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2 years ago

A tugboat is pulling a barge into a harbor. The barge is exerting a force of 3000 N against the tugboat.

Physics

2 answers:

laiz [17]2 years ago

5 0

Available options are:

A. –4500 N

B. –2500 N

C. 2500 N

D. 4500 N

Answer:

-4500 N

Explanation:

Given that we have F₁ = 3000N? (force of the barge)

the force of the tugboat is F '

To pull the barge, the force of the tugboat must be greater than or equal to the force of the barge.

Hence, according to Newton's third law which states that for every action or force in nature, there is an opposite reaction.

Therefore considering the available option the correct answer is "- 4500N, " this is because it is greater than the force of the barge and it's in opposition to the force exerted.

sleet_krkn [62]2 years ago

3 0

Answer:

The answer is -4500 N

Explanation:

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quester [9]

The acceleration of the quarterback and the lineman is 5.55m/s² and 4.16m/s² respectively in the same direction.

As, we know, the 120 Kg lineman is moving with a force of 500N.

His net acceleration will be in the same direction as his motion.

It is already known that, If M is the mass of the body and a is the acceleration of the body, then the force F on the body can be calculated by using the formula,

F = Ma.

The weight of the quarterback is 90 Kg. He is being hit by a force of 500N.

So, the acceleration can be calculated using the formula,

500N = 90kg x a

a = 5.55 m/s².

Now, the weight if the lineman is 120kg, the force applied by him is 500N.

So, from the formula, his acceleration A will be,

500N = 120Kg x A

A = 4.16 m/s².

both of them will have acceleration in the same direction,

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9 months ago

Nuclear power plants produce useful energy by controlling the process of?

Mandarinka [93]

Nuclear fusion is thed answer

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3 years ago

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a mass of 1.00 kg of water at temperature T is poured from a height of 0.100 km into a vessel containing water of the same tempe

Mariana [72]

Answer:

1.34352 kg

Explanation:

$m_w$ = Mass of water falling = 1 kg

h = Height of fall = 0.1 km

$\Delta T$ = Change in temperature = 0.1

c = Specific heat of water = 4186 J/kg K

g = Acceleration due to gravity = 9.81 m/s²

$m_v$ = Mass of water in the vessel

Here the potential energy will balance the internal energy

$m_wgh=m_wc\Delta T+m_vc\Delta T\\\Rightarrow m_v=\dfrac{m_wgh-m_wc\Delta T}{c\Delta T}\\\Rightarrow m_v=\dfrac{m_wgh}{c\Delta T}-m_w\\\Rightarrow m_v=\dfrac{1\times 9.81\times 100}{4186\times 0.1}-1\\\Rightarrow m_v=1.34352\ kg$

Mass of the water in the vessel is 1.34352 kg

6 0

3 years ago

the average american receives 2.28 mSv dose equivalent from radon each year. Assuming you receive this dose, and it all comes fr

Dmitry [639]

Answer:

The approximate number of decays this represent is $N= 23*10^{10}$

Explanation:

From the question we are told that

The amount of Radiation received by an average american is $I_a = 2.28 \ mSv$

The source of the radiation is $S = 5.49 MeV \ alpha \ particle$

Generally

$1 \ J/kg = 1000 mSv$

Therefore $2.28 \ mSv = \frac{2.28}{1000} = 2.28 *10^{-3} J/kg$

Also $1eV = 1.602 *10^{-19}J$

Therefore $2.28*10^{-3} \frac{J}{kg} = 2.28*10^{-3} \frac{J}{kg} * \frac{1ev}{1.602*10^{-19} J} = 1.43*10^{16} ev/kg$

An Average american weighs 88.7 kg

The total energy received is mathematically evaluated as

$1 kg ------> 1.423*10^{16}ev \\88.7kg --------> x$

Cross-multiplying and making x the subject

$x = 88.7 * 1.423*10^{16} eV$

$x = 126.2*10^{16}eV$

Therefore the total energy deposited is $x = 126.2*10^{16}eV$

The approximate number of decays this represent is mathematically evaluated as

N = $\frac{x}{S}$

Where n is the approximate number of decay

Substituting values

$N = \frac{126 .2*10^{16}}{5.49*10^6}$

$N= 23*10^{10}$

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3 years ago

What is your acceleration while sitting in your chair. the latitude of corvallis is 44.4˚.?

marta [7]

<span>You can start with the equations you know

a=v^2/r = (2pi*r/T)^2/r = 4pi^2r/T^2

Radius of earth (R) = 6378.1 km
Time in one day (T) = 86400 seconds
Latitude = 44.4 degrees

If you draw a circle and have the radius going out at a 44.4 degree angle above the center you can then find the r.

r=Rcos(44.4)
r=6378.1cos(44.4)
r= 4556.978198 km or 4556978 m

Now you can plug this value into the acceleration equation from above...

a= 1.8*10^8/7.47*10^9
a= .0241 m/s^2 </span>

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2 years ago