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Mila [183]
2 years ago
13

If the cannonball weighs 5kg and the acceleration at 400 m/s², what is the acceleration of the 500kg cannon?

Physics
1 answer:
Elina [12.6K]2 years ago
3 0
It is p d day we’ll be together if it weighs 5kg you divide by the accelerated pace 400 dividdd by 400 is 45 and 500 divided by 500 is 75 answer is 75.
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What is the correct path of electrons?
MrRa [10]

Electrons are transferred sequentially between the two photosystems, with photosystem I acting to generate NADPH and photosystem II acting to generate ATP. The pathway of electron flow starts at photosystem II, which is homologous to the photosynthetic reaction center of R. viridis already described.

3 0
2 years ago
A pulley system lifts a 1345 n weight a distance of 0.975m. Paul pulls the rope a distance of 3.90m, exerting a force of 375N. A
Rashid [163]

A. IMA: 4

The Ideal Mechanical Advantage (IMA) is given by:

IMA = \frac{d_i}{d_o}

where

d_i is the input distance

d_o is the output distance

For the pulley system in this problem, d_i = 3.90 m and d_o = 0.975 m, so the IMA is

IMA=\frac{3.90 m}{0.975 m}=4


B. MA: 3.59

The actual mechanical advantage (AMA), or simply the Mechanical Advantage (MA), is given by

MA=\frac{F_o}{F_i}

where F_o is the output force and F_i is the input force. For the pulley system in this problem, F_i = 375 N and F_o = 1345 N, so the MA is

MA=\frac{1345 N}{375 N}=3.59


C. Efficiency: 89.8 %

The efficiency of a machine is equal to the ratio between the MA and the AMA:

\eta = \frac{MA}{AMA} \cdot 100

Therefore, in this case,

\eta=\frac{3.59}{4}\cdot 100=0.898=89.8 \%

3 0
3 years ago
Which was least likely to have been a component of Earth’s atmosphere before life began?
Otrada [13]
The answer to this question is a) sulfur
4 0
3 years ago
Read 2 more answers
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nika2105 [10]

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6 0
2 years ago
1. Calcula la fuerza de atracción electrostática entre dos cuerpos de cargas q1 = -18 C y q2 = +5 mC, separados entre sí por una
romanna [79]

Answer:

A) F=-20.16×10⁹N

B) if the distance doubles, force is 4 times smaller.

Explanation:

q1=-28C

q2=5mC=0.005C

d=25cm=0.25m

Electrostatic force between charges: F=k×q1×q2/d², where k is a coefficient that has the value k=9 × 10⁹ N⋅m²⋅C^(-2) for air.

Thus:

F=9×10⁹×(-28)×0.005/0.25²

F=-20.16×10⁹N

The minus sign indicates attraction.

If distance doubles, d1=2×d, then we have 4d² at the denominator and the force is 4 times smaller.

6 0
3 years ago
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