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const2013 [10]
2 years ago
6

If you jump off a 7 m diving platform, how long will it take you to hit the

Physics
1 answer:
Allushta [10]2 years ago
3 0

Answer:

C.) T = 1.20s

Explanation:

From equation of motion,

S = ut + ½gt²

S = 7m

g = 9.8m/s²

U = 0m/s (since he starts from rest)

t = ?

S = ut + ½gt²

7 = 0 * t + ½ * 9.8 * t²

7 = 0 + 4.9t²

t² = 7 / 4.9

t² = 1.42857

t = √(1.42857)

t = 1.195s

t = 1.20s

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lidiya [134]

Answer:

1,3,5

Explanation:

i think maybe dont come at me

6 0
3 years ago
Read 2 more answers
A wheel has a constant angular acceleration of 4.5 rad/s2. during a certain 5.0 s interval, it turns through an angle of 128 rad
dalvyx [7]
The solution for this problem is through this formula:Ø = w1 t + 1/2 ã t^2 
where:Ø - angular displacement w1 - initial angular velocity t - time ã - angular acceleration 
128 = w1 x 4 + ½ x 4.5 x 5^2 128 = 4w1 + 56.254w1 = -128 + 56.25 4w1 = 71.75w1 = 71.75/4
w1 = 17.94 or 18 rad s^-1 
w1 = wo + ãt 
w1 - final angular velocity 
wo - initial angular velocity 
18 = 0 + 4.5t t = 4 s
3 0
3 years ago
As light from a star spreads out and weakens, do gaps form between the photons?
Sati [7]

Answer:

There are no gaps in space between the photons as they travel.

Yes, you can form the shadow of a fire

Explanation:

8 0
2 years ago
Read 2 more answers
A ball starts from rest and accelerates at a constant rate of 1.0m/s to a final velocity of
den301095 [7]

Answer:

Time taken to reach final velocity = 5.5 second

Explanation:

Given:

Initial velocity (Starting from rest)(u) = 0 m/s

Acceleration of ball (a) = 1 m/s²

Final velocity (v) = 5.5 m/s

Find:

Time taken to reach final velocity

Computation:

Using first equation of motion;

v = u + at

where,

v = final velocity

u = initial velocity

a = acceleration

t = time taken

5.5 = 0 + (1)(t)

5.5 = t

Time taken to reach final velocity = 5.5 second

8 0
2 years ago
A planet orbits a star, in a year of length 3.37 x 107 s, in a nearly circular orbit of radius 1.04 x 1011 m. With respect to th
PIT_PIT [208]

Answer

Given,

Time period of star,T = 3.37 x 10⁷ s

Radius of circular orbit,R = 1.04 x 10¹¹ m

a) Angular speed of the planet

   \omega = \dfrac{2\pi}{T}=\dfrac{2\pi}{3.37\times 10^{7}}

   \omega = 1.864\times 10^{-7}\ rad/s

b) tangential speed

   v = r \omega = 1.04\times 10^{11}\times 1.864 \times 10^{-7}

       v = 1.94 x 10⁴ m/s

c) centripetal acceleration magnitude

      a = \dfrac{v^2}{r}= \dfrac{(1.94\times 10^4)^2}{1.04\times 10^{11}}

          a = 3.62 x 10⁻³ m/s²

8 0
3 years ago
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