Atoms gain the same electron configuration as noble gases by

A group of students launches a model rocket in the vertical direction. Based on tracking data, they determine that the altidude

Serjik [45]

Answer:

a) 252 ft/s

b) 1076 ft

Explanation:

The equation for motion for uniform acceleration (we can use it because the rocket is affected only by gravity) is as follows:

Y(t) = Y0 + V0 * t + 1/2 * a * t^2

Where

Y(t): altitude at a given time

Y0: initial altitude

V0: initial speed

a: acceleration, in this case -32.2 ft/s^2 (negative because gravity points down)

We set a 1 dimensional coordinate system with Y pointing up and the origin of coordinates at ground level.

We consider t=0 as the moment where powered flight ended (motor ran oou of fuel), at this moment the altitude was

Y(0) = 89.6 ft

Therefore:

Y0 = 89.6 ft

We also know that the rocket fell to ground 16 seconds later, therefore

Y(16) = 0 ft

So we can write

Y(t=16) = Y0 + V0 * t + 1/2 * a * t^2

V0 * t = Y(t=16) - Y0 - 1/2 * a * t^2

V0 =( Y(t=16) - Y0 - 1/2 * a * t^2 )/t

V0 =( 0 - 89.6 - 1/2 * (-32.2) * 16^2 )/16 = 252 ft/s

In the highest point of flight the rocket will have a speed = 0

The first derivative of the equation of motion is the equation of speed:

V(t) = V0 + a * t

If we equate this to zero we eill find the time at which the rocket achieved it's highest altitude.

0 = V0 + a * t

a * t = 0 - V0

t = -V0/a

t = -252/(-32.2) = 7.83 s

Now, we can take this time value andd plug it back into the position equation

Y(7.83) = 89.6 + 252 * 7.83 + 1/2 * (-32.2) * 7.83^2 = 1076 ft

6 0

3 years ago

Oxygen is a reactant in a combination reaction, which always produces energy in the form of heat and light. A. True B. False

harkovskaia [24]

The answer is B.False

8 0

3 years ago

Read 2 more answers

Consider a bicycle wheel to be a ring of radius 30 cm and mass 1.5 kg. Neglect the mass of the axle and sprocket. If a force of

Charra [1.4K]

To solve the problem it is necessary to apply the Torque equations and their respective definitions.

The Torque is defined as,

$\tau = I \alpha$

Where,

I=Inertial Moment

$\alpha =$ Angular acceleration

Also Torque with linear equation is defined as,

$\tau = F*d$

Where,

F = Force

d= distance

Our dates are given as,

R = 30 cm = 0.3m

m = 1.5 kg

F = 20 N

r = 4.0 cm = 0.04 m

t = 4.0s

Therefore matching two equation we have that,

$d*F = I\alpha$

For a wheel the moment inertia is defined as,

I= mR2, replacing we have

$d*F= \frac{mR^2a}{R}$

$d*F= mRa$

$a = \frac{rF}{ mR}$

$a = \frac{0.04*20}{1.5*0.3}$

$a=1.77 m/s^2$

Then the velocity of the wheel is

$V = a *t \\V=1.77*4 \\V=7.11 m/s$

Therefore the correct answer is D.

4 0

3 years ago

The magnitude of the gravitational acceleration at the surface of a planet of mass M and radius R is g. What is the magnitude of

maksim [4K]

2M/4R2=.5M/R2= .0.5 g

4 0

3 years ago

What frequency is received by a stationary mouse just before being dispatched by a hawk flying at it at 24.7 m/s and emitting a

Montano1993 [528]

Answer:

f' = 3665.51 Hz

Explanation:

given,

speed of the hawk = 24.7 m/s

frequency of screech emitted by the hawk = 3400 Hz

speed of sound = 331 m/s

By Doppler's effect

$f' = (\dfrac{v}{v-v_s}})f$

f' is the frequency received by the mouse

v is the speed of the sound

v_s is the speed of the hawk

now,

$f' = \dfrac{341}{341-24.7}}\times 3400$

f' = 1.078 x 3400

f' = 3665.51 Hz

The frequency received by the stationary mouse is equal to 3665.51 Hz

6 0

3 years ago