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2 years ago

20 points

Physics

2 answers:

Karolina [17]2 years ago

8 0

Answer:

i beleive its A and B will both follow the rule

This is due to property of charges.

hope this helped

Explanation:

krok68 [10]2 years ago

3 0

Answer:

(a) and (b)

Both will follow the rule

This is due to property of charges.

As in case of gravitational potential energy as it increases as height increases.

In case of electrical potential energy it will increase as charges comes close.

Explanation:

have a good day:))

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Boxes A and B are in contact on a horizontal, frictionless surface. Box A has mass 21.0 kg and box B has mass 8.0 kg. A horizont

vfiekz [6]

Answer:

2.75 N

Explanation:

Given that,

Box A has a mass 21.0 kg and box B has a mass 8.0 kg.

A horizontal force of 100N is exerted on box A.

Let a be the acceleration of the system. Using second law of motion,

$F=(m_A+m_B)a\\\\a=\dfrac{F}{(m_A+m_B)}\\\\a=\dfrac{10}{(21+8)}\\\\a=0.344\ m/s^2$

Now applying Newton's second law to box B. So,

$F_A=m_Ba\\\\=8\times 0.344\\\\=2.75\ N$

So, 2.75 N is the force that box A exerts on box B.

8 0

3 years ago

One of the games at a carnival involves trying to ring a bell with a ball by hitting a lever that propels the ball into the air.

RoseWind [281]

Answer:

Ball will not hit bell.

Explanation:

The height of the ball is modeled by the equation h(t)= -16t²+39t

We need to find if the bell is 25 ft above the ground, will it be hit by the ball.

So we need to find maximum height of the ball.

For maximum height we have

$\frac{dh}{dt}=0\\\\\frac{d}{dt}\left ( -16t^2+39t\right )=0\\\\-32t+39=0\\\\t=1.22s$

Substituting in h(t) = -16t²+39t

Maximum height reached = h(1.22) = -16 x 1.22²+39 x 1.22 = 23.77 ft

So maximum height reached is less than height of bell.

Ball will not hit bell.

8 0

3 years ago

PLEASE HELP ME WITH THIS ONE QUESTION

shtirl [24]

Answer:

$\lambda=1.39\times 10^{-4}\ s^{-1}$

Explanation:

Given that,

The half-life of Barium-139 is $4.96\times 10^3$

A sample contains $3.21\times 10^{17}$ nuclei.

We need to find the decay constant for this decay. The formula for half life is given by :

$T_{1/2}=\dfrac{0.693}{\lambda}\\\\\lambda=\dfrac{0.693}{T_{1/2}}$

Put all the values,

$\lambda=\dfrac{0.693}{4.96\times 10^3}\\\\=1.39\times 10^{-4}\ s^{-1}$

So, the decay constant is $1.39\times 10^{-4}\ s^{-1}$ .

4 0

3 years ago

Consider atmospheric air at 20°C and a velocity of 30 m/s flowing over both surfaces of a 1‐m‐long flat plate that is maintained

andrew11 [14]

Answer:

q = 8472.2 W/m

Explanation:

The solving or solution is given in the attach documents.

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3 years ago

Which of the following statements about the cosmic background radiation is not true?

belka [17]

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<h2>It is the leftover radiation from the big bang. </h2>

Its frequency belongs to the microwaves range and is one of the main demonstrations of the of the Big Bang theory model.

It is important to note the Big Bang occurred 400,000 years before the events related to the microwave cosmic radiation, this means the Big Bang occurred first.

6 0

3 years ago