

Substitute equation 2 into equation 1

Ep = Wh

Make W the subject of the equation

W = Ep/h................... Equation 3

Given: Ep = 25000 J, h = 44 m

Substitute into equation 3

W = 25000/44

W = 568.18 N.

Hence the weight of the diver = 568.18 N

5 0
3 years ago
Which of the following compounds is not likely to have ionic bonds? Question 5 options: LiF NaCl CH4 MgF2
Harrizon [31]

Answer:

CH4

Explanation:

CH4 is joined together by a covalent bond, aka a bond between two non-metals. Non-metals are found on the right side of the periodic table and include Carbon (C) and Hydrogen. Although Hydrogen is technically on the left side of the table, it has the characteristics of a non-metal. Futhermore, Ionic bonds generally are between an element on the right joined with an element on the left. This is because ionic bonds want charges that will cancel out to create a neutral molecule.

example: LiF

Li→ Li+

F→F-

(Li+)+(F-)=charges cancel out.

7 0
3 years ago
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