Which constants and directions always apply when using gravity?

What was one result of the second great awakening

Brrunno [24]

The membership rose among the baptist and methodists.

4 0

4 years ago

A hot-air balloon of diameter 12 mm rises vertically at a constant speed of 14 m/s. A passenger accidentally drops his camera fr

fgiga [73]

Answer:

<em>The balloon is 66.62 m high</em>

Explanation:

<u>Combined Motion </u>

The problem has a combination of constant-speed motion and vertical launch. The hot-air balloon is rising at a constant speed of 14 m/s. When the camera is dropped, it initially has the same speed as the balloon (vo=14 m/s). The camera has an upward movement for some time until it runs out of speed. Then, it falls to the ground. The height of an object that was launched from an initial height yo and speed vo is

$\displaystyle y=y_o+v_o\ t-\frac{g\ t^2}{2}$

The values are

$\displaystyle y_o=15\ m$

$\displaystyle v_o=14\ m/s$

We must find the values of t such that the height of the camera is 0 (when it hits the ground)

$\displaystyle y=0$

$\displaystyle y_o+v_o\ t-\frac{g\ t^2}{2}=0$

Multiplying by 2

$\displaystyle 2y_o+2v_ot-gt^2=0$

Clearing the coefficient of $t^2$

$\displaystyle t^2-\frac{2\ V_o}{g}\ t-\frac{2\ y_o}{g}=0$

Plugging in the given values, we reach to a second-degree equation

$\displaystyle t^2-2.857t-3.061=0$

The equation has two roots, but we only keep the positive root

$\displaystyle \boxed {t=3.69\ s}$

Once we know the time of flight of the camera, we use it to know the height of the balloon. The balloon has a constant speed vr and it already was 15 m high, thus the new height is

$\displaystyle Y_r=15+V_r.t$

$\displaystyle Y_r=15+14\times3.69$

$\displaystyle \boxed{Y_r=66.62\ m}$

3 0

3 years ago

In the Olympic shot-put event, an athlete throws the shot with an initial speed of 12.0 m/s at a 40.0 ∘ angle from the horizonta

Gre4nikov [31]

Answer:

Explanation:

given,

initial speed of the shot = 12.0 m/s

angle = 40°

height at which shot leaves her hand = 1.80 m

v_x = 12 cos 40° = 9.19 m/s

v_y = 12 sin 40° = 7.71 m/s

time to reach maximum height =

= $\dfrac{v_y}{g}$

= $\dfrac{7.71}{9.8}$

= 0.787 s

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

h = 7.71 × 0.787 - 0.5 × 9.81 × 0.787²

h = 3.03 m

the maximum height attain = 3.03 + 1.8 = 4.83 m

now free fall from the maximum height

$h =\dfrac{1}{2}gt^2$

$4.83 = \dfrac{1}{2}\times 9.8 \times t^2$

t = 0.9928 s

total time = 0.9928 + 0.787 = 1.7798 s

range =

d = vₓ t

d = 16.36 m

8 0

4 years ago

What type of chemical equation is represented in the following 2KCl + Pb(NO3)2 → 2 KNO3 + PbCl2

UNO [17]

If your talking about the chemical equations like combustion, single displacement, etc. Then the equation would be double displacement:
AB + CD ---> AD + CB

6 0

3 years ago

Read 2 more answers

To practice Problem-Solving Strategy 27.1: Magnetic Forces. A particle with mass 1.81×10−3 kgkg and a charge of 1.22×10−8 CC has

Roman55 [17]

Answer:

The magnitude of the acceleration is $a = 0.33 m/s^2$