Which statement best describes the force of gravity

A cart travels down a ramp at an average speed of 5.00 centimeters/second. What is the speed of the cart in miles/hour? (Remembe

viva [34]

Set up the problem with the conversion rates as fractions where when you multiply the units cancel out leaving the desired units behind.

$( \frac{ 5 cm}{1 sec} ) *( \frac{1m }{100cm})*( \frac{1km}{1000m} )*( \frac{1mi}{1.6km} ) = 0.00003125 mi/sec \\ \\ ( \frac{0.00003125 mi}{1sec} )*( \frac{60sec}{1min} )*( \frac{60min}{1hr} )=0.1125 mi/hr$

7 0

3 years ago

Read 2 more answers

If a car has a kinetic energy of 40000 J and is moving at a velocity of 25 m/s, what is the mass of the car? KE=1/2mv

ohaa [14]

Answer:

3200 Kg

Explanation:

Just apply the formula:

40.000 = 1/2 . m . 25

80.000 = 25m

m = 80.000/25

m = 3200 Kg

7 0

2 years ago

Both objects A and D represent fixed, negatively-charged particles of equal magnitude and Object B represents a fixed, positivel

Kipish [7]

Answer:

Option A = 1.

Explanation:

So, in order to solve this question we are given the Important infomation or data or parameters in the question above as;

(1). First, Both objects A and D represent fixed.

(2). Both objects A and D are negatively-charged particles of equal magnitude.

(3). "Object B represents a fixed, positively-charged particle (equal, but opposite charge from A and D)."

(4). "Object C shows a moving, positively-charged particle."

So, our mission is to determine the arrow that would correctly show the force of attraction or repulsion on object C caused by the other two objects.

We can do that by drawing out the forces of attraction and the resultants. Therefore, CHECK THE ATTACHED FILE/PICTURE FOR THE DRAWINGS.

The forces of attraction due to objects A and B on on object C will be towards themselves. Hence, the resultant is ONE(1).

6 0

3 years ago

Ml(d^2θ/dt^2) =-mgθ

Nata [24]

The equation of motion of a pendulum is:

$\dfrac{\textrm{d}^2\theta}{\textrm{d}t^2} = -\dfrac{g}{\ell}\sin\theta,$

where $\ell$ it its length and $g$ is the gravitational acceleration. Notice that the mass is absent from the equation! This is quite hard to solve, but for <em>small</em> angles ( $\theta \ll 1$ ), we can use:

$\sin\theta \simeq \theta.$

Additionally, let us define:

$\omega^2\equiv\dfrac{g}{\ell}.$

We can now write:

$\dfrac{\textrm{d}^2\theta}{\textrm{d}t^2} = -\omega^2\theta.$

The solution to this differential equation is:

$\theta(t) = A\sin(\omega t + \phi),$

where $A$ and $\phi$ are constants to be determined using the initial conditions. Notice that they will not have any influence on the period, since it is given simply by:

$T = \dfrac{2\pi}{\omega} = 2\pi\sqrt{\dfrac{g}{\ell}}.$

This justifies that the period depends only on the pendulum's length.

4 0

3 years ago

Two sinusoidal waves, which are identical except for a phase shift, travel along in the same direction. The wave equation of the

Y_Kistochka [10]

Answer:

two sinusoidal waves, which are identical except for a phase shift, travel along in the same direction. The wave equation of the resultant wave is yR (x, t) = 0.70 m sin⎛ ⎝3.00 m−1 x − 6.28 s−1 t + π/16 rad⎞ ⎠ . What are the angular frequency, wave number, amplitude, and phase shift of the individual waves?

ω = 6.28 s − 1 ,

k = 3.00 m− 1 ,

φ = π rad,

A R = 2 A cos (φ 2 ) ,

A = 0.37 m

Explanation:

y1 ( x , t ) = A sin( k x − ω t +φ ) ,

y 2 ( x , t ) = A sin ( k x − ω t ) .

from the principle of superposition which states that when two or more waves combine, there resultant wave is the algebriac sum of the individual waves

y1 ( x , t ) = A sin( k x − ω t +φ ) , is generaL form of thw wave eqaution

A=amplitude

k=angular wave number

ω=angular frequency

φ =phase constant

k=2π/lambda

ω=2π/T

yR (x, t) = 0.70 m sin{3.00 m−1 x − 6.28 s−1 t + π/16 rad}....................*

two waves superposed to give the above, assuming they are moving in the +x direction

y1 ( x , t ) = A sin( k x − ω t +φ ) , .....................1

y 2 ( x , t ) = A sin ( k x − ω t ) ...........................2

adding the two equation will give

A sin( k x − ω t +φ )+A sin ( k x − ω t ) .................3

A( sin( k x − ω t +φ )+ sin ( k x − ω t ) ),......................4

similar to the following trigonometry identity

sina+sinb=2cos(a-b)/2sin(a+b)/2

let a= ( k x − ω t

b=k x − ω t +φ )

y(x,t)=2Acos(φ/2)sin(k x − ω t +φ/2)

k=3m^-1

lambda=2π/k=2.09m

ω=6.28= T=2π/6.28

T=1s

φ/2=π/16

φ=π/8rad

amplitude

2Acos(φ/2)=0.70 m

A=0.7/2cos(π/8)

A=0.37 m

6 0

3 years ago