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3 years ago

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Now assume that the stream flows east at 4.0 m/s. draw the vectors v⃗ w, representing the velocity of the stream, and v⃗ tot, re

presenting the velocity of your rowboat relative to the stream bank. be sure to draw both vectors. draw v⃗ tot and v⃗ w starting at the tail and tip of v⃗ still respectively. the location, orientation and length of the vectors will be graded. each vector's length is displayed in meters per second.

1 answer:

ELEN [110]3 years ago

3 0

Remember that the total velocity of the motion is the vector sum of the velocity you would have in still water and the stream. Always place the vectors carefully to be able to come up with an accurate sum vector.

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A current of 0.92 a flows in a wire. how many electrons are flowing past any point in the wire per second? the charge on one ele

Fantom [35]

The current is defined as the ratio between the charge Q flowing through a certain point of a wire and the time interval, $\Delta t$ :
$I= \frac{Q}{\Delta t}$
First we need to find the net charge flowing at a certain point of the wire in one second, $\Delta t=1.0 s$ . Using I=0.92 A and re-arranging the previous equation, we find
$Q=I \Delta t= (0.92 A)(1.0 s)=0.92 C$

Now we know that each electron carries a charge of $e=1.6 \cdot 10^{-19} C$ , so if we divide the charge Q flowing in the wire by the charge of one electron, we find the number of electron flowing in one second:
$N= \frac{Q}{q} = \frac{0.92 C}{1.6 \cdot 10^{-19} C}=5.75 \cdot 10^{18}$

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2 years ago

A ceiling fan with 90-cm-diameter blades is turning at 64 rpm . Suppose the fan coasts to a stop 28 s after being turned off. Wh

mixas84 [53]

Answer:

the speed of the tip of a blade 10 s after the fan is turned off is 16.889 m/s.

Explanation:

Given;

diameter of the ceiling fan, d = 90 cm = 0.9 m

angular speed of the fan, ω = 64 rpm

time taken for the fan to stop, t = 28 s

The distance traveled by the ceiling fan when it comes to a stop is calculated as;

$d = vt = \omega r\times t= ( \frac{64 \ rev}{\min} \times \frac{2 \pi \ rad}{rev} \times \frac{1 \min}{60 \ s} \times 0.9 \ m) \times 28 \ s\\\\d = 168.89 \ m$

The speed of the tip of a blade 10 s after the fan is turned off is calculated as;

$v = \frac{d}{t} \\\\v = \frac{168.89}{10} \\\\v = 16.889 \ m/s$

Therefore, the speed of the tip of a blade 10 s after the fan is turned off is 16.889 m/s.

4 0

3 years ago

I'll mark brainliest

irga5000 [103]

Answer:

A) 35 ft

B) 5 ft

C) Net displacement = distance covered by the dog to retrieve the stick - distance covered before the dog starts chewing the stick

Explanation:

A) Total distance covered by the dog = 20 + 15

= 35 ft

B) Since the other distance covered by the dog before chewing the stick, after the retrieval, was in an opposite direction to the initial direction, then;

total displacement of the dog = 20 - 15

= 5 ft

C) Net displacement = distance covered by the dog to retrieve the stick + distance covered before the dog starts chewing the stick

But, displacement involves a specified direction. The distance covered before the dog starts chewing the stick was in an opposite direction to the initial direction.

Thus,

Net displacement = distance covered by the dog to retrieve the stick - distance covered before the dog starts chewing the stick

7 0

2 years ago

Whats 6 3/7 ×1 5/9.

aniked [119]

6 3/7 * 1 5/9
45/7 * 14/9
630/63
10

7 0

3 years ago

Read 2 more answers

The engine in an imaginary sports car can provide constant power to the wheels over a range of speeds from 0 to 70 miles per hou

puteri [66]

Answer:

$t=5.3687\ s$ is the time taken by the car to accelerate the desired range of the speed from zero at full power.

Explanation:

Given:

Range of speed during which constant power is supplied to the wheels by the car is $0\ mph\ to\ 70\ mph$ .

Initial velocity of the car, $v_i=0\ mph$

final velocity of the car during the test, $v_f=32\ mph=14.3052\ m.s^{-1}$

Time taken to accelerate form zero to 32 mph at full power, $t=1.2\ s$

initial velocity of the car, $u_i=0\ mph$

final desired velocity of the car, $u_f=64\ mph=28.6105\ m.s^{-1}$

Now the acceleration of the car:

$a=\frac{v_f-v_i}{t}$

$a=\frac{14.3052-0}{1.2}$

$a=11.921\ m.s^{-1}$

Now using the equation of motion:

$u_f=u_i+a.t$

$64=0+11.921\times t$

$t=5.3687\ s$ is the time taken by the car to accelerate the desired range of the speed from zero at full power.

8 0

2 years ago