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snow_tiger [21]

2 years ago

12

Two men are standing on a frictionless ice surface holding opposite ends of a rope. One man (mass = 80 kg) pulls on the rope wit

h a force of 250 N. The other has a mass of 60 kg. What is the acceleration of each man?

1 answer:

ankoles [38]2 years ago

7 0

Answer:

The acceleration of man 1 and 2 is $3.125\ m/s^2$ and $4.167\ m/s^2$ .

Explanation:

Mass of man 1, m₁ = 80 kg

Mass of man 2, m₂ = 60 kg

One man pulls on the rope with a force of 250 N.

Let a₁ is acceleration of man 1,

F = m₁a₁

$a_1=\dfrac{F}{m_1}\\\\a_1=\dfrac{250}{80}\\\\a_1=3.125\ m/s^2$

Let a₂ is acceleration of man 1,

F = m₂a₂

$a_2=\dfrac{F}{m_2}\\\\a_2=\dfrac{250}{60}\\\\a_2=4.167\ m/s^2$

So, the acceleration of man 1 and 2 is $3.125\ m/s^2$ and $4.167\ m/s^2$ .

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Answer:

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A person throws a baseball from height of 7 feet with an initial vertical velocity of 50 feet per second. Use the vertical motio

Gnoma [55]

Answer:

3.25 seconds

Explanation:

It is given that,

A person throws a baseball from height of 7 feet with an initial vertical velocity of 50 feet per second. The equation for his motion is as follows :

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Where

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For the given condition, the equation becomes:

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Hence, for 3.25 seconds the baseball is in the air before it hits the ground.

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What is 3,278,000 as scientific notation?

jonny [76]

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2 years ago

lozanna [386]

Answer:

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7. A mother pushes her 9.5 kg baby in her 5kg baby carriage over the grass with a force of 110N @ an angle

jasenka [17]

Weight of the carriage $=(m+M)g =142.1\ N$

Normal force $=Fsin(\theta) + W = 197.1\ N$

Frictional force $=\mu N=27.59\ N$

Acceleration $=4.66\ m\ s^{-2}$

Explanation:

We have to look into the FBD of the carriage.

Horizontal forces and Vertical forces separately.

To calculate Weight we know that both the mass of the baby and the carriage will be added.

So Weight(W) $=(m+M)\times g =(9.5+5)\ kg \times 9.8 =142.1\ Newton\ (N)$

To calculate normal force we have to look upon the vertical component of forces, as Normal force is acting vertically.We have weight which is a downward force along with $F_x$ , force of $110\ N$ acting vertically downward.Both are downward and Normal is upward so Normal force $=Summation\ of\ both\ forces$

Normal force (N) $= Fsin(\theta)+W=110sin(30) + 142.1 =197.1\ N$

Frictional force (f) $=\mu N=0.14\times 197.1 =27.59\ N$

To calculate acceleration we will use Newtons second law.

That is Force is product of mass and acceleration.

We can see in the diagram that $F_y=Horizontal$ and $F_x=Vertical$ component of forces.

So Fnet = Fy(Horizontal) - f(friction) $= m\times a$

Acceleration (a) = $\frac{Fcos(\theta)-\mu N}{mass(m)} =\frac{(95.26-27.59)}{14.5}= 4.66\ m\ s{^2 }$

So we have the weight of the carriage, normal force,frictional force and acceleration.

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2 years ago