-- 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) = <u>181 kg-m/s</u>
<u>relative to the dock</u>. 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 = <em>0.12 m/s ,</em> <u>away</u> 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 <u>toward</u> 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 = <em>0.62 m/s</em> <u>away</u> 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 <u>that</u> at 0.62 m/s, and the person standing
on the dock sees you start to move away <u>from him</u> at 0.12 m/s, and <em><u>that's</u></em> 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.
Answer:
Explanation:
Project mass m=3.8 kg
Initial speed vi= 0m/s
Final speed vf= 9.3×10³ m/s
Force F=9.3×10⁵N
To find
Time t
Solution
From Newtons second law we know that
∑F=ma
Where m is mass
a is acceleration
We can write this equation as
∑F=m(Δv/Δt)
Rearrange this equation to find time t
So
Substitute the given values
<h2>Answers:</h2><h2 /><h2>a) Arrow B</h2><h2>b) Arrow E</h2>
Explanation:
Refraction is a phenomenon in which a wave (the light in this case) bends or changes its direction <u>when passing through a medium with a refractive index different from the other medium.</u> Where the Refractive index is a number that describes how fast light propagates through a medium or material.
According to this, if we observe the rays A an D passing throgh the biconcave lens, we will have two mediums:
1) The air
2)The material of the biconcave lens
This two mediums have different refractive indexes, hence the rays will change the direction.
-For the incident ray A, the corresponding refractive ray is B, because is the ray that bends after passing throgh the lens
-For the incident ray D, the refracted ray is E following the same principle.
Answer: MOTION
Explanation:
motion is defined as the displacement of an object with respect to time relative to a stationary object (reference point). A good example of an object that can serve as a reference point includes: a tree or a building. The movement of a body at constant speed towards a particular direction at regular intervals of time can be determined and it's called uniform motion.
There are different types of motion, these includes: simple harmonic motion,
linear motion,
circular motion,
Brownian motion,
Rotatory motion
