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zheka24 [161]
3 years ago
8

The United Nations was founded primarily __________.

Physics
2 answers:
Margaret [11]3 years ago
9 0
<span>The United Nations was founded primarily B. to help international diplomacy</span>
Mekhanik [1.2K]3 years ago
8 0

Answer;

B. to help international diplomacy

Explanation;

The United Nations is an international organization founded in 1945 after the Second World War by 51 countries committed to maintaining international peace and security, developing friendly relations among nations and promoting social progress, better living standards and human rights.

The UN Charter sets out four main purposes: Maintaining worldwide peace and security. Developing relations among nations. Fostering cooperation between nations in order to solve economic, social, cultural, or humanitarian international problems.

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A. In the building there are 133 steps from the ground floor to the sixth floor.
faust18 [17]

(a) The work done by the person is 18.83465 kJ.

(b) The average power performed by the person during the walk is 51.4 W.

(c) The amount of food calories burnt is 4.5 Cal.

<h3>Work done by the person</h3>

The work done by the person is calculated as follows;

W = Fd

W = mgh

W = (89.2 x 9.8) x (0.162 x 133)

W = 18,834.65 J

W = 18.83465 kJ

<h3>Average power of the person</h3>

P = Fv

where;

  • v is velocity

v = (d)/t

v = (133 x 0.162)/(6 x 60   +  6)

v = (133 x 0.162)/(366)

v = 0.0588 m/s

P = (89.2 x 9.8) x 0.0588

P = 51.4 W

<h3>Amount of food calories burnt</h3>

4.1868 kJ = 1 Cal

18.83465 kJ = ?

= 4.5 Cal

Learn more about work done here: brainly.com/question/8119756

#SPJ1

7 0
2 years ago
Ways that industry and agriculture use physical properties to separate substances
Nata [24]

They use size and weight

5 0
3 years ago
Four charges 7 × 10−9 C at (0 m, 0 m), −9 × 10−9 C at (3 m, 3 m), 7 × 10−9 C at (1 m, 3 m), and −8 × 10−9 C at (−3 m, 2 m), are
Ivanshal [37]

Answer:

Magnitude of the resulting force on the 7 nC charge at the origin:

Fn₁= 23.95*10⁻⁹ N

Explanation:

Look at the attached graphic:

Charges of positive signs exert repulsive forces on q₁ + and charges of negative signs exert attractive forces on q₁ +.

q₁ experiences three forces (F₂₁,F₃₁,F₄₁) and we calculate them with Coulomb's law:

F = (k*q₁*q)/(d)²

d_{12} = \sqrt{3^{2}+3^{2}  }  = \sqrt{18} m : distance from q₁ to q₂

(d₁₂)² = 18 m²

d_{13} =\sqrt{1^{2}+3^{2}  } = \sqrt{10} m  : distance from q₁ to q₃

(d₁₃)² = 10 m²

d_{14} =\sqrt{3^{2}+2^{2}  } = \sqrt{13} m  : distance from q₁ to q₄

(d₁₄)² = 13 m²

K=  8.98755 × 10⁹ N *m²/C²

q₁=  7*10⁻⁹C

k*q₁=8.98755*10⁹ *7*10⁻⁹= 62.9

F₂₁= (62.9)*(9* 10⁻⁹) /(18) = 31.45*10⁻⁹ C

F₃₁= (62.9)*(7* 10⁻⁹) /(10) = 44*10⁻⁹ C

F₄₁= (62.9)*(8* 10⁻⁹) /(13) = 38.7*10⁻⁹ C

x-y components of the net force on q₁ (Fn₁):

α= tan⁻¹(3/3)= 45°  ,  β= tan⁻¹(3/1)= 71.56° , θ= tan⁻¹(2/3)= 33.69°

Fn₁x = F₂₁x+ F₃₁x+F₄₁x

F₂₁x =+ F₂₁*cosα =+ (31.45*10⁻⁹)* (cos 45°) = +22.24 *10⁻⁹ N

F₃₁x= -F₃₁*cosβ = - ( 44*10⁻⁹)* (cos 71.56°) = -13.91 *10⁻⁹ N

F₄₁x= -F₄₁*cosθ = -(38.7*10⁻⁹)* (cos 33.69°) = -32.2*10⁻⁹ N

Fn₁x = (+22.24 - 13.91 - 32.2)*10⁻⁹ N

Fn₁x = -23.87 *10⁻⁹ N

Fn₁y = F₂₁y+ F₃₁y+F₄₁y

F₂₁x =+ F₂₁*sinα =+ (31.45*10⁻⁹)* (sin 45°) = +22.24 *10⁻⁹ N

F₃₁x= -F₃₁*sinβ = - ( 44*10⁻⁹)* (sin 71.56°) = -41.74 *10⁻⁹ N

F₄₁x= +F₄₁*sinθ = +(38.7*10⁻⁹)* (sin 33.69°) =+21.47*10⁻⁹ N

Fn₁y = (22.24 -41.74+21.47)*10⁻⁹ N  

Fn₁y = 1.97*10⁻⁹ N

Magnitude of the resulting force on the 7 nC charge at the origin (q₁):

F_{n1} =\sqrt{(Fn_{1x} )^{2}+(Fn_{1y} )^{2} }

F_{n1} =\sqrt{(23.87 )^{2}+(1.97 )^{2} }

Fn₁= 23.95*10⁻⁹ N

8 0
3 years ago
A 9.00 g bullet is fired horizontally into a 1.20 kg wooden block resting on a horizontal surface. The coefficient of kinetic fr
Zinaida [17]

Answer:

The initial speed of bullet is "164 m/s".

Explanation:

The given values are:

mass of bullet,

m'=9.00 \ g

or,

    =0.009 \ kg

mass of wooden block,

m=1.20 \ kg

speed,

s=0.390 \ m

Coefficient of kinetic friction,

\mu=0.20

As we know,

The Kinematic equation is:

⇒  v^2=u^2+2as

then,

Initial velocity will be:

⇒  u=v^2-2as

        =v^2-2 \mu gs

On substituting the given values, we get

⇒  u=\sqrt{0-2\times 0.20\times 9.8\times 0.390}

       =\sqrt{-1.5288}

       =1.23 \ m/s

As we know,

The conservation of momentum is:

⇒  mu=m'u'

or,

⇒ Initial speed, u'=\frac{mu}{m'}

On substituting the values, we get

⇒                            =\frac{1.20\times 1.23}{0.009}

⇒                            =\frac{1.476}{0.009}

⇒                            =164 \ m/s                              

3 0
3 years ago
A large helium filled balloon is used as the center piece for a graduation party. The balloon alone has a mass of 222 kg and it
inn [45]

Answer:

The buoyant force is 3778.8 N in upward.

Explanation:

Given that,

Mass of balloon = 222 Kg

Volume = 328 m³

Density of air = 1.20 kg/m³

Density of helium = 0.179 kg/m³

We need to calculate the buoyant force acting

Using formula of buoyant force

F_{b}=\rho_{air}\times V_{b}\times g

Where, \rho_{air} = density of air

V = Volume of balloon

g = acceleration due to gravity

Put the value into the formula

F_{b}=1.20\times321\times9.81

F_{b}=3778.8\ N

This buoyant force is in upward direction.

Hence, The buoyant force is 3778.8 N in upward.

4 0
3 years ago
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