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3 years ago

How do the weights of the tug of war teams affect the match

2 answers:

8 0

In Newton's Second Law of Motion, it simply states that mass x acceleration= force. So if there's more weight on one side and the same amount of acceleration on both side, then the side with more weight has more force. And with more force it tugs it more than the other side.

Shkiper50 [21]3 years ago

3 0

Because when the players lean towards their side, gravity pulls them backwards, it also makes it harder for the other side to pull someone so heavy. It's like a kid trying to pull their parent. Their parent is so much heavier that if the parent doesn't want to move they won't.

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8_murik_8 [283]

Answer:

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Explanation:

4 0

3 years ago

When the pressure at the bottom of a submerged object is greater than the pressure at the top of submerged object, buoyant force

ELEN [110]

The answer would most like be true!

4 0

3 years ago

Read 2 more answers

Ohm’s Law is represented by the equation I=V/R. Explain how the current would change if the amount of resistance decreased and t

vesna_86 [32]

Answer:

The current will increase with reduction in the resistance.

Explanation:

Electrical resistance reduces the flow of electricity through a conductor just like friction reduces our speed. The higher the resistance the harder it will be for the current to flow and vice versa, hence, higher resistance produces a smaller current if the voltage is held constant. The voltage is the electrical drive.

3 0

3 years ago

Read 2 more answers

Josh tests an unknown liquid substance using litmus paper. When he compared the used litmus paper to a pH scale, the color match

Hunter-Best [27]

Answer:

acid

Explanation:

because ph below 7 is acid

8 0

2 years ago

A nonconducting sphere of diameter 10.0 cm carries charge distributed uniformly inside with charge density of +5.50 µC/m3 . A pr

VLD [36.1K]

Answer:

t = 2.58*10^-6 s

Explanation:

For a nonconducting sphere you have that the value of the electric field, depends of the region:

$rR:\\\\E=k\frac{Q}{r^2}$

k: Coulomb's constant = 8.98*10^9 Nm^2/C^2

R: radius of the sphere = 10.0/2 = 5.0cm=0.005m

In this case you can assume that the proton is in the region for r > R. Furthermore you use the secon Newton law in order to find the acceleration of the proton produced by the force:

$F=m_pa\\\\qE=m_pa\\\\k\frac{qQ}{r^2}=m_pa\\\\a=k\frac{qQ}{m_pr^2}$

Due to the proton is just outside the surface you can use r=R and calculate the acceleration. Also, you take into account the charge density of the sphere in order to compute the total charge:

$Q=\rho V=(5.5*10^{-6}C/m^3)(\frac{4}{3}\pi(0.05m)^3)=2.87*10^{-9}C\\\\a=(8.98*10^9Nm^2/C^2)\frac{(1.6*10^{-19}C)(2.87*10^{-9}C)}{(1.67*10^{-27}kg)(0.05m)^2}=9.87*10^{11}\frac{m}{s^2}$

with this values of a you can use the following formula:

$a=\frac{v-v_o}{t}\\\\t=\frac{v-v_o}{a}=\frac{2550*10^3m/s-0m/s}{9.87*10^{11}m/s^2}=2.58*10^{-6}s$

hence, the time that the proton takes to reach a speed of 2550km is 2.58*10^-6 s

3 0

3 years ago