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Nikitich [7]
3 years ago
8

What does a light year measure

Physics
2 answers:
SVEN [57.7K]3 years ago
8 0
It’s used in measuring distance from earth with other celestial object. Hope this helps mark brainest
IRINA_888 [86]3 years ago
5 0
Light years measure the distance in space
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How long will it take a 2190 W motor to lift a 1.47 x 104 g box, 6.34 x 104 mm vertically.​
Rasek [7]

Answer:

t = 4.17 [s]

Explanation:

We know that work is defined as the product of force by distance.

W = F*d

where:

F = force [N] (units of Newtons)

d = distance = 6.34 x 10⁴ [mm] = 63.4 [m]

In order to find the force, we must determine the weight of the box, the weight can be determined by means of the product of mass by gravitational acceleration.

w = m*g

where:

m = mass = 1.47 x 10⁴ [g] = 14.7 [kg]

g = gravity acceleration = 9.81 [m/s²]

w = 14.7*9.81

w = 144.2 [N]

Therefore the work can be calculated.

W = w*d

W = 144.2*63.4

W = 9142.72 [J] (units of Joules)

Power is now defined in physics as the relationship of work at a given time

P = W/t

where:

P = power = 2190 [W]

t = time [s]

Now clearing t, we have.

t = W/P

t = 9142.72/2190

t = 4.17 [s]

6 0
3 years ago
i like honey but when u put it on your feet it sticks to the ground. how do i avoid the stickyness of honey when i want to clean
Gre4nikov [31]

Answer:

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Explanation:

6 0
3 years ago
Read 2 more answers
Gli iceberg nel Nord Atlantico rappresentano un rischio per la navigazione, costringendo ad allungare le rotte di navigazione di
nikklg [1K]

I DONT UNDERSTND IS ANOTHER LANGUAGR

6 0
3 years ago
A charge +Q is located at the origin and a second charge, +4Q, is at distance (d) on the x-axis.
tatiyna

Answer:

a)   x = ⅔ d , b) the charge must be negative, c) Q

Explanation:

a) In this exercise the force is electric between the charges, we are asked that the system of the three charges is in equilibrium, we use Newton's second law. Balance is on the third load that we are placing

         ∑ F = 0

        -F₁₂ + F₂₃ = 0

         F₁₂ = F₂₃

         

let's replace the values

        k Q Q / r₁₂² = k Q 4Q / r₂₃²

            Q² / r₁₂² = 4 Q² / r₂₃²

suppose charge 3 is placed at point x

        r₁₂ = x

        r₂₃ = d-x

             

we substitute

             1 / x² = 4 / (d-x) 2

             1 / x = 2 / (d-x)

             x = 2 (x-d)

             x = 2x -2d

            3x = 2d

              x = ⅔ d

b) The sign of the charge must be negative, to have an attractive charge on the two initial charges

c)  Q

5 0
3 years ago
The internal shear force V at a certain section of a steel beam is 80 kN, and the moment of inertia is 64,900,000 . Determine th
Luba_88 [7]

Here is the complete question

The internal shear force V at a certain section of a steel beam is 80 kN, and the moment of inertia is 64,900,000 . Determine the horizontal shear stress at point H, which is located L  = 20 mm below the centriod

The missing image which is the remaining part of this question is attached in the image below.

Answer:

The horizontal shear stress at point H is  \mathbf{\tau_H \approx  42.604 \ N/mm^2}

Explanation:

Given that :

The internal shear force V  =  80 kN = 80 × 10³ N

The moment of inertia = 64,900,000

The length = 20 mm below the centriod

The horizontal shear stress  \tau can be calculated by using the equation:

\tau = \dfrac{VQ}{Ib}

where;

Q = moment of area above or below the point H

b = thickness of the beam = 10  mm

From the centroid ;

Q = Q_1 + Q_{2}

Q = A_1y_1 + A_{2}y_{2}  

Q = ( ( 70 × 10) × (55) + ( 210 × 15) (90 + 15/2) ) mm³

Q = ( ( 700) × (55) + ( 3150 ) ( 97.5)  ) mm³

Q = ( 38500 +  307125 ) mm³

Q = 345625 mm³

\tau_H = \dfrac{VQ}{Ib}

\tau_H = \dfrac{80*10^3  * 345625}{64900000*10 }

\tau_H = \dfrac{2.765*10^{10}}{649000000 }

\tau_H = 42.60400616 \ N/mm^2

\mathbf{\tau_H \approx  42.604 \ N/mm^2}

The horizontal shear stress at point H is  \mathbf{\tau_H \approx  42.604 \ N/mm^2}

7 0
3 years ago
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