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dolphi86 [110]
3 years ago
9

A fire helicopter carries a 700-kg bucket of water at the end of a 20.0-m long cable. Flying back from a fire at a constant spee

d of 40.0 m/s, the cable makes an angle of 38.0° with respect to the vertical. Determine the force exerted by air resistance on the bucket.
Physics
1 answer:
rodikova [14]3 years ago
5 0

Answer:

F = 50636.873 N

Explanation:

given,

bucket of water =  700-kg

length of cable = 20 m

Speed  = 40 m/s

angle of the cable = 38.0°

let air resistance be = F

tension in rope be = T

T cos 38° = m×g..................(1)

T sin 38^0= \dfrac{mv^2}{l} + F..........(2)

equation (1)/(2)

tan 38^0 =\dfrac{\dfrac{mv^2}{l} + F}{mg}

       0.781=\dfrac{\dfrac{700\times 40^2}{20} + F}{700\times 9.8}

           F = 50636.873 N

Hence the force exerted on the bucket is equal to F = 50636.873 N

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hodyreva [135]

Answer:

d = 1700 meters

Explanation:

During a rainy day, as a result of colliding clouds an observer saw lighting and a heard thunder sound.  The time between seeing the lighting and hearing the sound was 5 second, t = 5 seconds

Speed of sound, v = 340 m/s (say)

Let d is the distance of the colliding cloud from the observer. The distance covered by the object. It is given by :

d=v\times t

d=340\ m/s\times 5\ s  

d = 1700 meters

So, the distance of the colliding cloud from the observer is 1700 meters. Hence, this is the required solution.    

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3 years ago
Ponds and lakes are part of marine ecosystem.<br> True or false ?
finlep [7]

Answer:

True

Explanation:

because they can hold marine organism inside

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3 years ago
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Consider a car travelling at 60 km/hr. If the radius of a tire is 25 cm, calculate the angular speed of a point on the outer edg
vlabodo [156]

To solve this problem it is necessary to apply the concepts given in the kinematic equations of movement description.

From the perspective of angular movement, we find the relationship with the tangential movement of velocity through

\omega = \frac{v}{R}

Where,

\omega =Angular velocity

v = Lineal Velocity

R = Radius

At the same time we know that the acceleration is given as the change of speed in a fraction of the time, that is

\alpha = \frac{\omega}{t}

Where

\alpha =Angular acceleration

\omega = Angular velocity

t = Time

Our values are

v = 60\frac{km}{h} (\frac{1h}{3600s})(\frac{1000m}{1km})

v = 16.67m/s

r = 0.25m

t=6s

Replacing at the previous equation we have that the angular velocity is

\omega = \frac{v}{R}

\omega = \frac{ 16.67}{0.25}

\omega = 66.67rad/s

Therefore the angular speed of a point on the outer edge of the tires is 66.67rad/s

At the same time the angular acceleration would be

\alpha = \frac{\omega}{t}

\alpha = \frac{66.67}{6}

\alpha = 11.11rad/s^2

Therefore the angular acceleration of a point on the outer edge of the tires is 11.11rad/s^2

5 0
3 years ago
In a $100$ meter track event, Alice runs at a constant speed and crosses the finish line $5$ seconds before Beatrice does. If it
Liono4ka [1.6K]

Answer:

10s

Explanation:

If it took Beatrice 25 seconds to complete the race

Distance = 100 meter

Beatrice speed = 100/25

                          = 4m/s

If Alice runs at a constant speed and crosses the finish line $5$ seconds, she must have completed the race in 20s (25 -5).

Her speed where constant

= 100/20

= 5 m/s

It would take Alice

= 50/5

= 10s

It would take Alice 10s to run $50$ meters.

5 0
3 years ago
A solid cylinder is released from the top of an inclined plane of height 0.81 m. From what height, in meters, on the incline sho
Jlenok [28]

Answer:

same 0.81m

Explanation:

in this problem if we assume there no resistance of any sort. and we apply the energy conservation

change in Potential energy = change in kinetic energy

mgh = 0.5mv^2

gh = 0.5v^2

the above relation suggests that the speed at the bottom is only depending on the height it is released from not on the shape, mass or radius.

so at the bottom

put h = 0.81m

9.81 * 0.81 * 2 = v^2

v=3.99 m/s

both CYLINDER and SPHERE will have same velocity at the bottom if released from the same height irrespective of shape and size

3 0
3 years ago
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