An isotope of carbon has 6 protons and 8 neutrons.

A skier starting from rest skis straight down a slope 50 meters long in 5.0 seconds. What is the magnitude of the acceleration.

3241004551 [841]

His average speed all the way is

   (50 meters) / (5 sec) = 10 m/s .

But, if the acceleration is uniform all the way down, then

Average speed = (1/2) (start speed + end speed)

Start speed = 0.
So
   Average speed = (1/2) (0 + end speed)

   = (1/2) of end speed .

   10 m/s = (1/2) of end speed

   End speed (at the bottom) = 20 m/s .

Magnitude of acceleration = (change in speed) / (time for the change)

      = (20 m/s) / (5 sec)

   =    4 m/s² .

6 0

3 years ago

PLEASE HELP I AM ON A TIMED QUIZ

White raven [17]

Answer:

d

Explanation:

I believe its d I thinm

3 0

3 years ago

Read 2 more answers

Which statements best describe science? Check all that apply. A)Science uses beliefs and opinions to construct explanations. B)S

Maru [420]

Answer:

Science is supported by facts and processes.

Science involves observation and experimentation.

Science continually changes and is constantly updated.

6 0

3 years ago

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Kathy 82 kg performer standing on a diving board at the carnival dive straight down into a small pool of water. Just before stri

mixas84 [53]

Solution :

Given weight of Kathy = 82 kg

Her speed before striking the water, $V_o$ = 5.50 m/s

Her speed after entering the water, $V_f$ = 1.1 m/s

Time = 1.65 s

Using equation of impulse,

$dP = F \times dT$

Here, F = the force ,

dT = time interval over which the force is applied for

= 1.65 s

dP = change in momentum

dP = m x dV

$= m \times [V_f - V_o]$

= 82 x (1.1 - 5.5)

= -360 kg

∴ the net force acting will be

$F=\frac{dP}{dT}$

$F=\frac{-360}{1.65}$

= 218 N

8 0

3 years ago

a projectile is fired in such away that its horizontal range is equal to three times its naximum height.what is the angle of pro

finlep [7]

Answer:

$\theta=53.13^o$

Explanation:

<u>2-D Projectile Motion</u>

In 2-D motion, there are two separate components of the acceleration, velocity and displacement. The horizontal component has zero acceleration, while the acceleration in the vertical direction is always the acceleration due to gravity. The basic formulas for this type of movement are

$V_x=V_{ox}=V_ocos\theta$

$V_y=V_{oy}-gt=V_osin\theta-gt$

$x=V_{ox}t$

$\displaystyle y=y_o+V_{oy}t-\frac{gt^2}{2}$

$\displaystyle x_{max}=\frac{2V_{ox}V_{oy}}{g}$

$\displaystyle y_{max}=\frac{V_{oy}^2}{2g}$

The projectile is fired in such a way that its horizontal range is equal to three times its maximum height. We need to find the angle \theta at which the object should be launched. The range is the maximum horizontal distance reached by the projectile, so we establish the base condition:

$x_{max}=3y_{max}$