T=2π√(l/g) is dimensionally correct

9.) The fastest land animal, the cheetah, can accelerate from 0 m/s to 33 m/s in 3 seconds. What is the cheetah's acceleration?

Montano1993 [528]

Answer:

11 m/s

Explanation:

8 0

4 years ago

Helen can write 15 postcards in one hour, while Kate can write

Akimi4 [234]

Answer:5 hours

Explanation:

300 divided by 20 = 15

300 divided by 15 = 20

the difference is 5

4 0

3 years ago

one car accelerates at half the rate of another how much longer does it take the first car to travel a quarter mile

Orlov [11]

Answer:

t=1/4v1

Explanation:

Given data

Car one

Speed =v1

Time =t

Distance =1/4 mile

Given data

Car two

Speed =v1/2

Time =t

Distance =d

Speed =distance/time

v1=1/4/t

v1t=1/4

t=1/4*1/v1

t=1/4v1 seconds

7 0

3 years ago

At sunset, red light travels horizontally through the doorway in the western wall of your beach cabin, and you observe the light

Nady [450]

Answer:

$9.8\cdot 10^{-6}m$

Explanation:

For light passing through a single slit, the position of the nth-minimum from the central bright fringe in the diffraction pattern is given by

$y=\frac{n \lambda D}{d}$

where

$\lambda$ is the wavelength

D is the distance of the screen from the slit

d is the width of the slit

In this problem, we have

$\lambda=700 nm = 7.00\cdot 10^{-7}m$ is the wavelength of the red light

D = 14 m is the distance of the screen from the doorway

d = 1.0 m is the width of the doorway

Substituting n=1 into the equation, we find the distance between the central bright fringe and the first-order dark fringe (the first minimum):

$y=\frac{(1)(7.00\cdot 10^{-7} m)(14 m)}{1.0 m}=9.8\cdot 10^{-6}m$

6 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:

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