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krek1111 [17]
3 years ago
5

The airplane hits even worse turbulence and you find yourself pinned against the ceiling of the airplane by a force of 100 N. Wh

at is the vertical acceleration vector of the airplane
Physics
1 answer:
brilliants [131]3 years ago
8 0

Answer:

The correct answer is "12 m/s²".

Explanation:

Given:

F_{app} = 100 \ N

As we know,

⇒ F_{app} = mg-ma

Or,

⇒ a = g-(\frac{F_{app}}{m} )

By substituting the values, we get

⇒    =10-(-\frac{100}{50} )

⇒    =10+2

⇒    =12 \ m/s^2

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A sample of gold has a mass of 30.94 grams and density of 19.32g/cm^3. What volume of space will this sample of gold occupy?
sdas [7]
Data:
mass, m = 30.94 g
density, d = 19.32 g/cm^3

Formula: d = m / v => v = m / d = 30.94 g / 19.32 g/cm^3 = 1.60 cm^3

Then, the answer is the option C. 
6 0
3 years ago
Read 2 more answers
Two long parallel wires are separated by forty centimeters and carry oppositely-directed currents of ten amperes. Find the magni
Softa [21]

Answer:

1.04μT

Explanation:

Due to both wires have opposite currents, the magnitude of the total magnetic field is given by

B_T=\frac{\mu_o I}{2 \pi r_1}-\frac{\mu_o I}{2 \pi r_2}

I: electric current = 10A

mu_o: magnetic permeability of vacuum = 4pi*10^{-7} N/A^2

r1: distance from wire 1 to the point in which B is measured.

r2: distance from wire 2.

The distance between wires is 40cm = 0.4m. Hence, r1=0.2m r2=0.6m

By replacing in the formula you obtain:

B_T=\frac{(4\pi *10^{-7}N/A^2)(10A)}{2\pi}(\frac{1}{0.4m}-\frac{1}{0.6m})=1.04*10^{-6}T =1.04\mu T

hence, the magnitude of the magnetic field is 1.04μT

4 0
4 years ago
Una caja de 5.0kg de masa se acelera desde el reposo a través del piso mediante una fuerza a una tasa de 2.0 /s2 durante 7.0s en
Nady [450]

Responder:

<h2>490 julios </h2>

Explicación:

Se dice que el trabajo se realiza cuando una fuerza aplicada a un objeto hace que el objeto se mueva a través de una distancia. El trabajo realizado por un cuerpo se expresa mediante la fórmula;

Workdone = Fuerza * Distancia

Como Fuerza = masa * aceleración,

Workdone = masa * aceleración * distancia

Masa dada = 5.0kg, aceleración = 2.0m / s² d =?

Para obtener d, usaremos una de las leyes del movimiento,

d = ut + 1 / 2at²

u = 0 (ya que el cuerpo acelera desde el reposo) yt = 7.0s

d = 0 + 1/2 (2) (7) ²

d = 49m

Workdone = 5 * 2 * 49

Workdone = 490 Julios

4 0
3 years ago
What is the minimum value of the friction coefficient between the boxes that will keep them from slipping when the 100 N force i
CaHeK987 [17]

Answer:

The friction coefficient's minimum value will be "0.173".

Explanation:

The given query seems to be incomplete. Below is the attached file of the complete question.

According to the question,

(a)

The net friction force's magnitude will be:

⇒ F_{net}=ma

           =5\times 1.7

           =8.5 \ N

(b)

For m₃,

⇒ ma=\mu m_3 g

Or,

⇒    \mu=\frac{a}{g}

          =\frac{1.7}{9.8}

          =0.173

5 0
3 years ago
how much power is needed to lift a box with a force of 780 newtons over a distance of 2 meters in 45 seconds
-BARSIC- [3]

Answer:

<h2>34.67 W</h2>

Explanation:

Power is the rate at which work is done and can be found by using the formula

p =  \frac{w}{t}  \\

p is the power in Watts (W)

w is the workdone in joules

t is time in s

but workdone = force × distance

From the question

force = 780 N

distance = 2 m

workdone = 780 × 2 = 1560 N

Since we now have the value of workdone we can find the power

We have

p =  \frac{1560}{45}  = 34.6666666... \\

We have the final answer as

<h3>34.67 W</h3>

Hope this helps you

3 0
3 years ago
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