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inna [77]
4 years ago
11

The height of a helicopter above the ground is given by h = 2.65t3, where h is in meters and t is in seconds. at t = 1.55 s, the

helicopter releases a small mailbag. how long after its release does the mailbag reach the ground?
Physics
1 answer:
lora16 [44]4 years ago
3 0
Free fall is the motion of the body such that it is only acted upon by the force of gravity. Gravitational acceleration is normally 10 m/s² ,for a body in free fall the initial velocity (u) is 0.
Considering an equation of linear motion S= 1/2at² + ut  but a=g and u=0
Therefore, S= 1/2gt², h (2.65 t³)= S = 2.65 ×1.55³ = 9.868
                t² = 2S/g
                t² = 9.868× 2/ 10
                   = 1.9736
                t = √1.9736
Hence, the time taken will be equivalent to the sqrt of 1.9736 (√1.9736) which is equal to 1.405 seconds.

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An object accelerates from rest to 85m/s over a distance of 36m. What acceleration did it experience?
Marizza181 [45]
Suvat
we have s, u, v and we want a
the suvat equation with these values in is: v^2 = u^2 - 2as
so a = (-v^2 + u^2)/-2s 
plug numbers in
a = (-85^2 + 0^2)/-2*36 = 7225/72 = 100.3... ms^-2
6 0
3 years ago
A pendulum is made up of a small sphere of mass 0.500 kg attached to a string of length 0.950 m. The sphere is swinging back and
Semenov [28]

Answer:

W = 0.842 J

Explanation:

To solve this exercise we can use the relationship between work and kinetic energy

         W = ΔK

In this case the kinetic energy at point A is zero since the system is stopped

         W = K_f                (1)

now let's use conservation of energy

starting point. Highest point A

          Em₀ = U = m g h

Final point. Lowest point B

         Em_f = K = ½ m v²

energy is conserved

         Em₀ = Em_f

         mg h = K

to find the height let's use trigonometry

at point A

            cos 35 = x / L

            x = L cos 35

so at the height is

            h = L - L cos 35

            h = L (1-cos 35)

we substitute

           K = m g L (1 -cos 35)

we substitute in equation 1

           W = m g L (1 -cos 35)

let's calculate

           W = 0.500 9.8 0.950 (1 - cos 35)

           W = 0.842 J

7 0
3 years ago
In the following diagram, the voltage is 1.5 volts and the resistance is 0.35 ohms. Use Ohm's Law to determine the current in th
Angelina_Jolie [31]

Answer:

I = 4.28 [amp]

Explanation:

To solve this type of problems we must have knowledge of the law of ohm, which tells us that the voltage is equal to the product of resistance by current.

Initial data:

v = 1.5 [volt]

R = 0.35 [ohms]

v = I * R

therefore:

I = 1.5 / 0.35

I = 4.28 [amp]

5 0
3 years ago
Why are nuclear power plants controversial
Sliva [168]
Because a nuclear meltdown can be caused if systems fail. 
6 0
3 years ago
A child slides down a snow‑covered slope on a sled. At the top of the slope, her mother gives her a push to start her off with a
kirill [66]

Answer:

θ = 13.16 °

Explanation:

Lets take mass of child = m

Initial velocity ,u= 1.1 m/s

Final velocity ,v=3.7 m/s

d= 22.5 m

The force due to gravity along the incline plane = m g sinθ

The friction force = (m g)/5

Now from work power energy

We know that

work done by all forces = change in kinetic energy

( m g sinθ - (m g)/5 ) d = 1/2 m v² - 1/2 m u²

(2  g sinθ - ( 2 g)/5 ) d = v² -  u²

take g = 10 m/s²

(20 sinθ - ( 20)/5 ) 22.5 = 3.7² -  1.1²

20 sinθ - 4 =12.48/22.5

θ = 13.16 °

5 0
3 years ago
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