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Anit [1.1K]
2 years ago
11

A boat is headed with a velocity of 18 meters/second toward the last with respect to the water in the river. if the river is Flo

wing with the velocity of 2.5 meters/second in the same direction as the boat. what would the magnitude of the boats velocity be?
Physics
2 answers:
scoray [572]2 years ago
8 0
In this case, the two vectors are in the same direction, so they simply add:

<span> total motion = 18m/s + 2.5m/s = 20.5m/s to the west </span>
muminat2 years ago
7 0
F the boat's velocity is 18m/sec relative to the water in the river and not the shore, it would need to be added the river speed of 2.5m/sec to get a total of 20.5m/sec. The 20.5m/sec would then be the total velocity of the boat relative to the shore. From personal experience, I know that when one runs with the tide, one is adding the tide flow speed to one's boat speed (what it would be in neutral waters) to get a sometimes much faster speed.


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A motorcyclist heading east through a small Iowa town accelerates after he passes a signpost at x=0 marking the city limits. His
adoni [48]

Answer:

1) v_2=23\ m.s^{-1}              &     x_2=43\ m east of sign post

2) x'=55\ m east of sign post

3) x_n=205\ m east of the signpost.

4) v_z=35\ m.s^{-1}

Explanation:

Given:

  • position of motorcyclist on entering the city at the signpost, x_0=0\ m
  • time of observation after being at x=5m east of the signpost, t_m=0\ s
  • constant acceleration of the on entering the city, a=4\ m.s^{-2}
  • distance of the motorcyclist moments later after entering, s_m=5\ m
  • velocity of the motorcyclist moments later after entering, u_m=15\ m.s^{-1}

<u>Now the initial velocity on at the sign board:</u>

u_m^2=u^2+2.a.x_m

where:

u= initial velocity of entering the city at the signpost

Putting respective values:

15^2=u^2+2\times 4\times 5

u=13.6015\ m.s^{-1}

1)

Position at time t_2=2\ s sec.:

Using equation of motion,

x_2=u_m.t_2+\frac{1}{2} a.(t_2)^2+5 because it has already covered 5m before that point

x_2=15\times 2+0.5\times 4\times 2^2+5

x_2=43\ m east of sign post

Velocity at time t_2=2\ s sec.:

v_2=u_m+a.t_2

v_2=15+4\times 2

v_2=23\ m.s^{-1}

2)

Position when the velocity is v'=25\ m.s^{-1}:

using equation of motion,

v'^2=u_m^2+2.a.x'+5

25^2=15^2+2\times 4\times x'+5

x'=55\ m east of sign post

3)

Given that:

acceleration be, a_n=2\ m.s^{-2}

time, t_n=5\ s

Position after the new acceleration and the new given time:

using equation of motion,

x_n=u_m.t_n+\frac{1}{2} a_n.(t_n)^2+5

x_n=15\times 5+0.5\times 2\times 5^2+5

x_n=205\ m east of the signpost.

4)

now time of observation, t_z=5\ s

v_z=u_m+a.t_z

v_z=15+4\times 5

v_z=35\ m.s^{-1}

8 0
2 years ago
A 480-kilogram horse runs across a field at a rate of 40 km/hr. What is the magnitude of the horse's momentum?
Andrej [43]
Momentum (p) = mass × velocity

so, 480×40 = 19,200 kg km/hr

so the answer is C !!
4 0
2 years ago
Read 2 more answers
An archer fires an arrow, which produces a muffled "thwok" as it hits a target. If the archer hears the "thwok" exactly 1 s afte
aniked [119]

Answer:

35,79 meters

Explanation:

So, we got an archer, and we got a target. Lets call the distance between this two d.

Now, the archer fires the arrow, that, in a time t_{arrow} travels the distance d with a speed v_{arrow} of 40 m/s and hits the target. We can see that the equation will be:

v_{arrow} * t_{arrow} = d\\ \\40 \frac{m}{s} * t_{arrow} = d

Immediately after this, the arrow produces a muffled sound, which will travel the distance d at  340 m/s in a time t_{sound}. Obtaining :

v_{sound} * t_{sound} = d\\ \\340 \frac{m}{s} * t_{sound} = d.

Finally, the sound reaches the archer, exactly 1 second after he fired the bow, so:

t_{arrow} + t _{sound} = 1 s.

This equation allows us to write:

t _{sound} = 1 s - t_{arrow}.

Plugging this  relationship in the distance equation for the sound:

340 \frac{m}{s} * t_{sound} = d \\ \\ 340 \frac{m}{s} * (1 s- t_{arrow}) = d.

Now, we can replace d from the first equation, and obtain:

40 \frac{m}{s} * t_{arrow} = d \\ 40 \frac{m}{s} * t_{arrow} = 340 \frac{m}{s} * (1 s- t_{arrow}).

Now, we can just work a little bit:

40 \frac{m}{s} * t_{arrow} = 340 \frac{m}{s} * 1 s - 340 \frac{m}{s} * t_{arrow} \\ \\ 40 \frac{m}{s} * t_{arrow} + 340 \frac{m}{s} * t_{arrow} = 340 m \\ \\ 380 \frac{m}{s} * t_{arrow} = 340 m \\ \\ t_{arrow} = \frac{340 m}{380 \frac{m}{s}} \\ \\ t_{arrow} = 0.8947 s.

Now, we can just plug this value into the first equation:

40 \frac{m}{s} * t_{arrow} = d

40 \frac{m}{s} * 340/380 s = 35,79 s = d

6 0
2 years ago
568 muons were counted by a detector on the top of Mount Washington in a one hour period of time. Assuming moving muons keep tim
AlladinOne [14]
The answer to this question is:

C-"That moving clocks run slower"

Your Welcome :)
6 0
2 years ago
Read 2 more answers
Consider a father pushing a child on a playground merry-go-round. the system has a moment of inertia of 84.4 kg · m2. the father
-Dominant- [34]
<span>At time t1 = 0 since the body is at rest, the body has an angular velocity, v1, of 0. At time t = X, the body has an angular velocity of 1.43rad/s2. Since Angular acceleration is just the difference in angular speed by time. We have 4.44 = v2 -v1/t2 -t1 where V and t are angular velocity and time. So we have 4.44 = 1.43 -0/X - 0. Hence X = 1.43/4.44 = 0.33s.</span>
6 0
2 years ago
Read 2 more answers
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