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lord [1]
3 years ago
14

URGENT

Physics
1 answer:
olga55 [171]3 years ago
3 0

If the boat's speed is s, and the river's speed is r, and the boat is traveling east (0 degrees),

(0,r) + (s cos297,s sin297) = (6,0)

now just solve for r and s.

Pls mark me as brainliest

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When the driver applies the brakes of a light truck traveling 40 km>h, it skids 3 m before stopping. how far will the truck s
saul85 [17]
I think mathematically it should be 6 m
6m I guess
6 0
4 years ago
A typical garden hose has an inner diameter of 5/8". Let's say you connect it to a faucet and the water comes out of the hose wi
castortr0y [4]
Since speed (v) is in ft/sec, let's convert our diameters from inches to feet:
1) 5/8in = 0.625in
0.625in × 1ft/12in = 0.0521ft
2) 0.25in × 1ft/12in = 0.021ft
Equation:
v = 4q \div ( {d}^{2} \pi) \: where \: q = flow \\ v = velocity \: (speed) \: and \:  \\ d = diameter \: of \: pipe \: or \: hose \\ and \: \pi = 3.142
we \: can \: only \: assume \:that \\  flow \: (q) \:stays \: same \: since \: it \\  isnt \: impeded \: by \:  anything \\ thus \:it  \: (q)\:  stays \: the \: same \:  \\ so \: 4q \: can \: be \: removed \: from \:  \\ the \: equation
then \: we \: can \: assume \: that \: only \\ v \: and \: d \: change \: leading \:us \: to >  >  \\ (v1 \times {d1}^{2} \pi) = (v2  \times   {d2}^{2}\pi)
both \: \pi \: will \: cancel \: each \: other \: out \:  \\ as \: constants \:since \: one \: is \: on \\ each \: side \: of \: the \:  =

(v1  \times   {d1}^{2}) = (v2 \ \times {d2}^{2}) \\ (7.0 \times   {0.052}^{2}) = (v2  \times   {0.021}^{2}) \\ divide \: both \: sides \: by \:  {0.021}^{2} \\ to \: solve \: for \: v2 >  >
v2 = (7.0)( {0.052}^{2} ) \div ( {0.021}^{2})  \\ v2 = (7.0)(.0027) \div (.00043) \\ v2 = 44 \: feet \: per \: second
new velocity coming out of the hose then is
44 ft/sec
4 0
3 years ago
What happens to the Connecticut Energy of a snowball as it rolls across the lawn and gains mass?
statuscvo [17]
I think it will go down like decrease minus or whatever you call it
6 0
4 years ago
A railroad car of mass 2.00 3 104 kg moving at 3.00 m/s collides and couples with two coupled railroad cars, each of the same ma
FrozenT [24]

Given:

Mass of the rail road car, m = 2 kg

velocity of the three cars coupled system, v' = 1.20 m/s

velocity of first car, v_{a} = 3 m/s

Solution:

a) Momentum of a body of mass 'm' and velocity 'v' is given by:

p = mv

Now for the coupled system according to law of conservation of momentum, total momentum of a system before and after collision remain conserved:

mv_{a} + 2mv_{b} = (m + 2m)v'                        (1)

where,

v_{a} = velocity of the first car

v_{b} = velocity of the 2 coupled cars after collision

Now, from eqn (1)

v' = \frac{v_{a} + 2v_{b}}{3}

v' = \frac{3.00 + 2\times 1.20}}{3}

v' = 1.80 m/s

Therefore, the velocity of the combined car system after collision is 1.80 m/s

7 0
3 years ago
Read 2 more answers
If an object accelerates at 40 m/s^2 in four minutes ("careful this is in minutes),
yaroslaw [1]

Change in speed = (acceleration) x (time)

4 minutes = 240 seconds

Change in speed = (40 m/s²) x (240 seconds)

Change in speed = <em>9,600 m/s</em>

What you're actually describing here is a car pulling 4 G's for 4 minutes, and ending up going 21,475 miles per hour.

The driver would definitely NOT get a speeding ticket, because nobody could catch him.

Also, his car would heat up and shoot flames from atmospheric friction.

(He could avoid this with some fancy steering, leave the atmosphere, and end up in low-Earth-orbit.)

Actually, I hope there's nobody in the car.  His experience wouldn't be pretty.

3 0
3 years ago
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