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rosijanka [135]

3 years ago

7

Sebuah benda dijatuhkan bebas dari ketinggian 200 m jika grafitasi setempat 10 m/s maka hitunglah kecepatan dan ketinggian benda

saat ek=4ep

1 answer:

Pie3 years ago

7 0

Please post in English so i or someone else can help you.

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A soccer ball is kicked from point Pi at an angle above a horizontal field. The ball follows an ideal path before landing on the

Nutka1998 [239]

Answer:

A. The horizontal velocity vector points to the right & equals v cos Î¸.

Explanation:

The motion describes a parabolic path, where the horizontal speed is constant and the horizontal velocity vector always points to the right and equals v*cos Î¸.

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

A leaky 10-kg bucket is lifted from the ground to a height of 11 m at a constant speed with a rope that weighs 0.9 kg/m. Initial

nalin [4]

Answer:

the work done to lift the bucket = 3491 Joules

Explanation:

Given:

Mass of bucket = 10kg

distance the bucket is lifted = height = 11m

Weight of rope= 0.9kg/m

g= 9.8m/s²

initial mass of water = 33kg

x = height in meters above the ground

Let W = work

Using riemann sum:

the work done to lift the bucket =∑(W done by bucket, W done by rope and W done by water)

= $\int\limits^a_b {(Mass of Bucket + Mass of Rope + Mass of water)*g*d} \, dx$

Work done in lifting the bucket (W) = force × distance

Force (F) = mass × acceleration due to gravity

Force = 9.8 * 10 = 98N

W done by bucket = 98×11 = 1078 Joules

Work done to lift the rope:

At Height of x meters (0≤x≤11)

Mass of rope = weight of rope × change in distance

= 0.8kg/m × (11-x)m

W done = integral of (mass×g ×distance) with upper 11 and lower limit 0

W done = $\int\limits^1 _0 {9.8*0.8(11-x)} \, dx$

Note : upper limit is 11 not 1, problem with math editor

W done = 7.84 (11x-x²/2)upper limit 11 to lower limit 0

W done = 7.84 [(11×11-(11²/2)) - (11×0-(0²/2))]

=7.84(60.5 -0) = 474.32 Joules

Work done in lifting the water

At Height of x meters (0≤x≤11)

Rate of water leakage = 36kg ÷ 11m = $\frac{36}{11}$ kg/m

Mass of rope = weight of rope × change in distance

= $\frac{36}{11}$ kg/m × (11-x)m = 3.27kg/m × (11-x)m

W done = integral of (mass×g ×distance) with upper 11 and lower limit 0

W done = $\int\limits^1 _0 {9.8*3.27(11-x)} \, dx$

Note : upper limit is 11 not 1, problem with math editor

W done = 32.046 (11x-x²/2)upper limit 11 to lower limit 0

W done = 32.046 [(11×11-(11²/2)) - (11×0-(0²/2))]

= 32.046(60.5 -0) = 1938.783 Joules

the work done to lift the bucket =W done by bucket+ W done by rope +W done by water)

the work done to lift the bucket = 1078 +474.32+1938.783 = 3491.103

the work done to lift the bucket = 3491 Joules

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4 years ago

The vertical displacement of the wave is measured from the

algol13

The vertical displacement of the wave is measured from
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