Ask question

Ask question

All categories

3 years ago

9

A car traveling at 27 m/s slams on its brakes to come to a stop. It decelerates at a rate of 8 m/s2 . What is the stopping dista

nce of the car?

1 answer:

qaws [65]3 years ago

5 0

<em>v</em>² - <em>u</em>² = 2 <em>a</em> ∆<em>x</em>

where <em>u</em> = initial velocity (27 m/s), <em>v</em> = final velocity (0), <em>a</em> = acceleration (-8 m/s², taken to be negative because we take direction of movement to be positive), and ∆<em>x</em> = stopping distance.

So

0² - (27 m/s)² = 2 (-8 m/s²) ∆<em>x</em>

∆<em>x</em> = (27 m/s)² / (16 m/s²)

∆<em>x</em> ≈ 45.6 m

You might be interested in

single goose sounds a loud warning when an intruder enters the farmyard. Some distance from the goose, you measure the sound lev

Goryan [66]

Answer:

The sound level of the 26 geese is $Z_{26}= 96.15 dB$

Explanation:

From the question we are told that

The sound level is $Z_1 = 81.0 \ dB$

The number of geese is $N = 26$

Generally the intensity level of sound is mathematically represented as

The intensity of sound level in dB for one goose is mathematically represented as

$Z_1 = 10 log [\frac{I}{I_O} ]$

Where I_o is the threshold level of intensity with value $I_o = 1*10^{-12} \ W/m^2$

$I$ is the intensity for one goose in $W/m^2$

For 26 geese the intensity would be

$I_{26} = 26 * I$

Then the intensity of 26 geese in dB is

$Z_{26} = 10 log[\frac{26 I }{I_o} ]$

$Z_{26} = 10 log (\ \ 26 * [\frac{ I }{I_o} ]\ \ )$

$Z_{26} = 10 log (\ \ 26 \ \ ) * (\ \ 10 log [\frac{ I }{I_o} ]\ \ )$

From the law of logarithm we have that

$Z_{26} = 10 log 26 + 10 log [\frac{I}{I_0} ]$

$= 14.15 + 82$

$Z_{26}= 96.15 dB$

4 0

3 years ago

What is the weight of a box with a mass of 150 kg on Earth?

Dmitry [639]

Answer:

W = M g = 150 kg * 9.81 m/s^2 = 1470 N

You were only given 3 significant figures in the question.

6 0

2 years ago

The mass of the uniform cantilever is 1100 kg. Determine the force on the beam at A. Determine the force on the beam at B. Use C

Mashcka [7]

Answer:

Force A=-−2,697.75 N

Force B=13, 488.75 N

Explanation:

Taking moments at point A, the sum of clockwise and anticlockwise moments equal to zero.

25 mg-20Fb=0

25*1100g=20Fb

Fb=25*1100g/20=1375g

Taking g as 9.81 then Fb=1375*9.81=13,488.75 N

The sum of upward and downward forces are same hence Fa=1100g-1375g=-275g

-275*9.81=−2,697.75. Therefore, force A pulls downwards

Note that the centre of gravity is taken to be half the whole length hence half of 50 is 25 m because center of gravity is always at the middle

8 0

3 years ago

Fear me or i am glory i give you free p oints <br> \ /<br> _| |_

8_murik_8 [283]

Answer:

kayyy

Explanation:

3 0

3 years ago

Read 2 more answers

Density is a physical property that relates the mass of a substance to its volume. A. Calculate the density, in g/mL , of a liqu

Angelina_Jolie [31]

Answer:

A) 0.660 g/ml

B) 1.297 ml

C) 0.272 g

Explanation:

Every substance, body or material has mass and volume, however the mass of different substances occupy different volumes. This is where density $D$ appears as a physical characteristic property of matter that establishes a relationship between the mass $m$ of a body or substance and the volume $V$ it occupies:

$D=\frac{m}{V}$ (1)

Knowing this, let's begin with the answers:

<h2 /><h2>Answer A:</h2>

Here the mass is $m=0.155g$ and th volume $V=0.000235L=0.235mL$

Solving (1) with these values:

$D=\frac{0.155g}{0.235mL}$ (2)

$D=0.660g/mL$ (3)

<h2>Answer B:</h2>

In this case the mass of a sample is $m=4.71g$ and its density is $D=3.63g/mL$ .

Isolating $V$ from (1):

$V=\frac{m}{D}$ (4)

$V=\frac{4.71g}{3.63g/mL}$ (5)

$V=1.297mL$ (5)

<h2>Answer C:</h2>

In this case the volume of a sample is $V=0.293mL$ and its density is $D=0.930g/mL$ .

Isolating $m$ from (1):

$m=D.V$ (6)

$m=(0.930g/mL)(0.293mL)$ (7)

$m=0.272g$ (8)

4 0

3 years ago

Read 2 more answers