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Alla [95]
3 years ago
10

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.

Physics
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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What is the only component of scalar quantities?
Bad White [126]

Answer:

Scalar quantities have a size or magnitude only and need no other information to specify them. Thus, 10 cm, 50 sec, 7 litres and 3 kg are all examples of scalar quantities.

Explanation:

7 0
3 years ago
A 0.046 kg golf ball hit by a driver can accelerate from rest to 67 m/s in 1 ms while the driver is in contact with the ball. Ho
Flauer [41]

Answer:

Average force = 67 mn

Explanation:

Given:

Initial velocity u = 0 m/s

Final velocity v = 67 m/s

Time t = 1 ms = 0.001 sec.

Computation:

Using Momentum theory

Change in momentum  = F × Δt

 (v-u)/t =  F × Δt

F × 0.001 = (67 - 0)/0.001

F= 67,000,000

Average force = 67 mn

3 0
2 years ago
Sound travels at a rate of 340 m/s in all directions through air. Matt rings a very loud bell at one location, and Steve hears i
Vika [28.1K]

Answer:

110 m/s

Explanation:

because if you subtract 450 from 340 you get 110

6 0
2 years ago
What has more kinetic energy 15 kg ball rolling north at 15 m/s or a 15 kg ball rolling backwards at 7m/s
Setler79 [48]

Answer:

15 kg ball

Explanation:

6 0
3 years ago
Salmon often jump waterfalls to reach their
natta225 [31]

Answer:

5.0 m/s

Explanation:

The horizontal motion of the salmon is uniform, so the horizontal component of the salmon's velocity is constant and it is

v_x = u cos \theta

where u is the initial speed and \theta=37.7^{\circ}. The horizontal distance travelled by the salmon is

d=v_x t = (ucos \theta)t

where d = 1.95 m and t is the time needed to reach the final point.

Re-arranging for t,

t=\frac{d}{v_x}=\frac{d}{u cos \theta} (1)

Along the vertical direction, the equation of motion is

y=h+u_y t -\frac{1}{2}gt^2

where:

y = 0.311 m is the final height reached by the salmon

h = 0 is the initial height

u_y = u sin \theta is the vertical component of the initial velocity of the salmon

g=9.81 m/s^2 is the acceleration of gravity

t is the time

Substituting t as found in eq.(1), we get the equation

y=(u sin \theta) \frac{d}{u cos \theta}- \frac{1}{2}g\frac{d^2}{u^2 cos^2 \theta}=d tan \theta - \frac{1}{2}g\frac{d^2}{u^2 cos^2 \theta}

and we can solve this formula for u, the initial speed of the salmon:

y=d tan \theta - \frac{1}{2}g\frac{d^2}{u^2 cos^2 \theta}\\\\u=\sqrt{\frac{gd^2}{2(dtan \theta -y)cos^2 \theta}}=\sqrt{\frac{(9.81)(1.95)^2}{2((1.95)(tan 37.7^{\circ}) -0.311)cos^2 37.7^{\circ}}}=5.0 m/s

5 0
3 years ago
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