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2 years ago

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A linebacker strikes a fullback with full force in a football game.The reaction force to the force of the linebacker striking th

e fullback is ____.
a. the fullback falls backwards to the ground
b. the force of the fullback colliding with the ground
c. the fullback fumbles
d. the fullback applies a force to the linebacker
e. the linebacker applies a force to himself.
f. the fullback applies a force to himself
g. the sound of the pads colliding
h. there is a 3-yard loss and the linebacker is applauded

1 answer:

Scrat [10]2 years ago

6 0

A. the back falls backwards to the ground

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A sample of gas has an initial volume of 4.5 L at a pressure of 754 mmHg . Part A If the volume of the gas is increased to 8.5 L

Firdavs [7]

Answer:

The pressure will be of 399.17 mmHg.

Explanation:

p1= 754 mmHg

V1= 4.5 L

p2= ?

V2= 8.5 L

p1*V1 = p2*V2

p2= (p1*V1)/V2

p2= 399.17 mmHg

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3 years ago

All of the noble gases, Group 18, have eight valence electrons in its outer shell (excluding helium which only has two). Which o

ikadub [295]

Answer:

Zero or +2

Explanation:

The noble gases already have a avplete outermost shell. They are the least reactive elements of earth?

Their normal oxidation number is zero but some have been shown to be reactive.

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3 years ago

Read 2 more answers

In an electronic transition an atom can not emit what?

iren [92.7K]

█ Question <span>█

</span><span>In an electronic transition, an atom cannot emit what?

</span>█ Answer █

When an electronic transition is occurring, an atom cannot emit ultra-violet light.

<span>Hope that helps! ★ <span>If you have further questions about this question or need more help, feel free to comment below or leave me a PM. -UnicornFudge aka Nadia</span></span>

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3 years ago

Two electrons are initially at rest separated by a distance of 2nm. At time t=0, they start to move apart due to Coulombic repul

Gnom [1K]

Answer:

$t=2.5\times 10^{-14}\ s$

Explanation:

We know that charge on electron

$q=1.6\times 10^{-19}\ C$

r= 2 nm

We know that force between two charge given

$F=K\dfrac{Q_1Q_2}{r^2}$

Now by putting the value

$F=9\times10^9\dfrac{1.6\times 10^{-19}\times 1.6\times 10^{-19}}{(2\times 10^{-9})^2}$

$F=5.67\times 10^{-11}\ N$

We know that mass of electron

The mass of electron

$m=9.1\times 10^{-31}\ kg$

F= m a

a= Acceleration of electron

a= F/m

$a=\dfrac{5.67\times 10^{-11}}{9.1\times 10^{-31}}\ m/s^2$

$a=6.2\times 10^{19} m/s^2$

$S=ut+\dfrac{1}{2}at^2$

initial velocity given that zero ,u=0

$20\times 10^{-9}=\dfrac{1}{2}\times 6.2\times 10^{19} t^2$

$t=\sqrt {\dfrac{40\times 10^{-9}}{6.2\times 10^{19}}}$

$t=2.5\times 10^{-14}\ s$

3 0

3 years ago

1. When you have different masses for each sphere, how does the force that the larger mass sphere exerts on the smaller mass sph

aleksandrvk [35]

1) The forces are equal (Newton's third law of motion)

2) The force between the spheres will quadruple

3) The force of gravity exerted by the notebook on you is negligible

Explanation:

1)

In this part of the problem, we want to compare the gravitational force exerted by the larger mass sphere on the smaller mass sphere to the force exerted by the smaller mass sphere to the larger mass sphere.

We can do this by using Newton's third law of motion, which states that:

<em>"When an object A exerts a force (called </em><em>action</em><em>) on an object B, then object B exerts an equal and opposite force (called </em><em>reaction</em><em>) on object A"</em>

In this problem, we can identify the larger mass sphere as object A and the smaller mass sphere as object B. This law tells us that the two forces are equal in magnitude and opposite in direction: therefore, the gravitational force exerted by the larger mass sphere on the smaller mass sphere is equal to the force exerted by the smaller mass sphere to the larger mass sphere.

2)

The magnitude of the gravitational force between the two spheres is given by

$F=G\frac{m_1 m_2}{r^2}$

where

G is the gravitational constant

$m_1, m_2$ are the masses of the two spheres

r is the separation between the two spheres

In this problem, we are asked to find what happens when the distance between the spheres is halved, therefore when the new distance is

$r'=\frac{r}{2}$

Substituting into the equation, we find

$F'=G\frac{m_1 m_2}{r'^2}=G\frac{m_1 m_2}{(r/2)^2}=4(\frac{Gm_1 m_2}{r^2})=4F$

So, the force between the two spheres will quadruple.

3)

We can give an estimate for the gravitational force exerted by your notebook on you.

As we said, the magnitude of the gravitational force is

$F=G\frac{m_1 m_2}{r^2}$

Where:

$G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2}$ is the gravitational constant

Let's estimate the following:

$m_1 = 60 kg$ is your mass

$m_2 = 2 kg$ is the mass of the notebook

$r=1 m$ , assuming the notebook is at 1 metre from you

Substituting,

$F=(6.67\cdot 10^{-11})\frac{(60)(2)}{1^2}=8.0\cdot 10^{-9} N$

We see that this force has an extremely small value: therefore, it is almost negligible in daily life, where other much stronger forces act on you.

Learn more about gravity:

brainly.com/question/1724648

brainly.com/question/12785992

#LearnwithBrainly

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3 years ago