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Oksana_A [137]
3 years ago
6

A heavy boy and a lightweight girl are balanced on a mass-less seesaw. If they both move forward so that they are one-half their

original distance from the pivot point, what will happen to the seesaw? Assume that both people are small enough compared to the length of the seesaw to be thought of as point masses.A. It is impossible to say without knowing the masses.
B. It is impossible to say without knowing the distances.
C. The side the boy is sitting on will tilt downward.
D. Nothing will happen; the seesaw will still be balanced.
E. The side the girl is sitting on will tilt downward.
Physics
2 answers:
Rom4ik [11]3 years ago
4 0

Answer:

b) Nothing will happen,  the sea saw will still be balanced.

Explanation:

b) Nothing will happen,  the sea saw will still be balanced.

Reason:-

When two kids are balanced, the sum of torques on the seesaw will be zero.

if each kid, reduces their distances by half, then the torque of each kid will be half and the sum of torque of each on the seesaw will be zero.

Therefore the seesaw is balanced

Katyanochek1 [597]3 years ago
4 0

Answer:D. Nothing will happen; the seesaw will still be balanced

Explanation: According to the principles of moment which states that for a body to be at equilibrium, the sum of clockwise moments must be equal to the sum of anticlockwise moments.

if the mass-less seesaw was initially balanced at a particular distance, it will still be balanced if the boy and the girl move forward to cover half their initial distances from the pivot.

D perfectly depicts the outcome.

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nataly862011 [7]

Answer:

The results have not been through the rigorous process of peer review

Explanation:

When a scientist conducts a study and obtains results, those results ought to be submitted to a reputable journal where the results would go through the rigorous protocol of peer review.

During this process, the reliability of the data presented is ascertained before the results are published for other scientists to see.

If the results are hurriedly published on the internet, many researchers who come in contact with the work may be fed with inaccurate information.

7 0
2 years ago
Which statement describes the effects of forces on an object
lukranit [14]

Answer:

Sorry this isn’t going to be any help. You don’t have any statement that I’m able to see.

Explanation:

8 0
3 years ago
The Huka Falls on the Waikato River is one of New Zealand's most visited natural tourist attractions. On average, the river has
OverLord2011 [107]

Answer:

(a) V = 0.75 m/s

(b) V = 0.125 m/s

Explanation:

The speed of the flow of the river can be given by following formula:

V = Q/A

V = Q/w d

where,

V = Speed of Flow of River

Q = Volume Flow Rate of River

w = width of river

d = depth of river

A = Area of Cross-Section of River = w d

(a)

Here,

Q = (300,000 L/s)(0.001 m³/1 L) = 300 m³/s

w = 20 m

d = 20 m

Therefore,

V = (300 m³/s)/(20 m)(20 m)

<u>V = 0.75 m/s</u>

<u></u>

(b)

Here,

Q = (300,000 L/s)(0.001 m³/1 L) = 300 m³/s

w = 60 m

d = 40 m

Therefore,

V = (300 m³/s)/(60 m)(40 m)

<u>V = 0.125 m/s</u>

8 0
3 years ago
1) An ice skater with a moment of inertia of 2.2 kg m^2 rotates at a frequency of 0.8 rotations per second. The ice skater tucks
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Answer:

Explanation:

2.3 kg·m/s²

4 0
3 years ago
At locations A and B, the electric potential has the values VA = 1.83 V and VB = 5.17 V, respectively. A proton released from re
densk [106]

Answer:

a. It starts at point B.

vp = 2.53*10⁴ m/s

a. it starts at point A.

ve= 1.08*10⁶ m/s

Explanation:

a)  As the proton is a positive charge, when released from rest, it will be accelerated due to the potential difference, from the higher potential to the lower one, so it is at the point B when released.

Once released, as the total energy must be conserved, the increase in kinetic energy must be equal (in magnitude) to the change in the electric potential energy, as follows:

ΔK + ΔUe = 0 ⇒ ΔK = -ΔUe =- (e*ΔV)

⇒ -( e* (VA-VB) ) = \frac{1}{2}*mp*v^{2}

where e= elementary charge= 1.6*10⁻¹⁹ C,  VA = 1.83 V, VB= 5.17V, and mp= mass of proton = 1.67*10⁻²⁷ kg.

Replacing by these values, and solving for v, we have:

v = \sqrt{\frac{2*1.6e-19C*3.34 V}{1.67e-27kg} } = 2.53e4 m/s

⇒ vp = 2.53*10⁴ m/s

b) If, instead of a proton, the charge realeased from rest, had been an electron, a few things would change:

First, as the electrons carry negative charges, they move from the lower potentials to the higher ones, which means that it would have started at point A.

Second, as its charge is (-e) the change in electric potential energy had been negative also:

ΔUe = -e*ΔV = -e* (VB-VA)

In order to find the speed of the electron when it is just passing point B, we can apply the conservation of energy principle as for the proton, as follows:

-( (-e)* (VB-VA) ) = \frac{1}{2}*me*v^{2}

where e= elementary charge= 1.6*10⁻¹⁹ C,  VA = 1.83 V, VB= 5.17V, and me= mass of electron = 9.1*10⁻³¹ kg.

Replacing by these values, and solving for v, we have:

v = \sqrt{\frac{2*1.6e-19C*3.34 V}{9.1e-31kg} } = 1.08e6 m/s

⇒ ve = 1.08*10⁶ m/s

4 0
3 years ago
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