Sarah walks WEST a distance of 5 meters then turns NORTH and walks 8

You collect some more data on that horse at a later time interval, but now you are measuring thehorse’s velocity, not its positi

Monica [59]

Answer:

a) x(t) = 10t + (2/3)*t^3

b) x*(0.1875) = 10.18 m

Explanation:

Note: The position of the horse is x = 2m. There is a typing error in the question. Otherwise, The solution to cubic equation holds a negative value of time t.

Given:

- v(t) = 10 + 2*t^2 (radar gun)

- x*(t) = 10 + 5t^2 + 3t^3 (our coordinate)

Find:

-The position x of horse as a function of time t in radar system.

-The position of the horse at x = 2m in our coordinate system

Solution:

- The position of horse according to radar gun:

v(t) = dx / dt = 10 + 2*t^2

- Separate variables:

dx = (10 + 2*t^2).dt

- Integrate over interval x = 0 @ t= 0

x(t) = 10t + (2/3)*t^3

- time @ x = 2 :

2 = 10t + (2/3)*t^3

0 = 10t + (2/3)*t^3 + 2

- solve for t:

t = 0.1875 s

- Evaluate x* at t = 0.1875 s

x*(0.1875) = 10 + 5(0.1875)^2 + 3(0.1875)^3

x*(0.1875) = 10.18 m

3 0

3 years ago

3.25 kcal is the same amount of energy as

PIT_PIT [208]

8 0

3 years ago

Read 2 more answers

An object is located 50 cm from a converging lens having a focal length of 15 cm. Which of the following is true regarding the i

crimeas [40]

Answer:

It is real, inverted, and smaller than the object.

Explanation:

Let's start by using the lens equation to find the location of the image:

$\frac{1}{q}=\frac{1}{f}-\frac{1}{p}$

where we have:

q = ? is the distance of the image from the lens

f = 15 cm is the focal length (positive for a converging lens)

p = 50 cm is the distance of the object from the lens

Solving the equation for q, we find

$\frac{1}{q}=\frac{1}{15 cm}-\frac{1}{50 cm}=0.047 cm^{-1}$

$q=\frac{1}{0.047 cm^{-1}}=+21.3 cm$

The sign of q is positive, so the image is real.

Now let's also write the magnification equation:

$h_i = - h_o \frac{q}{p}$

where

$h_i, h_o$ are the size of the image and of the object

By substituting p = 50 cm and q = 21.3 cm, we find

$h_i = - h_o \frac{21.3 cm}{50 cm}=-0.43 h_o$

So we notice that:

$|h_i| < |h_o|$ : this means that the image is smaller than the object

$h_i < 0$ : this means that the image is inverted

so, the correct option is:

It is real, inverted, and smaller than the object.

7 0

2 years ago

Costal residents must do many things to prepare for hurricanes

tangare [24]

The answer is A it’s more safer that way

3 0

2 years ago

Two 4.3546 cm x 4.3546 cm square aluminum electrodes, spaced 0.6408 mm apart are connected to a 73.68 V battery. What is the cap

Helen [10]

Answer:

Capacitance, C = 26.1 picofarad

Explanation:

It is given that,

Side of square, x = 4.3546 cm = 0.043546 m

Distance between electrodes, d = 0.6408 mm = 0.0006408 m

Voltage, V = 73.68 V

Capacitance of parallel plates is given by :

$C=\dfrac{\epsilon_oA}{d}$

$C=\dfrac{8.85\times 10^{-12}\times (0.043546)^2}{0.0006408}$

$C=2.61\times 10^{-11}\ F$

or

C = 26.1 picofarad

So, the capacitance of the capacitor is 26.1 picofarad. Hence, this is the required solution.

3 0

3 years ago