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3 years ago

Nasaâ€™s goal is to help industry reduce emissions from aircraft by how much by 2050 compared to 2005?

Physics

1 answer:

lys-0071 [83]3 years ago

8 0

Answer:

50%

Explanation:

Commercial aviation is responsible for 2% of global carbon emission. In the year 2009, the members of IATA (International Air Transport Association) had drawn a pledge to:

1. Halve the carbon emission due to aircraft by 2050, relative to emission of 2005

2. To make growth of industry carbon neutral by 2020

3. To cut CO₂ emission by 1.5% per year till 2020

To achieve these targets, a four pillar action plan was created.

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How many excess electrons must be present on each sphere if the magnitude of the force of repulsion between them is 4.57×10−21 n

hichkok12 [17]

Answer:

891 excess electrons must be present on each sphere

Explanation:

One Charge = q1 = q

Force = F = 4.57*10^-21 N

Other charge = q2 =q

Distance = r = 20 cm = 0.2 m

permittivity of free space = eo =8.854×10−12 C^2/ (N.m^2)

Using Coulomb's law,

F=[1/4pieo]q1q2/r^2

F = [1/4pieo]q^2 / r^2

q^2 =F [4pieo]r^2

q = r*sq rt F[4pieo]

q=0.2* sq rt[ 4.57 x 10^-21]*[4*3.1416*8.854*10^-12]

q = 1.42614*10^ -16 C

number of electrons = n = q/e=1.42614*10^ -16 /1.6*10^-19

n =891

891 excess electrons must be present on each sphere

5 0

3 years ago

A 0.45-kilogram football traveling at a speed of 22 meters per second is caught by an 84-kilogram stationary receiver. if the fo

Troyanec [42]

Impulse = Change in momentum.
The ball was moving with a momentum of 0.45 * 22 = 9.9
The ball comes to rest in the receivers arm; this means the ball's final velocity = 0. So mv2 = 0.45 * 0
The magnitude of the impact is just the change in momentum. 9.9 - (0.45 * 0) = 9.9

3 0

3 years ago

To practice Problem-Solving Strategy 11.1 Equilibrium of a Rigid Body. A horizontal uniform bar of mass 2.7 kg and length 3.0 m

Zanzabum

Answer:

$T_{1}$ = 14.88 N

Explanation:

Let's begin by listing out the given variables:

M = 2.7 kg, L = 3 m, m = 1.35 kg, d = 0.6 m,

g = 9.8 m/s²

At equilibrium, the sum of all external torque acting on an object equals zero

τ(net) = 0

Taking moment about $T_{1}$ we have:

(M + m) g * 0.5L - $T_{2}$ (L - d) = 0

⇒ $T_{2}$ = [(M + m) g * 0.5L] ÷ (L - d)

$T_{2}$ = [(2.7 + 1.35) * 9.8 * 0.5(3)] ÷ (3 - 0.6)

$T_{2}$ = 59.535 ÷ 2.4

$T_{2}$ = 24.80625 N ≈ 24.81 N

Weight of bar(W) = M * g = 2.7 * 9.8 = 26.46 N

Weight of monkey(w) = m * g = 1.35 * 9.8 = 13.23 N

Using sum of equilibrium in the vertical direction, we have:

$T_{1}$ + $T_{2}$ = W + w ------- Eqn 1

Substituting T2, W & w into the Eqn 1

$T_{1}$ + 24.81 = 26.46 + 13.23

$T_{1}$ = <u>14.88</u> N

8 0

3 years ago

Calculate the number of free electrons per cubic meter for some hypothetical metal, assuming that there are 1.3 free electrons p

boyakko [2]

Answer:

The number of free electrons per cubic meter is $7.61\times 10^{28}\ m^{-3}$

Explanation:

It is given that,

The number of free electrons per cubic meter is, 1.3

Electrical conductivity of metal, $\sigma=6.8\times 10^7\ \Omega^{-1}m^{-1}$

Density of metal, $\rho=10.5\ g/cm^3$

Atomic weight, A = 107.87 g/mol

Let n is the number of free electrons per cubic meter such that,

$n=1.3\ N$

$n=1.3(\dfrac{\rho N_A}{A})$

Where

$\rho$ is the density of silver atom

$N_A$ is the Avogadro number

A is the atomic weight of silver

$n=1.3\times (\dfrac{10.5\ g/cm^3\times 6.02\times 10^{23}\ atoms/mol}{107.87\ g/mol})$

$n=7.61\times 10^{22}\ cm^{-3}$

$n=7.61\times 10^{28}\ m^{-3}$

Hence, this is the required solution.

6 0

3 years ago

A child with a mass of 20 kg sits at a distance of 2 m from the pivot point of a seesaw. where should a 16-kg child sit to balan

Maksim231197 [3]

In the above problem, we need to find mass of the second child, so that the Center of Mass remains at the origin( pivot).

CM= m1r1+m2r2/m1+m2
0= 20*-2+16*r2/20+16
r2= 40/16
r2= +2.5 m

4 0

3 years ago