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Ksenya-84 [330]
3 years ago
12

I want the answer pls

Physics
1 answer:
storchak [24]3 years ago
3 0

Answer:

<em>Percentage error = 2.63%</em>

<em></em>

Explanation:

We are given the following

Original volume = 15.20mL

Student measurement = 14.8mL

Error = 15.20-14.80

Error = 0.40mL

percentage = Error/Original volume * 100

Percentage error = 0.4/15.20 * 100

Percentage error = 40/15.20

<em>Percentage error = 2.63%</em>

<em></em>

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If the outside air temperature increases during a flight at constant power and at a constant indicated altitude, the true airspe
Readme [11.4K]

The true airspeed will increase and true altitude will increase.

<h3>What is true air speed?</h3>

True airspeed is the airspeed of an aircraft relative to undisturbed air.

It's the aircraft speed relative to the airmass in which it's flying.

<h3>How does outside air temperature affect air speed?</h3>

If the outside air temperature increases during a flight at constant power and at a constant indicated altitude, the true airspeed will increase and true altitude will increase.

Thus, the true airspeed will increase and true altitude will increase.

Learn more about true airspeed here: brainly.com/question/13257916

#SPJ1

6 0
2 years ago
Una atracción de feria consiste en lanzar un trineo de 2 kg por una rampa ascendente que forma un ángulo de 30º con la horizonta
Arada [10]

Answer:

La velocidad con la que se debe lanzar el trineo es aproximadamente 9.96 m/s

Explanation:

La masa del trineo, m = 2 kg

El ángulo de inclinación del trineo, θ = 30 °

El coeficiente de fricción, μ = 0,15

La altura a la que debe ascender el trineo, h = 4 m

La reacción normal del trineo en la rampa, N = m · g · cos (θ)

La fuerza de fricción F_f = N × μ

Dónde;

g = Th aceleración debida a la gravedad ≈ 9.81 m/s²

∴ F_f  = 0.15 × 2 kg × 9.81 m/s² × cos (30°) ≈ 2.55 N

La longitud de la rampa que se mueve el trineo, l = h/(sin(θ))

∴ l = 4/(sin(30°)) = 8

La longitud de la rampa que se mueve el trineo, l = 8 m

El trabajo realizado sobre la fricción, W_f = F_f × l

W_f = 2.55 × 8 ≈ 20.4

El trabajo realizado sobre la fricción, W_f ≈ 20.4 Julios

El trabajo realizado para levantar el trineo a 4 metros de altura, P.E. = m·g·h

∴ P.E. = 2 kg × 9.81 m/s² × 4 m ≈ 78.48 Joules

El trabajo realizado para levantar el trineo a 4 metros de altura, P.E. ≈ 78.48 Joules

El trabajo total requerido para levantar el trineo 4 metros por la rampa, W_t = W_f + P.E.

W_t = 20.4 J + 78.84 J ≈ 99.24 J

Energía cinética, K.E. = 1/2 × m × v²

La energía cinética necesaria para mover el trineo por la rampa, K.E. se da de la siguiente manera;

K.E. = 1/2 × 2 kg × v² ≈ 99.24 J

∴ v² ≈ 99.24 J/(1/2 × 2 kg)

La velocidad con la que se debe lanzar el trineo para que ascienda 4 metros por la rampa, v ≈ 9.96 m/s

4 0
3 years ago
How much would a 25 kg suitcase weigh on the surface of…?
photoshop1234 [79]

Answer:

A. 95N

B. ?

C.225N

D.?

Explanation:

For Mars and Pluto im not so sure about those too but for A and C I am positive those are correct.

Sorry I could not help you all the way, please dont be mad ;(

8 0
3 years ago
A U-tube is open to the atmosphere at both ends. Water is poured into the tube until the water column on the vertical sides of t
zmey [24]

Answer:

The value is  \Delta h  =  0.003 \  m

Explanation:

From the question we are told that

   The  height of the water is  h_1  =  10 \ cm  =  0.10 \  m

    The  density of  oil is \rho_o  =  950 \  kg/m^3

  The  height of  oil  is  h_2  =  6 \ cm  =  0.06 \  m

Given that both arms of the tube are open then the pressure on both side is the same

So  

      P_a =  P_b

=>   Here  

             P_a  =  P_z + \rho_w  *  g *  h

where  \rho_w is the density of water with value  \rho_w =  1000 \ kg/m^3

and  P_z is the atmospheric pressure

and  

        P_b  =  P_z + \rho_o  *  g *  h_2

=>   P_z + \rho_w  *  g * h =    P_z + \rho_o  *  g *  h_2

=>    \rho_w   * h  =    \rho_o  *  h_2

=>      h  =  \frac{950 * 0.06 }{1000}

=>      h  = 0.057 \ m

The  difference in height is evaluated as    

           \Delta h  =  0.06 - 0.057

          \Delta h  =  0.003 \  m

     

6 0
4 years ago
the net force on a vehicle is accelerating at a rate of 1.5 milliseconds is 1800 Newtons. What is the mass of the vehicles neare
Goryan [66]

Answer:

The mass is 1200 kilograms

Explanation:

Because Force is equal to mass times acceleration (F=m×a)

F=m×a

1800N=?×1.5

1800÷1.5=1200

1800N=1200Kg×1.5

7 0
4 years ago
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