F(g)= Gm1m2/ r^2
If mass is increased, so will the force of gravity because it is in direct relationship with the gravitational force, but if distance is increased, the force of gravity will decrease because it is indirectly related ( since it is on the bottom of the equation)
Answer: D
Experiment 1 has a confounding variable related to the mass of the rockets. Any variation in mass may cause a discrepancy in the distance traveled.
This is the answer to the question because:
- Both experiments do have a confounding variable.
- Experiment 1 doesn't have to stay constant.
- A double-blind experiment will not do anything to the placebo.
- High blood pressure people will not make the results confusing.
The answer has to be the option D. Hope this helps you!
The frequency of a photon with an energy of 4.56 x 10⁻¹⁹ J is 6.88×10¹⁴ s⁻¹.
<h3>What is a frequency?</h3>
The number of waves that travel through a particular point in a given length of time is described by frequency. So, if a wave takes half a second to pass, the frequency is 2 per second.
Given that the energy of the photon is 4.56 x 10⁻¹⁹ J. Therefore, the frequency of the photon can be written as,
Hence, the frequency of a photon with an energy of 4.56 x 10⁻¹⁹ J is 6.88×10¹⁴ s⁻¹.
Learn more about Frequency:
brainly.com/question/5102661
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-- 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.
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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:
Le calcul du courant se fait avec deux éléments : la tension et la valeur de la résistance. Courant (A) = tension (V) / résistance (Ohm) ce qui donne la formule I = U/R.
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