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viva [34]
3 years ago
5

gAn Olympic diver is on a diving platform 5.40 m above the water. To start her dive, she runs off of the platform with a speed o

f 1.17 m/s in the horizontal direction. What is the diver's speed, in m/s, just before she enters the water
Physics
1 answer:
lina2011 [118]3 years ago
4 0

Answer:

The diver's speed, in m/s, just before she enters the water = 10.19 m/s

Explanation:

Assuming the horizontal velocity of the diver remains constant by neglecting the air resistance.

V^2 = U^2 + 2as

Where, V^2 = the diver's final velocity before impact with the water

            U^2 = Initial diver's velocity as she leaves the diving platform

                a = acceleration due to gravity

                s = the displacement

V =  √U^2 + √2as

= √(5.40 m/s)^2 + √2(9.81 m/s/s) (1.17 m)  

= √ 29.16 + √ 22.9554

=  5.40 + 4.79

= 10.19 m/s

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A closely wound search coil has an area of 3.13 cm2, 135 turns, and a resistance of 61.1 Ω. It is connected to a charge-measurin
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Answer:

Explanation:

Let the magnitude of magnetic field be B .

flux passing through the coil's  = area of coil x field x no of turns

Φ = 3.13 x 10⁻⁴ x B x 135 = 422.55 x 10⁻⁴ B .

emf induced = dΦ / dt , Φ is magnetic flux.

current i = dΦ /dt x 1/R

charge through the coil = ∫ i dt

= ∫   dΦ /dt x 1/R dt

= 1 / R ∫ dΦ

= Φ / R

Total resistance R = 61.1 + 44.4 = 105.5 ohm .

3.44 x 10⁻⁵ = 422.55 x 10⁻⁴ B / 105.5

B = 3.44 x 10⁻⁵ x 105.5  / 422.55 x 10⁻⁴

= .86 x 10⁻¹

= .086 T .

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2 years ago
PLZ help this should be super easy
Ugo [173]
The answer is number (3) if am wrong am sry
4 0
3 years ago
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What does this same experiment tell you about light waves? Explain the evidence that supports your claim.
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Answer:

Light wave is an EM wave that can only be seen by humans New questions in Physics Engineers at the Space Centre must determine the net force needed for a rockets engine to achieve an acceleration of 70 m/s2.

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3 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

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