Answer:
v = 11.0 m/s at 198.6° (18.6° south of west)
ΔKE = -145 kJ
Explanation:
I assume you want to find the final velocity and the change in kinetic energy.
Take east to be +x and north to be +y.
Momentum is conserved in the x direction:
(1050 kg) (0 m/s) + (750 kg) (-25.0 m/s) = (1050 kg + 750 kg) vₓ
vₓ = -10.4 m/s
Momentum is conserved in the y direction:
(1050 kg) (-6.00 m/s) + (750 kg) (0 m/s) = (1050 kg + 750 kg) vᵧ
vᵧ = -3.50 m/s
The magnitude of the final velocity is:
v² = (-10.4 m/s)² + (-3.50 m/s)²
v = 11.0 m/s
The direction of the final velocity is:
θ = atan(-3.50 m/s / -10.4 m/s)
θ = 198.6°
The initial kinetic energy is:
KE₀ = ½ (1050 kg) (6.00 m/s)² + ½ (750 kg) (25.0 m/s)²
KE₀ = 253,275 J
The final kinetic energy is:
KE = ½ (1800 kg) (11.0 m/s)²
KE = 108,682 J
The change in kinetic energy is:
ΔKE = 108,682 J − 253,275 J
ΔKE ≈ -145,000 J
At 1.70 atm, a gas sample occupies 4.25 liters. If the pressure in the gas increases to 2.40 atm, what will the new volume be?
Answer:
3.01L
Explanation:
Given parameters:
Initial pressure, P1 = 1.7atm
Initial volume, V1 = 4.25L
Final pressure, P2 = 2.4atm
Unknown:
Final or new volume, V2 = ?
Solution:
To solve this problem, we use Boyle's law which states that "the volume of a fixed mass of a gas varies inversely as the pressure changes, if the temperature is constant".
P1 V1 = P2 V2
P1 is the initial pressure
V1 is the initial volume
P2 final pressure
V2 final volume
1.7 x 4.25 = 2.4 x V2
V2 = 3.01L
Answer:
C.) The slinky particles move up and down
Explanation:
<u>Transverse Wave</u>-
<em>A wave that has a disturbance perpendicular to the wave motion</em>
<em></em>
<em>Hello! This is the correct answer! Have a blessed day! :)</em>
<em>If you are in K12, please review the lesson! :) It will give you some very helpful definitions! I hope this helped!</em>
<u />
-- 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.
