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mars1129 [50]
3 years ago
12

Brainliest for correct answer :)

Physics
2 answers:
kolezko [41]3 years ago
5 0

Answer:

Acceleration a =-1.75 m/s²

Explanation:

Given:

Initial speed u = 24 m/s

Final speed v = 10 m/s

Time taken t = 8 sec

Find:

Acceleration a

Computation:

a = (v-u)/t

a = (10-24)/8

a = -14 / 8

Acceleration a =-1.75 m/s²

sashaice [31]3 years ago
3 0

Answer:

Accelaration =(v-u)/t

40-10/20m/s^2

30/20m/s^2

1.5m/s^2=accelaeration

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When a sound wave first enters the auricle, which part of the ear does it travel through on its way to the eardrum?
ddd [48]

Ear Canal.

Explanation:

Sound waves enter the auricle and travel through a narrow passageway called the ear canal, which leads to the eardrum.

5 0
2 years ago
When the moon is between the earth and the sun what moon phase will this be
Aleksandr [31]

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3 years ago
The mass of your new motorcycle is 250kg. What is the weight on the moon in newtons
Hunter-Best [27]

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Did you understand??


8 0
3 years ago
To take off from the ground, an airplane must reach a sufficiently high speed. The velocity required for the takeoff, the takeof
BlackZzzverrR [31]
<h2>Answer: 26,8 s</h2>

Explanation:

If we are talking about an acceleration at a constant rate , we are dealing with constant acceleration, hence we can use the following equations:

{V_{f}}^{2}={V_{o}}^{2}+2ad (1)

V_{f}=V_{o}+at (2)

Where:

V_{f} is the final velocity of the plane (the takeoff velocity in this case)

V_{o}=0 is the initial velocity of the plane (we know it is zero because it starts from rest)

a=5m/s^{2} is the constant acceleration of the plane to reach the takeoff velocity

d=1800m is the distance of the runway

t is the time

Knowing this, let's begin with (1):

{V_{f}}^{2}=0+2(5m/s^{2})(1800m) (3)

{V_{f}}^{2}=18000m^{2}/s^{2} (4)

V_{f}=134.164 m/s (5)

Substituting (5) in (2):

134.164 m/s=0+(5m/s^{2})t (6)

Finding t:

t=26.8 s This is the time needed to take off

6 0
3 years ago
Determine the minimum angle at which a roadbedshould be banked
poizon [28]

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The horizontal force component is equivalent to the weight of the car, while the vertical component is linked to the centripetal force exerted on the car, therefore,

T cos\theta = mg \rightarrow T = \frac{mg}{cos\theta}

Tsin\theta = \frac{mv^2}{r} \rightarrow T = \frac{mv^2/r}{sin\theta}

Equating both equation we have that,

\frac{mv^2}{r} = mgtan\theta

tan\theta = \frac{v^2}{rg}

Rearranging to find the angle we have that,

\theta = tan^{-1} (\frac{v^2}{rg})

Our values are given as,

r = 2.00*10^2m

v = 20m/s

\theta = tan^{-1}(\frac{20^2}{(2*10^{2})(9.8)})

\theta = 11.53 \°

Therefore the minimum angle will be 11.53°

5 0
3 years ago
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