Ask question

Ask question

All categories

bagirrra123 [75]

3 years ago

11

Passengers on an airplane move from rest to 86 meters per second before the airplane takes off. If the airplane takes 100 second

s to take off what is the acceleration, in m/s^2, of the airplane? Show your answer to one decimal point. Make sure you include the proper unit. Do not put a space between the number and the unit!

1 answer:

VikaD [51]3 years ago

4 0

Utilize the formula: $V _{f} = V _{i} + a\Delta t$

$V _{f}$ = Final Velocity (86 m/s)

$V _{f}$ = Initial Velocity (0 m/s)

a = acceleration (m/s²)

$\Delta$ t = Time (100 seconds)

As a result,

86 m/s = 0 + (a)(100 seconds)

Using algebra, divide 86 m/s by 100 seconds:

86 m/s = 100a

a = 0.86 m/s²

Rounded to one decimal place: 0.9 m/s²

Let me know if you have any questions!

You might be interested in

Some elements are highly unstable and break apart within seconds, making them difficult to study. How can the periodic table hel

rusak2 [61]

You can look at groups in the same group (the columns), since they tend to have similar properties. For example, the alkali metals in group one react aggressively with water and form white compounds.

8 0

2 years ago

Read 2 more answers

A heater has resistance of 10 ohms. It operates at 120 volts. a. What is the current through the heater?

Natali5045456 [20]

Answer:

I = 12 A

Explanation:

For this exercise we use the relationship that voltage is proportional to the product of current and resistance

V = I R

I = V / R

let's calculate

I = 120/10

I = 12 A

4 0

2 years ago

Particle A of charge 2.70 10-4 C is at the origin, particle B of charge -6.36 10-4 C is at (4.00 m, 0), and particle C of charge

Serhud [2]

Answer:

FC vector representation

$F_{C} =(19.03i+13.78j)N$

Magnitude of FC

$F_{C}=23.495N$

Vector direction FC

$\beta=35.91$ degrees: angle that forms FC with the horizontal

Explanation:

Conceptual analysis

Because the particle C is close to two other electrically charged particles, it will experience two electrical forces and the solution of the problem is of a vector nature.

The directions of the individual forces exerted by qA and qB on qC are shown in the attached figure; The force (FAC) of qA over qC is repulsive because they have equal signs and the force (FBC) of qB over qC is attractive because they have opposite signs.

The FAC force is up in the positive direction and the FBC force forms an α angle with respect to the x axis.

$\alpha = tan^{-1}(\frac{3}{4}) = 36.86$ degrees

To calculate the magnitudes of the forces we apply Coulomb's law:

$F_{AC} = \frac{k*q_{A}*q_{C}}{r_{AC}^2}$ Equation (1): Magnitude of the electric force of the charge qA over the charge qC

$F_{BC} = \frac{k*q_{B}*q_{C}}{r_{BC}^2}$ Equation (2) : Magnitude of the electric force of the charge qB over the charge qC

Known data

$k=8.99*10^9 \frac{N*m^2}{C^2}$

$q_{A}=2.70*10^{-4} C$

$q_{B}=-6.36*10^{-4} C$

$q_{C}=1.04*10^{-4} C$

$r_{AC} =3$

$r_{BC}=\sqrt{4^2+3^2} = 5$

Problem development

In the equations (1) and (2) to calculate FAC Y FBC:

$F_{AC} =8.99*10^9*(2.70*10^{-4}* 1.04*10^{-4})/(3)^2=28.05N$

$F_{BC} =8.99*10^9*(6.36*10^{-4}* 1.04*10^{-4})/(5)^2=23.785N$

Components of the FBC force at x and y:

$F_{BCx}=23.785 *Cos(36.86)=19.03N$

$F_{BCy}=23.785 *Sin(36.86)=14.27N$

Components of the resulting force acting on qC:

$F_{Cx} = F_{ACx}+ F_{BCx}=0+19.03=19.03N$

$F_{Cy} = F_{ACy}+ F_{BCy}=28.05-14.27=13.78N$

FC vector representation

$F_{C} =(19.03i+13.78j)N$

Magnitude of FC

$F_{C}= \sqrt{19.03^2+13.78^2} =23.495N$

Vector direction FC

$\beta = tan^{-1} (\frac{13.78}{19.03})=35.91$ degrees: angle that forms FC with the horizontal

7 0

3 years ago

yarga [219]

Answer:

Please reframe your questions

4 0

2 years ago

The nucleus of the polonium isotope 214 Po (mass 214 u) is radioactive and decays by emitting an alpha particle (a helium nucleu

amid [387]

Answer:

368224.29906 m/s

Explanation:

M = Mass of Polonium nucleus = 214 u

V = Velocity of nucleus

m = Mass of Helium nucleus = 4 u

v = Velocity of alpha particle = $1.97\times 10^7\ m/s$

In this system the momentum is conserved

$MV+mv=0\\\Rightarrow MV=-mv\\\Rightarrow 214V=-4\times 1.97\times 10^7\\\Rightarrow V=\dfrac{-4\times 1.97\times 10^7}{214}\\\Rightarrow V=368224.29906\ m/s$

The recoil speed of the nucleus that remains after the decay is 368224.29906 m/s

6 0

2 years ago

Read 2 more answers