All categories

2 years ago

A 750 kg car moving at 25 m/s has Kinetic Energy. To stop the car requires work. On snowy roads, the force of friction that can

Physics

1 answer:

liraira [26]2 years ago

4 0

a. By the work-energy theorem, the total work required to stop the car is equal to the change in its kinetic energy,

W = 0 - 1/2 (750 kg) (25 m/s)² ≈ -230 kJ

b. The car covers a distance x as it stops such that

W = (-725 N) x ==> x ≈ 320 m

You might be interested in

Why do scientists not use US customary units when reporting their data?

Marianna [84]

From britaññica it said time and money. They didn’t have either to switch over from the industrial period and never did. Also from my own person reasoning i think most of the world uses not US customary, so to make stuff more accessible. hope this helps!

8 0

3 years ago

Problem 1: Two sources emit waves that are coherent, in phase, and have wavelengths of 26.0 m. Do the waves interfere constructi

Anton [14]

1) Destructive interference

The condition for constructive interference to occur is:

$\delta = m\lambda$ (1)

where

$\delta =|d_1 -d_2|$ is the path difference, with

$d_1$ is the distance of the point from the first source

$d_2$ is the distance of the point from the second source

m is an integer number

$\lambda$ is the wavelength

In this problem, we have

$d_1 = 78.0 m\\d_2 = 143 m\\\lambda=26.0 m$

So let's use eq.(1) to see if the resulting m is an integer

$\delta =|78.0 m-143 m|=65 m\\m=\frac{\delta }{\lambda}=\frac{65 m}{26.0 m}=2.5$

It is not an integer so constructive interference does not occur.

Let's now analyze the condition for destructive interference:

$\delta = (m+\frac{1}{2})\lambda$ (2)

If we apply the same procedure to eq.(2), we find

$m=\frac{\delta}{\lambda}-\frac{1}{2}=\frac{65.0 m}{26.0 m}-0.5=2$

which is an integer: so, this point is a point of destructive interference.

2) Constructive interference

In this case we have

$d_1 = 91.0 m\\d_2 =221.0 m$

So the path difference is

$\delta =|91.0 m-221.0 m|=130.0 m$

Using the condition for constructive interference:

$m=\frac{\delta }{\lambda}=\frac{130.0 m}{26.0 m}=5$

Which is an integer, so this is a point of constructive interference.

3) Destructive interference

In this case we have

$d_1 = 44.0 m\\d_2 =135.0 m$

So the path difference is

$\delta =|44.0 m-135.0 m|=91.0 m$

Using the condition for constructive interference:

$m=\frac{\delta }{\lambda}=\frac{91.0 m}{26.0 m}=3.5$

This is not an integer, so this is not a point of constructive interference.

So let's use now the condition for destructive interference:

$m=\frac{\delta}{\lambda}-\frac{1}{2}=\frac{91.0 m}{26.0 m}-0.5=3$

which is an integer: so, this point is a point of destructive interference.

3 0

3 years ago

The purpose of the fraud fighter/ black light machine is to:

Aleks [24]

The purpose of the fraud fighter or also known as black light machine is that it has the ability to check counterfeited bills. It is designed to be a detector with the counterfeited bills, plus, it is a way of verification. It is composed of a UV light, where in, making it have the ability to detect the counterfeited bills which is its task to do.

3 0

2 years ago

To apply Problem-Solving Strategy 12.2 Sound intensity. You are trying to overhear a most interesting conversation, but from you

Ivenika [448]

Answer:

r₂ = 0.316 m

Explanation:

The sound level is expressed in decibels, therefore let's find the intensity for the new location

β = 10 log $\frac{I}{I_o}$

let's write this expression for our case

β₁ = 10 log \frac{I_1}{I_o}

β₂ = 10 log \frac{I_2}{I_o}

β₂ -β₁ = 10 ( $log \frac{I_2}{I_o} - log \frac{I_1}{I_o}$ )

β₂ - β₁ = 10 $log \frac{I_2}{I_1}$

log \frac{I_2}{I_1} = $\frac{60 - 20}{10}$ = 3

$\frac{I_2}{I_1}$ = 10³

I₂ = 10³ I₁

having the relationship between the intensities, we can use the definition of intensity which is the power per unit area

I = P / A

P = I A

the area is of a sphere

A = 4π r²

the power of the sound does not change, so we can write it for the two points

P = I₁ A₁ = I₂ A₂

I₁ r₁² = I₂ r₂²

we substitute the ratio of intensities

I₁ r₁² = (10³ I₁ ) r₂²

r₁² = 10³ r₂²

r₂ = r₁ / √10³

we calculate

r₂ = $\frac{10.0}{\sqrt{10^3} }$

r₂ = 0.316 m

8 0

3 years ago

A satellite circles the Earth in an orbit whose radius is twice the Earth’s radius. The Earth’s mass is 5.98 x 1024 kg, and its

gavmur [86]

Hello!

Recall the period of an orbit is how long it takes the satellite to make a complete orbit around the earth. Essentially, this is the same as 'time' in the distance = speed * time equation. For an orbit, we can define these quantities:

$d = 2\pi r$ ← The circumference of the orbit