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3 years ago

1. Which will speed up the rate of a chemical

1 answer:

8 0

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a 300kg motorboat is turned off as it approaches a dock and coasts towards it at .5 m/s. Isaac, whose mass is 62 kg jumps off th

Zolol [24]

-- Before he jumps, the mass of (Isaac + boat) = (300 + 62) = 362 kg,
their speed toward the dock is 0.5 m/s, and their linear momentum is

Momentum = (mass) x (speed) = (362kg x 0.5m/s) = 181 kg-m/s

relative to the dock. So this is the frame in which we'll need to conserve
momentum after his dramatic leap.

After the jump:

-- Just as Isaac is coiling his muscles and psyching himself up for the jump,
he's still moving at 0.5 m/s toward the dock. A split second later, he has left
the boat, and is flying through the air at a speed of 3 m/s relative to the boat.
That's 3.5 m/s relative to the dock.

His momentum relative to the dock is (62 x 3.5) = 217 kg-m/s toward it.

But there was only 181 kg-m/s total momentum before the jump, and Isaac
took away 217 of it in the direction of the dock. The boat must now provide
(217 - 181) = 36 kg-m/s of momentum in the opposite direction, in order to
keep the total momentum constant.

Without Isaac, the boat's mass is 300 kg, so

(300 x speed) = 36 kg-m/s .

Divide each side by 300: speed = 36/300 = 0.12 m/s , away from the dock.
=======================================

Another way to do it . . . maybe easier . . . in the frame of the boat.

In the frame of the boat, before the jump, Isaac is not moving, so
nobody and nothing has any momentum. The total momentum of
the boat-centered frame is zero, which needs to be conserved.

Isaac jumps out at 3 m/s, giving himself (62 x 3) = 186 kg-m/s of
momentum in the direction toward the dock.

Since 186 kg-m/s in that direction suddenly appeared out of nowhere,
there must be 186 kg-m/s in the other direction too, in order to keep
the total momentum zero.

In the frame of measurements from the boat, the boat itself must start
moving in the direction opposite Isaac's jump, at just the right speed
so that its momentum in that direction is 186 kg-m/s.
The mass of the boat is 300 kg so
(300 x speed) = 186

Divide each side by 300: speed = 186/300 = 0.62 m/s away from the jump.

Is this the same answer as I got when I was in the frame of the dock ?
I'm glad you asked. It sure doesn't look like it.

The boat is moving 0.62 m/s away from the jump-off point, and away from
the dock.
To somebody standing on the dock, the whole boat, with its intrepid passenger
and its frame of reference, were initially moving toward the dock at 0.5 m/s.
Start moving backwards away from that at 0.62 m/s, and the person standing
on the dock sees you start to move away from him at 0.12 m/s, and that's the
same answer that I got earlier, in the frame of reference tied to the dock.

yay !

By the way ... thanks for the 6 points. The warm cloudy water
and crusty green bread are delicious.

4 0

3 years ago

a current of 180 mini amphere passes through a conductor for 5minute calculate the quantity of electricity transported

oksano4ka [1.4K]

Answer:

Explanation:

You can calculate the total electric charge that passes through the conductor as $q=It=(180\times 10^{-3})(5\times 60)= 54 C$ . It means that the number of electron that passes through the conductor is:

$n=\frac{q}{e}=\frac{54}{1.6\times 10^{-19}}=33.75\times 10^{19}$

8 0

2 years ago

I'm confusion, can I please get help???

kotykmax [81]

Answer:

Anything below 7.0 is acidic, so the range would be 0 to 7.

Neutral is simply 7, in the middle of the scale.

Lastly, anything above 7.0 is basic or alkaline, so that would be 7 to 14.

Good luck, I hope this helps

7 0

3 years ago

A shell is fired with 500km/h on a target half kilometer away.In what time in which shell will hit the target.

Drupady [299]

It depends on the direction in which the shell is launched. The time can be anything from 3.6 seconds to never.

6 0

3 years ago

Calculate, for the judge, how fast you were going in miles per hour when you ran the red light because it appeared Doppler-shift

sammy [17]

Answer:

The doppler effect equation is:

$f' = \frac{v +v0}{v - vs}*f$

In the equation we have frequencies, but then we have the wavelengths of the lights, remember the relation:

v = f*λ

then:

f = v/λ

and v is the speed of light, then:

f = c/λ

where:

f' is the observed frequency, in this case, is equal to f = (3*10^17nm/s)/550 nm

f is the real frequency, in this case, is (3*10^17nm/s)/650 nm

vs is the speed of the source, in this case, the source is not moving, then vs = 0 m/s.

v is the speed of the wave, in this case, is equal to the speed of light, v = 3*10^8 m/s

v0 is your speed, this is what we want to find.

Replacing those quantities in the equation, we get:

(3*10^17nm/s)/550 = (3*10^8 m/s + v0)/(3*10^8 m/s)*(3*10^17nm/s)/650 nm

(650nm)/(550nm) = (3*10^8 m/s + v0)/(3*10^8 m/s)

1.182*(3*10^8 m/s) = (3*10^8 m/s + v0)

1.182*(3*10^8 m/s) - (3*10^8 m/s) = v0 = 54,600,000 m/s

So your speed was 54,600,000 m/s, which is a lot.

6 0

3 years ago