Ask question

Ask question

All categories

3 years ago

8

A crew team rows a boat at a rate of 20 km/h in still water. In practice on a river, the team rows for 30 minutes up the river (

against the current), and then for 30 minutes down the river (with the current). The speed of the river current is 1.5 km/h. How much farther did they travel in the second 30 minutes?

1 answer:

sweet-ann [11.9K]3 years ago

6 0

In the second 30 mins, the speed should be 20 + 1.5 = 21.5 km/h

So S = 21.5 * 30/60 = 10.75 km

You might be interested in

Which of the following are vector quantities? Check all that apply.

Dennis_Churaev [7]

It’s distance for my answer but tell me if anyone tells u this answer thanks

8 0

3 years ago

Technician A says that a MAF sensor is a high-authority sensor and is responsible for determining the fuel needs of the engine b

xz_007 [3.2K]

Answer:

a. Technician A

Explanation:

Technician A says that a MAF sensor is a high-authority sensor and is responsible for determining the fuel needs of the engine based on the measured amount of air entering the engine. Technician B says that a cold wire MAF sensor uses the electronics in the sensor itself to heat a wire 20°C below the temperature of the air entering the engine. Who is right

MAF wich stands for mass airflow sensor determines the mass of air flowing into the engine's air intake system. ... , the wire cools When air flows past the wire, decreasing its resistance, thereby more current flows through the circuit. When the MAf goes bad, it can not sense the amount of air intake into the engine.

8 0

3 years ago

The tape in a videotape cassette has a total

xxMikexx [17]

Answer:

$w=19.76 \ rad/s$

Explanation:

<u>Circular Motion </u>

Suppose there is an object describing a circle of radius r around a fixed point. If the relation between the angle of rotation by the time taken is constant, then the angular speed is also constant. If that relation increases or decreases at a constant rate, the angular speed is given by:

$w=w_o+\alpha \ t$

Where $\alpha$ is the angular acceleration and t is the time. If the object was instantly released from the circular path, it would have a tangent speed of:

$\displaystyle v_t=w.r$

We have two reels: one loaded with the tape to play and the other one empty and starting to fill with tape. They both rotate at different angular speeds, one is increasing and the other is decreasing as the tape goes from one to the other. We'll assume the tangent speed is constant for both (so the tape can play correctly). Let's call $w_1$ the angular speed of the loaded reel and $w_2$ that from the empty reel. We have

$w_1=w_{o1}+\alpha_1 \ t$

$w_2=w_{o2}+\alpha_2 \ t$

If $r_f=35\ mm=0.035\ m$ is the radius of the reel when it's full of tape, the angular speed for the loaded reel is computed by

$\displaystyle w_{01}=\frac{v_t}{r_f}$

The tangent speed is computed by knowing the length of the tape and the time needed to fully play it.

$t=1.8\ h=1.8*3600=6480\ sec$

$\displaystyle v_t=\frac{x}{t}=\frac{249\ m}{6480\ sec}=0.0384 \ m/s$

$\displaystyle w_{01}=\frac{0.0384}{0.035}=1,098 \ rad/s$

If $r_e=10\ mm=0.001\ m$ is the radius of the reel when it's empty, the angular speed for the empty reel is computed by

$\displaystyle w_{02}=\frac{0.0384}{0.001}=38.426 \ rad/s$

The full reel goes from $w_{01}$ to $w_{02}$ in 6480 seconds, so we can compute the angular acceleration:

$\displaystyle \alpha_1=\frac{w_{02}-w_{01}}{6480}$

$\displaystyle \alpha_1=\frac{38.426-1.098}{6480}=0.00576 \ rad/sec^2$

The empty reel goes from $w_{02}$ to $w_{01}$ in 6480 seconds, so we can compute the angular acceleration:

$\displaystyle \alpha_2=\frac{1.098-38.426}{6480}=-0.00576 \ rad/sec^2$

So the equations for both reels are

$w_1=1.098+0.00576 \ t$

$w_2=38.426-0.00576 \ t$

They will be the same when

$1.098+0.00576 \ t=38.426-0.00576 \ t$

Solving for t

$\displaystyle t=\frac{38.426-1.098}{0.0115}$

$t=3240 \ sec$

The common angular speed is

$w_1=1.098+0.00576 \ 3240=19.76 \ rad/s$

$w_2=38.426-0.00576 \ 3240=19.76 \ rad/s$

They both result in the same, as expected

$\boxed{w=19.76 \ rad/s}$

6 0

3 years ago

A long uniformly charged thread (linear charge density λ= 2.5 C/m) lies along the x axis in the figure.(Figure 1) A small charge

Kamila [148]

Answer 1) The electric field at distance r from the thread is radial and has magnitude

E = λ / (2 π ε° r)

The electric field from the point charge usually is observed to follow coulomb's law:

E = Q / (4 π ε° $r^{2}$ )

Now, adding the two field vectors:

$E_{thread}$ = {2.5 / (22 π ε° X 0.07 ) ; 0}

Answer 2) $E_{q}$ = {2.3 / (4 2 π ε°) ( - 7/ (√(84); -12 / (√84))

Adding these two vectors will give the length which is magnitude of the combined field.

The y-component / x-component gives the tangent of the angle with the positive x-axes.

Please refer the graph and the attachment for better understanding.

5 0

3 years ago

NEED HELP ASAP, WILL GIVE BRAINLIEST

alexira [117]

a) El Niño is defined as an abnormal weather pattern caused by the warming of the Pacific Ocean near the equator, off the coast of South America. The sun warms the water near the equator, which can make more clouds and, therefore, more rain. It has detrimental effects on biodiversity leading to its large-scale loss by
warmer sea temperatures leading to plankton and fish kills in coastal waters
lower sea levels leading to exposure of underwater coral reefs, causing their loss.

6 0

2 years ago

Read 2 more answers