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Find the acceleration of the system and the tension in the ropes for the system shown. The table mass is 30 kg and the hanging m

marusya05 [52]

The system's tension is 616 N and acceleration is 5.6 $m / s^{2}$

<u>Explanation:</u>

From newton’s second law of motion which state that net force acting on a body is product of mass of a body and acceleration of a body which is given as,

$F_{n e t}=m_{t o t} \times a$

Where,

$F_{n e t}$ is net force acting on body

$m_{\mathrm{tot}}$ is mass of body

a is acceleration of body

Given values

Table mass (m) = 30 kg

Hanging mass (m) = 40 kg

$a=\frac{F_{n e t}}{m_{\mathrm{tot}}}=\frac{m \times g}{m_{\mathrm{tot}}}$

Put the value for m = hanging mass = 40 kg and $g=9.8 \mathrm{m} / \mathrm{s}^{2}$ , we get

$a=\frac{40 \times 9.8}{30+40}=\frac{392}{70}=5.6 \mathrm{m} / \mathrm{s}^{2}$

The tension in the ropes, $T=(m \times g)+(m \times a)$

Here, m as hanging mass

T = tension, N or $k g m / s^{2}$

m = mass, kg

g = gravitational force, $9.8 \mathrm{m} / \mathrm{s}^{2}$

a = acceleration, $m / s^{2}$

$T = (40 \times 9.8)+(40 \times 5.6) = 392+224 = 616 N$

3 0

2 years ago

Calculate the force necessary to accelerate a 10 kg table from<br> O m/s to 4 m/s in 2 seconds.

yanalaym [24]

Answer:

a= v/t

a = 4/2

a = 2 m/s^2

And F = M a

F = 10 × 2

F = 20 N

6 0

3 years ago

Consider the space between a point charge and the surface of a neutral spherical conducting shell. If the charge sits at the cen

Furkat [3]

Answer:

True

Explanation:

If a thin, spherical, conducting shell carries a negative charge, We expect the excess electrons to mutually repel one another, and, thereby, become uniformly distributed over the surface of the shell. The electric field-lines produced outside such a charge distribution point towards the surface of the conductor, and end on the excess electrons. Moreover, the field-lines are normal to the surface of the conductor. This must be the case, otherwise the electric field would have a component parallel to the conducting surface. Since the excess electrons are free to move through the conductor, any parallel component of the field would cause a redistribution of the charges on the shell. This process will only cease when the parallel component has been reduced to zero over the whole surface of the shell

According to Gauss law

∅ = EA =-Q/∈₀

Where ∅ is the electric flux through the gaussian surface and E is the electric field strength

If the gaussian surface encloses no charge, since all of the charge lies on the shell, so it follows from Gauss' law, and symmetry, that the electric field inside the shell is zero. In fact, the electric field inside any closed hollow conductor is zero

8 0

2 years ago

Read the given list of organisms.

Nadya [2.5K]

They are all herbivores

3 0

2 years ago

Read 2 more answers

Three beads are placed on the vertices of an equilateral triangle of side d = 3.4cm. The first bead of mass m1=140gis placed on

Vlad [161]

Answer:

Xcm = 1.95 cm and Ycm = 1.76 cm

Explanation:

The very useful concept of mass center is

R cm = 1/M ∑ $m_{i}$ $r_{i}$

Where ri, mi are the mass positions of the bodies from some reference point by selecting and M is the total mass of the body.

Let's look for the total mass

M = m₁ + m₂ + m₃

M = 140 + 45 + 85

M = 270 g

Let's look for the position of each point

Point 1. top vertex, if the triangle has as side d

R₁ = d / 2 i ^ + d j ^

R₁ = (1.7 cm i ^ + 3.4 j ^) cm

Point 2. left vertex. What is the origin of the system?

R₂ = 0

Point 3. Right vertex

R₃ = d i ^

R₃ = 3.4 i ^ cm

a) The x component of the massage center

Xcm = 1 / M (m₁ x₁ + m₂ x₂ + m₃ x₃)

Xcm = 1 / M (m₁ d / 2 + 0 + m₃ d)

Xcm = d / M (m₁ / 2 + m₃)

b) Let's write the mass center component x

Xcm = 1/270 (1.7 140 + 0 + 3.4 85)

Xcm = 238/270

Xcm = 1.95 cm

c) let's find the component and center of mass

Ycm = 1 / M (m₁ y₁ + m₂ y₂ + m₃ y₃)

Ycm = 1 / M (m₁ d + 0 + 0)

Ycm = m₁ / M d

d) let's calculate

Y cm = 1/270 (140 3.4 + 0 + 0)

Ycm = 1.76 cm

8 0

3 years ago