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V125BC [204]
3 years ago
15

Two wooden boxes of equal mass but different density are held beneath the surface of a large container of water. Box A has a sma

ller average density than box B. When the boxes are released, they accelerate upward to the surface. Which box has the greater acceleration?
(A) It depends on the cotent of the boxes
(B) Box B
(C) They are the same
(D) We need to know the actual densities of the boxes in order to answer the question
(E) Box A
Physics
1 answer:
sweet [91]3 years ago
4 0

Answer:

As a box A has a smaller average density than box B, then it will have a greater acceleraton towards the surface upon releasing. So the option E) Box A is a correct option.

Explanation:

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Answer:

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Explanation:

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3 0
3 years ago
In the most common isotope of Hydrogen the nucleus is made out of a single proton. When this Hydrogen atom is neutral, a single
FinnZ [79.3K]

Answer:

The ratio of electric force to the gravitational force is 2.27\times 10^{39}

Explanation:

It is given that,

Distance between electron and proton, r=4.53\ A=4.53\times 10^{-10}\ m

Electric force is given by :

F_e=k\dfrac{q_1q_2}{r^2}

Gravitational force is given by :

F_g=G\dfrac{m_1m_2}{r^2}

Where

m_1 is mass of electron, m_1=9.1\times 10^{-31}\ kg

m_2 is mass of proton, m_2=1.67\times 10^{-27}\ kg

q_1 is charge on electron, q_1=-1.6\times 10^{-19}\ kg

q_2 is charge on proton, q_2=1.6\times 10^{-19}\ kg

\dfrac{F_e}{F_g}=\dfrac{kq_1q_2}{Gm_1m_2}

\dfrac{F_e}{F_g}=\dfrac{9\times 10^9\times (1.6\times 10^{-19})^2}{6.67\times 10^{-11}\times 9.1\times 10^{-31}\times 1.67\times 10^{-27}}

\dfrac{F_e}{F_g}=2.27\times 10^{39}

So, the ratio of electric force to the gravitational force is 2.27\times 10^{39}. Hence, this is the required solution.

3 0
3 years ago
Someone please help really confused
Taya2010 [7]

Answer:

answer c: a mass and volume

Explanation:

no matter what elements are chosen, they all have a mass and volume measurement I believe

5 0
2 years ago
A projectile fired up into the air at an angle has a range of 235 m and a flight time of 47 s.
madam [21]
<h2>Answer:5ms^{-1},133.6m,51.18ms^{-1}</h2>

Explanation:

Let v_{x},v_{y} be the horizontal and vertical components of velocity.

Question a:

Horizontal component of velocity is the ratio of range and time of flight.

So,horizontal component of velocity is \frac{235}{47}=5ms^{-1}

So,v_{x}=5ms^{-1}

Question b:

Time of flight=\frac{2v_{y}}{g}

So,v_{y}=\frac{47\times 9.8}{2}=51.18ms^{-1}

Maximum height is given by \frac{v_{y}^{2}}{2g}

So,maximum height is \frac{51.18^{2}}{2\times 9.8}=133.6m

Question c:

The vertical velocity is already calculated in Question b.

v_{y}=51.18ms^{-1}

7 0
3 years ago
PLEASE IM TIMED! GIVE BRAINLIEST OF U ANSWER THESE 2 QUESTION!!
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Answer:

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Explanation:

hope this helps!

7 0
3 years ago
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