All categories

3 years ago

C: Calculate the kinetic energy of a car with a mass of 1563 kilograms that is traveling at 42 kilometers per hour.

Physics

1 answer:

kkurt [141]3 years ago

6 0

Hello!

C. KE ≈ 9120.105 J

D. m = 151.45 kg

Question C:

Calculate kinetic energy using the formula:

$KE = \frac{1}{2}mv^{2}$

Substitute in the given mass. We have to convert kilometers to meters in order to solve.

$\frac{42km}{1hr} * \frac{1hr}{3600sec} *\frac{1000m}{1km} = 11.67 m/s$

Use the equation above:

KE = 1/2(1563)(11.67)

KE ≈ 9120.105 J

Question D:

Plug in the given Kinetic Energy and velocity to solve. Convert kilometers/hour to meters/second:

$\frac{7km}{1hr} * \frac{1hr}{3600sec} *\frac{1000m}{1km} = 1.94 m/s$

$KE = \frac{1}{2}mv^{2}$

285 = 1/2(m)(1.94²)

570 = (1.94²)m

151.45kg = m

You might be interested in

Petroleum is a nonrenewable resource. true or false

Nata [24]

Coal, petroleum, and natural gas are considered nonrenewable because they can not be replenished in a short period of time

5 0

3 years ago

Which of the following is true when an object of mass m moving on a horizontal frictionless surface hits and sticks to an object

zhannawk [14.2K]

Answer:

Explanation:

A collision is said to be elastic when the total kinetic energy is the same after the collision. The speed of objects that are stuck together will always be less than the initial speed of the object that was in motion given that the other particle was at rest. It is because the kinetic energy of the system was due to the moving object. The objects have a greater overall mass when they are stuck. If the kinetic energy is the same and the mass increases, the velocity must decrease.

3 0

3 years ago

An airplane wing is designed so that the speed of the air across the top of the wing is 255 m/s when the speed of the air below

grin007 [14]

<h2>Answer:442758.96N</h2>

Explanation:

This problem is solved using Bernoulli's equation.

Let $P$ be the pressure at a point.

Let $p$ be the density fluid at a point.

Let $v$ be the velocity of fluid at a point.

Bernoulli's equation states that $P+\frac{1}{2}pv^{2}+pgh=constant$ for all points.

Lets apply the equation of a point just above the wing and to point just below the wing.

Let $p_{up}$ be the pressure of a point just above the wing.

Let $p_{do}$ be the pressure of a point just below the wing.

Since the aeroplane wing is flat,the heights of both the points are same.

$\frac{1}{2}(1.29)(255)^{2}+p_{up}= \frac{1}{2}(1.29)(199)^{2}+p_{do}$

So, $p_{up}-p_{do}=\frac{1}{2}\times 1.29\times (25424)=16398.48Pa$

Force is given by the product of pressure difference and area.

Given that area is $27ms^{2}$ .

So,lifting force is $16398.48\times 27=442758.96N$

6 0

3 years ago

Which scientific development has given rise to the most debate about the ethics of tampering with nature?

Burka [1]

<span>Good Morning!

The scientific advance that launched an ethical alert in the scientific community was the possibility of cloning animals, since it presents important questions, especially if such a project is extended to humans.

Hugs!</span>

7 0

3 years ago

Read 2 more answers

Light propagate faster through medium “a” than medium “b”

dangina [55]

1) Medium "b" has more optical density

2) Light must hit the interface between the two mediums perpendicularly

Explanation:

Refraction occurs when light propagates from a medium into a second medium.

The optical density of a medium is given by its index of refraction, which is defined as:

$n=\frac{c}{v}$

where

c is the speed of light in a vacuum

v is the speed of light in a medium

Higher index of refraction means higher optical density, and light propagater slower into a medium with higher optical density.

In this problem, light propagates faster through medium "a" than medium "b": this means that medium "a" has lower refractive index of medium "b", and so "b" has more optical density.

We can answer this part by referring to Snell's law, which gives the relationship between the direction of the incident ray and of the refracted ray when light passes through the interface between two media:

$n_1 sin \theta_1 = n_2 sin \theta_2$

where

$n_1, n_2$ are the index of refraction of the two mediums

$\theta_1, \theta_2$ are the angle of incidence and of refraction (the angle that light makes with the normal to the surface in medium 1 and medium 2)