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alexandr402 [8]
3 years ago
14

A hot air balloon has just lifted off and is rising at the constant rate of 1.60 m/s. suddenly, one of the passengers realizes s

he has left her camera on the ground. a friend picks it up and tosses it straight upward with an initial speed of 10.2 m/s. if the passenger is 2.60 m above her friend when the camera is tossed, how much time does it take for the camera to reach her?

Physics
2 answers:
yKpoI14uk [10]3 years ago
8 0

It takes 0.388 seconds for the camera to reach her

<h3>Further explanation </h3>

These are the formulas that we have to remember before solving the problem.

Speed is the rate of change of distance.

\large { \boxed {v = \frac{d}{t}}}

<em>v = speed ( m/s )</em>

<em>d = distance ( m )</em>

<em>t = time ( s )</em>

Acceleration is the rate of change of velocity.

\large { \boxed {a = \frac{\Delta v}{t}}}

<em>a = acceleration ( m/s² )</em>

<em>Δv = change in speed ( m/s )</em>

<em>t = time ( s )</em>

Let us now tackle the problem!

<u>Given:</u>

Speed of Air Balloon = u = 1.60 m/s

Initial Speed of Camera = vo = 10.2 m/s

Initial Distance of Passenger and Camera = d = 2.60 m

Gravitational Acceleration = g = 9.80 m/s²

<u>Unknown:</u>

Time Required = t = ?

<u>Solution:</u>

When camera reaches her :

displacement of camera = 2.60 + displacement of passenger

v_o ~ t - \frac{1}{2} ~ g ~ t^2 = 2.60 + u ~ t

10.2 ~ t - \frac{1}{2} ~ 9.80 ~ t^2 = 2.60 + 1.60 ~ t

4.9 ~ t^2 - 8.60 ~ t + 2.60 = 0

To solve this quadratic equation, we can use the following formula:

t= \frac{-b - \sqrt{b^2-4ac}}{2a}

t = \frac{8.60 - \sqrt{(-8.60)^2 - 4(4.9)(2.60)} }{2(4.9)}

\boxed {t \approx 0.388 ~seconds}

<h3>Learn more</h3>
  • Velocity of A Runner : brainly.com/question/3813437
  • Kinetic Energy : brainly.com/question/692781
  • Acceleration : brainly.com/question/2283922

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Kinematics

Keywords: Speed , Distance , Acceleration , Time , Velocity , Jet , Plane , TakeOff , Runway

Wewaii [24]3 years ago
5 0

speed of balloon which is rising upwards with constant speed is

v_1 = 1.6 m/s

speed of the camera which is thrown up from the ground

v_2 = 10.2 m/s

the distance between camera and balloon when it is thrown from the ground

d = 2.60 m

now let say after time "t" the camera is catch by her friend in the balloon

So the position of balloon and camera must be same

y_{balloon} = y_{camera}

2.6 + 1.60 t = 10.2* t + \frac{1}{2}(-g)t^2

2.6 = 8.6 *t - 4.9 t^2

Solving above quadratic equation we got

t = 0.39 seconds

So after t = 0.39 s her friend will get her camera on the balloon.

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Explanation:

the tangential speed of the HST is given by

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In our problem, we know the tangential speed: v=7,750 m/s. The radius of the orbit is the sum of the Earth's radius and the distance of the HST above Earth's surface:

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So, we can re-arrange equation (1) to find the orbital period:

T=\frac{2 \pi r}{v}=\frac{2 \pi (6.95\cdot 10^6 m/s)}{7,750 m/s}=5,640 s

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A certain digital camera having a lens with focal length 7.50 cmcm focuses on an object 1.70 mm tall that is 4.70 mm from the le
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Answer:

a. 7.62cm

b. Real and inverted

c. 2.76 cm

d. 3450

Explanation:

We proceed as follows;

a. the lens equation that relates the object distance to the image distance with the focal length is given as follows;

1/f = 1/p + 1/q

making q the subject of the formula;

q = pf/p-f

From the question;

p = 4.70m

f = 7.5cm = 0.075m

Substituting these, we have ;

q = (4.7)(0.075)/(4.7-0.075) = 0.3525/4.625 = 0.0762 = 7.62 cm

b. The image is real and inverted since the image distance is positive

c. We want to calculate how tall the image is

Mathematically;

h1 = (q/p)h0

h1 = (7.62/4.70)* 1.7

h1 = 2.76 cm

d. We want to calculate the number of pixels that fit into this image

Mathematically:

n = h1/8 micro meter

n = 2.76cm/8 micro meter = 2.76 * 10^-2/8 * 10^-6 = 3450

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A vector → A has a magnitude of 56.0 m and points in a direction 30.0° below the negative x axis. A second vector, → B , has a m
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Answer:

  • The magnitude of the vector \vec{C} is 107.76 m

Explanation:

To find the components of the vectors we can use:

\vec{A} = | \vec{A} | \ ( \ cos(\theta) \ , \ sin (\theta) \ )

where | \vec{A} | is the magnitude of the vector, and θ is the angle over the positive x axis.

The negative x axis is displaced 180 ° over the positive x axis, so, we can take:

\vec{A} = 56.0 \ m \ ( \ cos( 180 \° + 30 \°) \ , \ sin (180 \° + 30 \°) \ )

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\vec{B} = 82.0 \ m \ ( \ cos( 180 \° - 49 \°) \ , \ sin (180 \° - 49 \°) \ )

\vec{B} = 82.0 \ m \ ( \ cos( 131 \°) \ , \ sin (131 \°) \ )

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Now, we can perform vector addition. Taking two vectors, the vector addition is performed:

(a_x,a_y) + (b_x,b_y) = (a_x+b_x,a_y+b_y)

So, for our vectors:

\vec{C} = ( \ -48.497 \ m \ , \ - 28 \ m \ ) + ( \ -53.797 \ m \ ,  ) = ( \ -48.497 \ m \ -53.797 \ m , \ - 28 \ m \ + \ 61.886\ m \ )

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To find the magnitude of this vector, we can use the Pythagorean Theorem

|\vec{C}| = \sqrt{C_x^2 + C_y^2}

|\vec{C}| = \sqrt{(- 102.294 \ m)^2 + (\ 33.886 m \)^2}

|\vec{C}| =107.76 m

And this is the magnitude we are looking for.

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