When a motorist passes a bicyclist, what is the minimum legal distance (clearance) to leave between vehicle and bike? *

Vector A extends from the origin to a point having polar coordinates (7, 70ᵒ ) and vector B extends from the origin to a point h

yaroslaw [1]

Answer:

13.95

Explanation:

Given :

Vector A polar coordinates = ( 7, 70° )

Vector B polar coordinates = ( 4, 130° )

To find A . B we will

A ( r , ∅ ) = ( 7, 70 )

A = rcos∅ + rsin∅

therefore ; A = 2.394i + 6.57j

B ( r , ∅ ) = ( 4, 130° )

B = rcos∅ + rsin∅

therefore ; B = -2.57i + 3.06j

Hence ; A .B

( 2.394 i + 6.57j ) . ( -2.57 + 3.06j ) = 13.95

8 0

3 years ago

HW6P2 (20 points) The recorded daily temperature (°F) in New York City and in Denver, Colorado during the month of January 2014

Maurinko [17]

Answer & Explanation:

function Temprature

NYC=[33 33 18 29 40 55 19 22 32 37 58 54 51 52 45 41 45 39 36 45 33 18 19 19 28 34 44 21 23 30 39];

DEN=[39 48 61 39 14 37 43 38 46 39 55 46 46 39 54 45 52 52 62 45 62 40 25 57 60 57 20 32 50 48 28];

%AVERAGE CALCULATION AND ROUND TO NEAREST INT

avgNYC=round(mean(NYC));

avgDEN=round(mean(DEN));

fprintf('\nThe average temperature for the month of January in New York city is %g (F)',avgNYC);

fprintf('\nThe average temperature for the month of January in Denvar is %g (F)',avgDEN);

%part B

count=1;

NNYC=0;

NDEN=0;

while count<=length(NYC)

if NYC(count)>avgNYC

NNYC=NNYC+1;

end

if DEN(count)>avgDEN

NDEN=NDEN+1;

end

count=count+1;

end

fprintf('\nDuring %g days, the temprature in New York city was above the average',NNYC);

fprintf('\nDuring %g days, the temprature in Denvar was above the average',NDEN);

%part C

count=1;

highDen=0;

while count<=length(NYC)

if NYC(count)>DEN(count)

highDen=highDen+1;

end

count=count+1;

end

fprintf('\nDuring %g days, the temprature in Denver was higher than the temprature in New York city.\n',highDen);

end

%output

check the attachment for additional Information

8 0

3 years ago

A sample of sand weighs 490 g in stock and 475 in Oven Dry (OD) condition, respectively. If absorption capability of the sand is

Ivahew [28]

The weight of the specimen in SSD condition is 373.3 cc

<u>Explanation</u>:

a) Apparent specific gravity = $\frac{A}{A-C}$

Where,

A = mass of oven dried test sample in air = 1034 g

B = saturated surface test sample in air = 1048.9 g

C = apparent mass of saturated test sample in water = 975.6 g

apparent specific gravity = $\frac{A}{A-C}$

= $\frac{1034}{1034-675 \cdot 6}$

Apparent specific gravity = 2.88

b) Bulk specific gravity $G_{B}^{O D}=\frac{A}{B-C}$

$G_{B}^{O D}=\frac{1034}{1048.9-675 \cdot 6}$

= 2.76

c) Bulk specific gravity (SSD):

$G_{B}^{S S D}=\frac{B}{B-C}$

$=\frac{1048 \cdot 9}{1048 \cdot 9-675 \cdot 6}$

$G_{B}^{S S D}$ = 2.80

d) Absorption% :

$=\frac{B-A}{A} \times 100 \%$

$=\frac{1048 \cdot 9-1034}{1034} \times 100$

Absorption = 1.44 %

e) Bulk Volume :

$v_{b}=\frac{\text { weight of dispaced water }}{P \omega t}$

$=\frac{1048 \cdot 9-675 \cdot 6}{1}$

= $373.3 cc$

5 0

3 years ago

An inductor has a 50.0-Ω reactance when connected to a 60.0-Hz source. The inductor is removed and then connected to a 45.0-Hz s

nignag [31]

Given:

$X_{L} = 50.0 \ohm$

frequency, f = 60.0 Hz

frequency, f' = 45.0 Hz

$V_rms} = 85.0 V$

Solution:

To calculate max current in inductor, $I_{L(max)$ :

At f = 60.0 Hz

$X_{L} = 2\pi fL$

$50.0 = 2\pi\times 60.0\times L$

L = 0.1326 H

Now, reactance $X_{L}$ at f' = 45.0 Hz:

$X'_{L} = 2\pi f'L$

$X'_{L} = 2\pi\times 45.0\times 0.13263 = 37.5\ohm$

Now, $I_{L(max)$ is given by:

$I_{L(max) = \sqrt {\frac{2V_{rms}}{X'_{L}}}$

$I_{L(max) = \sqrt {\frac{2\times 85.0}{37.5}} = 2.13 A$

Therefore, max current in the inductor, $I_{L(max)$ = 2.13 A

7 0

3 years ago

Two rods, with masses MA and MB having a coefficient of restitution, e, move along a common line on a surface, figure 2. a) Find

ahrayia [7]

Answer:

A.) Find the answer in the explanation

B.) Ua = 7.33 m/s , Vb = 7.73 m/s

C.) Impulse = 17.6 Ns

D.) 49%

Explanation:

Let Ua = initial velocity of the rod A

Ub = initial velocity of the rod B

Va = final velocity of the rod A

Vb = final velocity of the rod B

Ma = mass of rod A

Mb = mass of rod B

Given that

Ma = 2kg

Mb = 1kg

Ub = 3 m/s

Va = 0

e = restitution coefficient = 0.65

The general expression for the velocities of the two rods after impact will be achieved by considering the conservation of linear momentum.

Please find the attached files for the solution

6 0

3 years ago