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Ilia_Sergeevich [38]

3 years ago

14

If one of the masses of the Atwood's machine below is 3.2 kg, what should be the other mass so that the displacement of either m

ass during the first second following release is 0.23 m? Assume a massless, frictionless pulley and a massless string

1 answer:

nignag [31]3 years ago

5 0

Answer:

m₂ = 2.91 kg

Explanation:

Let's analyze the exercise, ask us for the mass of the body we could find with Newton's second law and we need the acceleration that we can calculate with kinematics

Let's start looking for acceleration with kinematics

y = vo t + ½ a t²

They indicate for the first second (t = 1 s) it descends y = 0.23 m

y = ½ a t²

a = 2 y / t²

a = 2 0.23 / 1²

a = 0.46 m/s²

Let's look for the mass of the Atwood machine with Newton's second law, let's write the equations for each mass

W₁- T = m₁ a

T - W₂ = m₂ a

Let's add the two equations

W₁ -W₂ = (m₁ + m₂) a

m₁ g - m₂ g = m₁ a + m₂ a

m₂ (a + g) = m₁ (g-a)

m₂ = m₁ (g-a) / (g + a)

m₂ = 3.2 (9.8 -0.46) / (9.8 + 0.46)

m₂ = 3.2 9.34 / 10.26

m₂ = 2.91 kg

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If Mrs. Reichelt throws a chromebook, because it won't login correctly, with a force of 8N, and the chromebook accelerates at 5m

suter [353]

Answer:

1.6 kg

Step-by-step Solution:

Since Force = mass × acceleration we have:

F = 8N

a= 5 m/s^2

m = ?

By plugging the values above into F=ma we obtain:

$F=ma\\\\8=m(5)\\\\\frac{8}{5}=\frac{m(5)}{5}\\\\m=\frac{8}{5}=1.6$

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7 0

3 years ago

Someone help please by providing work and answers please :)

cupoosta [38]

There are two ways to solve this. The longer way is to use those equations to calculate numbers for total distance.

The easier way is to find the area under the graph. That's right, AREA UNDER VELOCITY-TIME graph is the TOTAL DISTANCE travelled!

it's a shortcut.

Let's split up the area into a triangle and rectangle:

Triangle = 0.5(4-0)(10-0) = 20 m
Rectangle = (6-4)(10-0) = 20 m

Total distance = 40 m!

6 0

3 years ago

A light-year is _______________

kozerog [31]

The correct answer would be A "<span>A light-year is the distance light travels in a year.

This is considered a unit of distance connected to the distance that light can travel in one year. It is proved that light travels at 300,000 km per second so, in 1 year, it might travel 10 trillion km.

</span>

3 0

3 years ago

Read 2 more answers

A helicopter lifts a 72 kg astronaut 15 m vertically from the ocean by means of a cable. The acceleration of the astronaut is g/

NemiM [27]

Answer:

a) $F_H=776.952\ N$

b) $F_g=706.32\ N$

c) $v=5.4249\ m.s^{-1}$

d) $KE=1059.48\ J$

Explanation:

Given:

mass of the astronaut, $m=72\ kg$

vertical displacement of the astronaut, $h=15\ m$

acceleration of the astronaut while the lift, $a=\frac{g}{10} =0.981\ m.s^{-2}$

a)

<u>Now the force of lift by the helicopter:</u>

Here the lift force is the resultant of the force of gravity being overcome by the force of helicopter.

$F_H-F_g=m.a$

where:

$F_H=$ force by the helicopter

$F_g=$ force of gravity

$F_H=72\times 0.981+72\times9.81$

$F_H=776.952\ N$

b)

The gravitational force on the astronaut:

$F_g=m.g$

$F_g=72\times 9.81$

$F_g=706.32\ N$

d)

Since the astronaut has been picked from an ocean we assume her initial velocity to be zero, $u=0\ m.s^{-1}$

using equation of motion:

$v^2=u^2+2a.h$

$v^2=0^2+2\times 0.981\times 15$

$v=5.4249\ m.s^{-1}$

c)

Hence the kinetic energy:

$KE=\frac{1}{2} m.v^2$

$KE=0.5\times 72\times 5.4249^2$

$KE=1059.48\ J$

8 0

3 years ago

Read 2 more answers

A 222 kg bumper car moving right at 3.10 m/s collides with a 165 kg bumper car moving 1.88 m/s left Find the total momentum of t

Svetlanka [38]

Answer:

Total momentum of the system is 378 kg-m/s

Explanation:

It is given that,

Mass of first bumper car, m₁ = 222 kg

Velocity of first bumper car, v₁ = 3.10 m/s (in right)

Mass of other bumper car, m₂ = 165 kg

Velocity of second bumper car, v₂ = -1.88 m/s (in left)

Momentum of the system is given by the product of its mass and velocity. So, the total momentum of this system is given by :

$p=m_1v_1+m_2v_2$

$p=222\ kg\times 3.10\ m/s+165\ kg\times (-1.88\ m/s)$

p = 378 kg-m/s

Hence, the total momentum of the system is 378 kg-m/s

5 0

3 years ago