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MissTica
3 years ago
5

A river flows from south to north at 5.4 km/hr. on the west bank of this river, a boat launches and travels perpendicular to the

current with a velocity of 7.6 km/hr due east. if the river is 1.4 km wide at this point, how far downstream does the boat land on the east bank of the river relative to the point it started at on the west bank?
Physics
1 answer:
Aleks [24]3 years ago
7 0
The motion of the boat is basically a uniform motion on both directions, north-south (NS) and east-west (EW), with two constant velocities: v_{SN}=5.4~km/h and v_{EW}=7.6~km/h. 
From the law of motion on the EW direction we can find the total time of the motion:
S_{EW}(t) = v_{EW} t
since we know S_{EW}=1.4~km, we find
t= \frac{S_{EW}}{v_{EW}} =0.18~h
and then, we can find how far the boat went on the NS direction during this time:
S_{NS}=v_{NS} t =5.4~km/h \cdot 0.18~h = 1~km
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Two kids are playing on a newly installed slide, which is 3 m long. John, whose mass is 30 kg, slides down into William (20 kg),
yuradex [85]

Answer:

v=3.564\ m.s^{-1}

\Delta v =2.16\ m.s^{-1}

Explanation:

Given:

  • mass of John, m_J=30\ kg
  • mass of William, m_W=30\ kg
  • length of slide, l=3\ m

(A)

height between John and William, h=1.8\ m

<u>Using the equation of motion:</u>

v_J^2=u_J^2+2 (g.sin\theta).l

where:

v_J = final velocity of John at the end of the slide

u_J = initial velocity of John at the top of the slide = 0

Now putting respective :

v_J^2=0^2+2\times (9.8\times \frac{1.8}{3})\times 3

v_J=5.94\ m.s^{-1}

<u>Now using the law of conservation of momentum at the bottom of the slide:</u>

<em>Sum of initial momentum of kids before & after collision must be equal.</em>

m_J.v_J+m_w.v_w=(m_J+m_w).v

where: v = velocity with which they move together after collision

30\times 5.94+0=(30+20)v

v=3.564\ m.s^{-1} is the velocity with which they leave the slide.

(B)

  • frictional force due to mud, f=105\ N

<u>Now we find the force along the slide due to the body weight:</u>

F=m_J.g.sin\theta

F=30\times 9.8\times \frac{1.8}{3}

F=176.4\ N

<em><u>Hence the net force along the slide:</u></em>

F_R=71.4\ N

<em>Now the acceleration of John:</em>

a_j=\frac{F_R}{m_J}

a_j=\frac{71.4}{30}

a_j=2.38\ m.s^{-2}

<u>Now the new velocity:</u>

v_J_n^2=u_J^2+2.(a_j).l

v_J_n^2=0^2+2\times 2.38\times 3

v_J_n=3.78\ m.s^{-1}

Hence the new velocity is slower by

\Delta v =(v_J-v_J_n)

\Delta v =5.94-3.78= 2.16\ m.s^{-1}

8 0
3 years ago
Estimate the volume of a piece of molecular cloud that has the same amount of water as your body.
noname [10]

Question:

The water molecules now in your body were once part of a molecular cloud. Only about onemillionth of the mass of a molecular cloud is in the form of water molecules, and the mass density of such a cloud is roughly 2.0×10−21 g/cm^3.

Estimate the volume of a piece of molecular cloud that has the same amount of water as your body.

Answer:

The volume of cloud that has the same density as the amount of water in our body is 1.4×10²⁵ cm³

Explanation:

Here, we have mass density of cloud  =  2.0×10⁻²¹ g/cm^3

Density = Mass/Volume

Volume = Mass/Density =   If the mass is 40 kg and the body is made up of 70% by mass of water, we have

28 kg water = 28000 g

Therefore the Volume = 28 kg/ 2.0×10⁻²¹ g/cm^3 = 1.4×10¹⁹ m³ = 1.4×10²⁵ cm³.

Therefore, the volume of cloud that has the same density as the amount of water in our body = 1.4×10²⁵ cm³.

6 0
3 years ago
Help please..................
Shtirlitz [24]
<h3><em>If two objects with the same charge are brought towards each other the force produced will be repulsive, it will push them apart. If two objects with opposite charges are brought towards each other the force will be attractive, it will pull them towards each other.</em></h3><h3><em>hope it helps.... thank you....</em></h3>
7 0
2 years ago
Which indicates that light travels in straight lines?
aleksandr82 [10.1K]

Answer:

A

Explanation:

8 0
3 years ago
Read 2 more answers
.
beks73 [17]

Answer:

2 m/s²

Explanation:

the equations of motion are

S= ut +½at²

v² = u²+ 2as

v = u + at

s = (u+v)/2 × t

From the parameters given

u = 0m/s this is because it starts from rest

Distance (s)  = 9m

Time (t)  = 3s

Based on this the first equation would be used

s = ut + ½at²

Input values

9 = 0×3 + ½ × a x 3²

9 = 0 + 9a/2

9 = 4.5a

Divide both sides by 4.5

a = 9 / 4.5 m/s²

a = 2 m/s²

I hope this was helpful, please mark as brainliest

3 0
3 years ago
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