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timofeeve [1]
2 years ago
11

Which one is itttttttttttt

Physics
1 answer:
Vlada [557]2 years ago
6 0

Answer:

Red

Explanation:

You might be interested in
crowbar of 5 metre is used to lift an object of 800 metre if the effort arm is 200cm calculate the force applied​
Klio2033 [76]

Answer:

 F = 5226.6 N

Explanation:

To solve a lever, the rotational equilibrium relation must be used.

We place the reference system on the fulcrum (pivot point) and assume that the positive direction is counterclockwise

         F d₁ = W d₂

where F is the applied force, W is the weight to be lifted, d₁ and d₂ are the distances from the fulcrum.

In this case the length of the lever is L = 5m, t the distance desired by the fulcrum from the weight to be lifted is

d₂ = 200 cm = 2 m

therefore the distance to the applied force is

          d₁ = L -d₂

         d₁ = 5 -2

         d₁= 3m

we clear from the equation

          F = W d₂ / d₁

          W = m g

          F = m g d₂ / d₁

we calculate

      F = 800 9.8 2/3

      F = 5226.6 N

4 0
2 years ago
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
An object at rest on a flat, horizontal surface explodes into two fragments, one seven times as massive as the other. The heavie
Veronika [31]

Answer:

Explanation:

Given that,

One fragment is 7 times heavier than the other

Let one fragment mass be M

Let this has a velocity v

And the other 7M

And this a velocity V

Initially the fragment is at rest u = 0

Applying conservation of momentum

Momentum is given as p=mv

Initial momentum = final momentum

Po = Pf

(M+7M) × 0 = 7M •V − Mv

0 = 7M•V - Mv

Divide both sides by M

0 = 7V -v

v = 7V

Since friction decelerates the masses to zero speed, we can calculate the NET work on the individual blocks and relate this quantity to the change in kinetic energy of each block

The workdone by the 7M mass is

Distance moved by 7M mass is 6.8m, Then, d =6.8m

W = fr × d

Where fr = µkN

When N=W =mg, where m=7M

N= 7Mg

fr = −µk × 7mg

Then, W(7m) = −7µk•Mg×d

W(7m) = −7µk•Mg×6.8

W(7m) = −47.6 µk•Mg

Then, same procedure,

Let distance move by the small mass be m

Work done by M mass

W(m) = −µk•Mg×d'

Since it is a wordone by friction, that is why we have a negative sign.

Using conservation of energy

Work done by 7M mass is equal to work done by M mass

W(7m) = W(m)

−47.6 µk•Mg = −µk•Mg×d

Then, M, g and µk cancels out

We are left with

-46.7 = -d

Then, d = 46.7m

7 0
3 years ago
Read 2 more answers
If the distance between the Earth and Moon were half what it is now, by what factor would the force of gravity between them be c
hichkok12 [17]

Answer:

4

Explanation:

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

m_1 = Mass of Earth

m_2 = Mass of Moon

r = Distance between Earth and Moon

Old gravitational force

F_o=\dfrac{Gm_1m_2}{r^2}

New gravitational force

F_n=\dfrac{Gm_1m_2}{(\dfrac{1}{2}r)^2}

Dividing the equations

\dfrac{F_n}{F_o}=\dfrac{\dfrac{Gm_1m_2}{(\dfrac{1}{2}r)^2}}{\dfrac{Gm_1m_2}{r^2}}\\\Rightarrow \dfrac{F_n}{F_o}=\dfrac{\dfrac{Gm_1m_2}{\dfrac{1}{4}r^2}}{\dfrac{Gm_1m_2}{r^2}}\\\Rightarrow \dfrac{F_n}{F_o}=4

The ratio is \dfrac{F_n}{F_o}=4

The new force would be 4 times the old force

7 0
2 years ago
A radio wave has a frequency of 5.5 × 104 hertz and travels at a speed of 3.0 × 108 meters/second. What is its wavelength
Ne4ueva [31]
Use v=fλ
3x10^8=5.5x 10^4 λ
λ=5.45x10^3m
4 0
3 years ago
Read 2 more answers
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