Ask question

Ask question

All categories

4 years ago

5

Semi-trailer trucks have an odometer on one hub of a trailer wheel. The hub is weighted so that it does not rotate, but it conta

ins gears to count the number of wheel revolutions—it then calculates the distance traveled. If the wheel has a 1.30 m diameter and goes through 300,000 rotations, how many kilometers should the odometer read?

1 answer:

yKpoI14uk [10]4 years ago

7 0

Answer: 1,224.6km

S=r*theta

r=d/2=1.30m/2=.650m

Theta=(2pi/rev)rev

Theta=300,000rev(2pi/rev)=1,884,000rads

S=0.650m*1,844,000rad=1,224,600m

S=1,224,600m(1km/1000m)=1,224.6km

You might be interested in

The Moon is made up mostly of _______, similar to minerals on Earth.

yawa3891 [41]

The moon is made up mostly of Silicates, hope this answer helps.

3 0

3 years ago

Read 2 more answers

two forces P and Q pass through a point A which is 4 ft to the right of and 3 ft above a moment center O. force P is 200 lb dire

valentinak56 [21]

Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions here.

The moment of the resultant of these two forces with respect to O 376 lb-ft CCW which is <span>about moment center point O.</span>

6 0

3 years ago

A particle (mass = 2.0 mg, charge = −6.0 μC) moves in the positive direction along the x axis with a velocity of 3.0 km/s. It en

Virty [35]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The acceleration would be $a = 0.003* 6 =0.018\ m/s^2$

Explanation:

The objective of this solution is to obtain the acceleration of the particle

Now looking at Newton law which is mathematically represented as

$F = ma$

Where F is the force experience by a particle

m is the mass of the particle

a is the acceleration of the particle

And also this force is equivalent to magnetic force in a magnetic field which is mathematically represented as

$F = qvB$

Where q is the charge of the particle

v is the velocity of the charge

B is the magnetic field the charge is under it influence

Now equating this two formulas

$ma = qvB$

Making a the subject we have

$a = \frac{qvB}{m}$

In the question the direction of the is in the positive x-axis which is $i$ hence the direction would be in the $i$ direction

So substituting $(2.0i +3.0j+4.0k)mT = (2.0i +3.0j+4.0k)*10^{-3}T$ for B

$a = \frac{q}{m} * v (2.0i +3.0j +4.0k)*10^{-3}$

Substituting $3.0 Km /s = 3.0*10^{3}\ m/s$ for v and $-6.0 \muC = -6.0*10^{-6} C$ for q

$a = \frac{-6.0*10^{-6}}{2.0*10^{-3}} * 3.0*10^{3} *(2i+3j+4k) *10^{-3}$

$a = 0.003 * 3i(2i+3j+4k)$

$a = 0.003 *((3*2)i \ \cdot i \ +(3*3) i \ \cdot \ j \ + (3*4)i \ \cdot \ k)$

According to vector multiplication

$i \cdot i = j \cdot j = k\cdot k = 1\\\\and \ i\cdot j = i\cdot k = 0$

So

$a = 0.003* 6 =0.018\ m/s^2$

8 0

3 years ago

You leave on a 450 miles trip in order to attend a meeting that will start 10.8 hours after you begin your trip. Along the way y

konstantin123 [22]

Answer:

c. 2.6 h

Explanation:

The longest time spent over dinner is the time that you have available minus the minimum possible time spent in the trip.

The time of the trip is found using:

t = $\frac{d}{v}$

Where distance is d and velocity is v. The time will be minimum at maximum velocity. Replacing with the data we have:

Ttrip = $\frac{450}{55}$ = 8.1818 h

Tdinner = 10.8h - 8.1818 h = 2.6181h

that aproximates 2.6 h.

4 0

3 years ago

a vector has an x-component of 19.5m and a y-component of 28.4m. Find the magnitude and direction of the vector

Delicious77 [7]

Answer:

magnitude=34.45 m

direction= $55.52\°$

Explanation:

Assuming the initial point P1 of this vector is at the origin:

P1=(X1,Y1)=(0,0)

And knowing the other point is P2=(X2,Y2)=(19.5,28.4)

We can find the magnitude and direction of this vector, taking into account a vector has a initial and a final point, with an x-component and a y-component.

For the magnitude we will use the formula to calculate the distance $d$ between two points:

$d=\sqrt{{(Y2-Y1)}^{2} +{(X2-X1)}^{2}}$ (1)

$d=\sqrt{{(28.4 m - 0 m)}^{2} +{(19.5 m - 0m)}^{2}}$ (2)

$d=\sqrt{1186.81 m^{2}}$ (3)

$d=34.45 m$ (4) This is the magnitude of the vector

For the direction, which is the measure of the angle the vector makes with a horizontal line, we will use the following formula:

$tan \theta=\frac{Y2-Y1}{X2-X1}$ (5)

$tan \theta=\frac{24.8 m - 0m}{19.5 m - 0m}$ (6)

$tan \theta=\frac{24.8}{19.5}$ (7)

Finding $\theta$ :

$\theta= tan^{-1}(\frac{24.8}{19.5})$ (8)

$\theta= 55.52\°$ (9) This is the direction of the vector

3 0

3 years ago