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3 years ago

An airplane flew 1000 m in 400 seconds. What is the airplane's speed?

Physics

2 answers:

Eddi Din [679]3 years ago

7 0

the answer is 2.5 m/s^2

step by step:

1. speed= distance/time

so the speed of the plane = 1000/400

(cancel the two zero's of the 400 with two of the three zero's of the 1000)

you'll have 10/4 left

10/4= 2.5

weqwewe [10]3 years ago

6 0

Answer:

$speed = \frac{distance}{time} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{1000}{400} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 2.5m {s}^{ - 1}$

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A ball is thrown horizontally from the top of a 55 m building and lands 150 m from the base of the building. Ignore air resistan

PtichkaEL [24]

Answer:

a) t =3.349 s

b) V_x,i = 44.8 m/s

c) V_y,f = 32.85 m/s

d) V = 55.55 m/s

Explanation:

Given:

- Total throw in x direction x(f) = 150 m

- Total distance traveled down y(f) = 55 m

Find:

a) How long is the rock in the air in seconds.

b) What must have been the initial horizontal component of the velocity, in meters per second?

c) What is the vertical component of the velocity just before the rock hits the ground, in meters per second?

d) What is the magnitude of the velocity of the rock just before it hits the ground, in meters per second?

Solution:

- Use the second equation of motion in y direction:

y(f) = y(0) + V_y,i*t + 0.5*g*t^2

- V_y,i = 0 (horizontal throw)

55 = 0 + 0 + 0.5*(9.81)*t^2

t = sqrt ( 55 * 2 / 9.81 )

t =3.349 s

- Use the second equation of motion in x direction:

x(f) = x(0) + V_x,i*t

150 = 0 + V_x,i*3.349

V_x,i = 150 / 3.349 = 44.8 m/s

- Use the first equation of motion in y direction:

V_y,f = V_y,i + g*t

V_y,f = 0 + 9.81*3.349

V_y,f = 32.85 m/s

- The magnitude of velocity of ball when it hits the ground is:

V^2 = V_y,f^2 + V_x,i^2

V = sqrt (32.85^2 + 44.8^2)

V = 55.55 m/s

5 0

3 years ago

Calculate the total resistance in a series circuit made up of resistances of 3Ω, 4Ω, and 5Ω.

-BARSIC- [3]

The total resistance in a series circuit is equal to the sum of all resistors (R total = ΣRi).

R total = R1 + R2 + R3 = (3 + 4 + 5) Ω = 12 Ω

5 0

3 years ago

A 125kg bumper car going 12m/s hits a 235kg bumper car going -13m/s.if the first car bounces back at -12.5m/s what is the veloci

vovikov84 [41]

According to the law of conservation of momentum:

$m_{1}v_{1}+m_{2}v_{2}=m_{1}v_{1}'+m_{2}v_{2}'$

m1 = mass of first object
m2 = mass of second object
v1 = Velocity of the first object before the collision
v2 = Velocity of the second object before the collision
v'1 = Velocity of the first object after the collision
v'2 = Velocity of the second object after the collision

Now how do you solve for the velocity of the second car after the collision? First thing you do is get your given and fill in what you know in the equation and solve for what you do not know.

m1 = 125 kg v1 = 12m/s v'1 = -12.5m/s
m2 = 235kg v2 = -13m/s v'2 = ?

$m_{1}v_{1}+m_{2}v_{2}=m_{1}v_{1}'+m_{2}v_{2}'$
$(125kg)(12m/s)+(235kg)(-13m/s)=(125kg)(-12.5m/s)+(235kg)(v_{2}'$
$1,500kg.m/s+(-3055kg.m/s)=(-1562.5kg.m/s)+(235kg)(v_{2}')$
$-1,555kg.m/s=(-1562.5kg.m/s)+(235kg)(v_{2}')$

Transpose everything on the side of the unknown to isolate the unknown. Do not forget to do the opposite operation.

$-1,555kg.m/s + 1562.5kg.m/s=(235kg)(v_{2}')$
$7.5kg.m/s=(235kg)(v_{2}')$
$(7.5kg.m/s)/(235kg)=(v_{2}')$
$0.03m/s=(v_{2}')$

The velocity of the 2nd car after the collision is 0.03m/s.

5 0

3 years ago

Select Light for the type of wave, adjust the wavelength so that the light is red, and increase the amplitude of the light to th

Sergeu [11.5K]

Answer:

here as we increase the distance the intensity will decrease and hence the amplitude of the electric field will decrease and vice-versa

Explanation:

As wee know that the amplitude of the wave will decide the energy of the wave

Here we know that energy density of electromagnetic wave is given as

$u = \frac{1}{2}\epsilon_0E_0^2$