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Fiesta28 [93]
2 years ago
6

A space shuttle is flying north at 6,510 m/s. It

Physics
1 answer:
Maksim231197 [3]2 years ago
7 0

Answer:

7808 m/s

Explanation:

Find NE velocity after 60 s  of acceleration in that direction:

  = a t  =  28.4 m/s^2 * 60 s = 1704  m/s

    Vertical component = 1704 sin 45 = 1204.9 m/s

     Horiz component = 1704 cos 45 = 1204.9  m/s

Add the two vertical components

  6510 + 1204.9 = 7714.9 m/s = vertical velocity

Pythagorean theorem to find resultant of vertical and horiz  v's

Vf ^2 = 1204.9^2   + 7714.9^2 0

Vf = 7808. m/s

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5 0
2 years ago
8. A turtle crawls along a straight line, which we will call the x-axis with the positive direction to the right. The equation f
pychu [463]

(a) The turtle's initial velocity is 4 m/s, initial position of the turtle is 5 cm, and initial acceleration is -1.25 m/s².

(b) The time when the velocity of the turtle is zero is 3.2 s.

(c) The time taken for the turtle to return to its starting point is 6.4 s.

(d) The time taken for the turtle to travel 30 cm is 0.08 s.

<h3>Initial velocity of the turtle</h3>

The initial velocity of the turtle is calculated as follows;

v = \frac{dx}{dt} \\\\x = 5 + 4t -0.625t^2\\\\v(t) = 4 - 1.25t\\\\v(0) = 4-0\\\\v(0) = 4 \ m/s

<h3>Initial acceleration of the turtle</h3>

The initial acceleration of the turtle is calculated as follows;

a = \frac{dv}{dt} \\\\v(t) = 4 - 1.25t\\\\a = -1.25\ m/s^2

<h3>Initial position of the turtle</h3>

x(t) = 5 + 4t - 0.625t²

x(0) = 5 cm

<h3>Time when the velocity becomes zero</h3>

v(t) = 4 - 1.25t

0 = 4 - 1.25t

1.25t = 4

t = 4/1.25

t = 3.2 s

<h3>Time taken to return to starting point</h3>

The total distance traveled is calculated as follows

v² = u² + 2ad

0 = (4)² + 2(-1.25)d

0 = 16 - 2.5d

2.5d = 16

d = 16/2.5

d = 6.4 m

Time to travel the given distance;

d = ut + ¹/₂at²

6.4 = (4)t + ¹/₂(-1.25)t²

6.4 = 4t - 0.625t²

0.625t² - 4t + 6.4 = 0

solve the quadratic equation using formula method;

t = 3.2 s

The time travel the distance two times, = 2 x 3.2 s = 6.4 s

<h3>Time taken for the turtle to travel 30 cm</h3>

d = ut + ¹/₂at²

0.3 = (4)t + ¹/₂(-1.25)t²

0.3 = 4t - 0.625t²

0.625t² - 4t + 0.3 = 0

solve the quadratic equation using formula method;

t = 0.08 s

Learn more about velocity here: brainly.com/question/6504879

7 0
2 years ago
A stoplight with weight 100 N is suspended at the midpoint of a cable strung between two posts 200 m apart. The attach points fo
Tasya [4]

There are 3 forces acting on the stoplight:

• its weight <em>W</em>, with magnitude <em>W</em> = 100 N, pointing directly downward

• two tension forces <em>T</em>₁ and <em>T</em>₂ with equal magnitude <em>T</em>₁ = <em>T</em>₂ = <em>T</em> = 1000 N, both making an angle of <em>θ</em> with the horizontal, but one points left and the other points right

The stoplight is in equilibrium, so by Newton's second law, the net vertical force acting on it is 0, such that

∑ <em>F</em> = <em>T</em>₁ sin(<em>θ</em>) + <em>T</em>₂ sin(180° - <em>θ</em>) - <em>W</em> = 0

We have sin(180° - <em>θ</em>) = sin(<em>θ</em>) for all <em>θ</em>, so the above reduces to

2<em>T</em> sin(<em>θ</em>) = <em>W</em>

2 (1000 N) sin(<em>θ</em>) = 100 N

sin(<em>θ</em>) = 0.05

<em>θ</em> ≈ 2.87°

If <em>y</em> is the vertical distance between the stoplight and the ground, then

tan(<em>θ</em>) = (15 m - <em>y</em>) / (100 m)

Solve for <em>y</em> :

tan(2.87°) = (15 m - <em>y</em>) / (100 m)

<em>y</em> = 15 m - (100 m) tan(2.87°)

<em>y</em> ≈ 9.99 m

3 0
2 years ago
Why do you think football players are usually large in size?
kiruha [24]
Football players in the NFL are large in size due to their eating habits since as players they have to work out constantly to stay and shape and eat to gain weight and be strong, so its important for them to be strong against their opponents in gameplay. Football players sometimes take steroids which make their muscles stronger and harder, and also gives them a growth spurt but the effects are worse and makes them sicker, high blood pressure, and even more. 
8 0
3 years ago
Read 2 more answers
A rock thrown straight up takes 4.2 s to reach its maximum height what was its initial velocity
liubo4ka [24]

Consider the upward direction of motion as positive and downward direction of motion as negative.

a = acceleration due to gravity in downward direction = - 9.8 \frac{m }{s^{2}}

v₀ = initial velocity of rock in upward direction = ?

v = final velocity of rock at the highest point = 0 \frac{m }{s}

t = time to reach the maximum height = 4.2 sec

Using the kinematics equation

v = v₀ + a t

inserting the values

0 = v₀ + (- 9.8) (4.2)

v₀ = 41.2 \frac{m }{s}


8 0
2 years ago
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