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Lena [83]
3 years ago
11

an airplane flies at a speed of 100 m/s and starts to accelerate constantly at a rate of 50 m/s2. how fast is the plane flying a

fter traveling a distance of 1 kilometer?
Physics
1 answer:
mart [117]3 years ago
4 0

Answer:

331.7m/s

Explanation:

Given parameters:

Initial velocity  = 100m/s

Acceleration  = 50m/s²

Distance  = 1km   = 1000m

Unknown:

Final velocity = ?

Solution:

To solve this problem, we have to apply the right motion equation shown below;

     v²  = u²  + 2aS

 v is the final velocity

 u is the initial velocity

 a is the acceleration

 S is the distance

 Now insert the parameters and solve;

     v² = 100² + (2 x 50 x 1000)

     v² = 110000

     v = √110000  = 331.7m/s

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2. The components of vector A are given as follows:
Stella [2.4K]

Answer:

50 degree.

Explanation:

Given that the components of vector A are given as follows: Ax = 5.6 Ay = -4.7

The angle between vector A and B in the positive direction of x-axis will be achieved by using the formula:

Tan Ø = Ay/Ax

Substitute Ay and Ax into the formula above.

Tan Ø = -4.7 / 5.6

Tan Ø = -0.839

Ø = tan^-1(-0. 839)

Ø = - 40 degree

Therefore, the angle between vector A and B positive direction of x-axis will be

90 - 40 = 50 degree.

3 0
3 years ago
A vertical spring gun is used to launch balls into the air. If the spring is compressed by 4.9 cm, the ball of mass 5.5 g is lau
AleksandrR [38]

We know, by conservation of energy :

\dfrac{kx^2}{2}=mgh

Therefore,

\dfrac{x_1^2}{x_2^2}=\dfrac{h_1}{h_2}

Putting given values, we get :

\dfrac{x_1^2}{x_2^2}=\dfrac{h_1}{h_2}\\\\\dfrac{4.9^2}{x_2^2}=\dfrac{50.2}{2\times 50.2}\\\\x_2^2=2\times 4.9^2\\\\x_2 = 4.9\times \sqrt{2}\\\\x_2=6.93\ cm

Therefore, the spring be compressed to 6.93 cm to send the ball twice as high.

Hence, this is the required solution.

6 0
3 years ago
Fbejfsjnvfdjbesjkbf dsjlbc dsjlc
pav-90 [236]

Answer:

???

Explanation:

7 0
4 years ago
Read 2 more answers
Assuming that the limits of the visible spectrum are approximately 380 and 700 nm, find the angular range of the first-order vis
ch4aika [34]

Answer:

angular range is ( 0.681 rad , 0.35 rad )

Explanation:

given data

wavelength λ = 380 nm = 380 × 10^{-9} m

wavelength λ  = 700 nm =  700 × 10^{-9} m

to find out

angular range of the first-order

solution

we will apply here slit experiment equation that is

d sinθ = m λ    ...........1

here m is 1 for single slit and d is = \frac{1}{900*10^3 m}

so put here value in equation 1 for 380 nm

we get

d sinθ = m λ

\frac{1}{900*10^3} sinθ = 1 × 380 × 10^{-9}

θ = 0.35 rad

and for 700 nm

we get

d sinθ = m λ

\frac{1}{900*10^3} sinθ = 1 × 700 × 10^{-9}

θ = 0.681 rad

so angular range is ( 0.681 rad , 0.35 rad )

3 0
3 years ago
Help ASAP.
Temka [501]

Answer:

Depending on where people are located in the world (Northern hemisphere, Southern hemisphere, etc) depends on the difference in direction (North, South, east, West) which is most likely why it'd look different.

Explanation:

I dunno if this is along the lines of an answer you're looking for, but hope this helps :)

7 0
3 years ago
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