Ask question

Ask question

All categories

3 years ago

15

A hot-air balloon is descending at a rate of 2.3 m/s when a pas- senger drops a camera. If the camera is 41 m above the ground w

hen it is dropped, (a) how much time does it take for the cam- era to reach the ground, and (b) what is its velocity just before it lands? Let upward be the positive direction for this problem.

1 answer:

monitta3 years ago

3 0

Answer:

a) time taken = 2.66 s

b) v = 28.34

Explanation:

given,

rate of descending = 2.3 m/s

height of camera above ground = 41 m

using equation of motion

$h = u t + \dfrac{1}{2}gt^2$

$41 =2.3t + \dfrac{1}{2}\times 9.8\times t^2$

4.9 t² + 2.3 t - 41 =0

t = 2.66 ,-3.13

time taken = 2.66 s

b) v² = u² + 2 g h

v² = 0 + 2× 9.8 × 41

v = 28.34

You might be interested in

If vector C is added to vector D, the result is a third vector that is perpendicular to D and has a magnitude equal to 3D. What

Jet001 [13]

Answer:

(e) 3.2

Explanation:

We are given that vector C and D.

Let R be the magnitude of C+D.

According to question

R=3D

We have to find the ratio of the magnitude of C to that of D.

By using right triangle property

$C^2=R^2+D^2$

$C^2=(3D)^2+D^2$

$C^2=9D^2+D^2$

$C^2=10D^2$

$C=\sqrt{10D^2}=3.2D$

$\frac{C}{D}=3.2$

Hence, the ratio of the magnitude of C to that of D=3.2

(e) 3.2

5 0

3 years ago

A single polarizer will stop _____ of the incoming light.

True [87]

A single polarizer will stop 50% of the incoming light.

8 0

3 years ago

Read 2 more answers

Why do we get dizzy when we spin?

iris [78.8K]

The body senses whether it is upright or lying down or whether it is moving or standing still through the vestibular system, which is in the upper portion of the inner ear.

3 0

3 years ago

A ball is thrown directly downward with an initial speed of 7.70 m/s, from a height of 30.2 m. After what time interval does it

KatRina [158]

Answer:

$t = 1.82$

Explanation:

Given

$u = 7.70m/s$ -- initial velocity

$s = 30.2m$ --- height

Required

Determine the time to hit the ground

This will be solved using the following motion equation.

$s = ut + \frac{1}{2}gt^2$

Where

$g = 9,8m/s^2$

So, we have:

$30.2 = 7.70t + \frac{1}{2} * 9.8 * t^2$

$30.2 = 7.70t + 4.9 * t^2$

Subtract 30.2 from both sides

$30.2 -30.2 = 7.70t + 4.9 * t^2 - 30.2$

$0 = 7.70t + 4.9 * t^2 - 30.2$

$0 = 7.70t + 4.9t^2 - 30.2$

$7.70t + 4.9t^2 - 30.2 = 0$

$4.9t^2 + 7.70t - 30.2 = 0$

Solve using quadratic formula:

$t = \frac{-b\±\sqrt{b^2 - 4ac}}{2a}$

Where

$a = 4.9;\ b = 7.70;\ c = -30.2$

$t = \frac{-7.70\±\sqrt{7.70^2 - 4*4.9*-30.2}}{2*4.9}$

$t = \frac{-7.70\±\sqrt{651.21}}{9.8}$

$t = \frac{-7.70\±25.52}{9.8}$

Split the expression

$t = \frac{-7.70+25.52}{9.8}$ or $t = \frac{-7.70-25.52}{9.8}$

$t = \frac{17.82}{9.8}$ or $t = -\frac{33.22}{9.8}$

Time can't be negative; So, we have:

$t = \frac{17.82}{9.8}$

$t = 1.82$

Hence, the time to hit the ground is 1.82 seconds

7 0

3 years ago

Mumz [18]

No se ha da han dicho nada más de lo dicho y han ido de vuelta y han dicho nada más de que se pueda hacer el favor del niño

4 0

3 years ago

Read 2 more answers