Most metals have...

A man starts walking north at 3 ft/s from a point P. Five minutes later a woman starts walking south at 4 ft/s from a point 500

mrs_skeptik [129]

Answer:

ds/dt = 6.98 ft/s

Explanation:

Given:

- The speed of man due north Vm = 3 ft/s

- The speed of woman due south Vw = 4 ft/s

- Woman starts walking 5 mins later than man

Find:

At what rate are the people moving apart 15 min after the woman starts walking?

Solution:

- The total time for which the man is walking due north from P, is Tm:

Tm = 5 + 15 = 20 mins

- The total distance traveled by man in Tm mins is:

Dm = Tm*Vm

Dm = 20*60*3

Dm = 3,600 ft

- The total time for which the woman is walking due south from 500 ft due east from P, is Tw:

Tw = 15 = 15 mins

- The total distance traveled by man in Tw mins is:

Dw = Tw*Vw

Dw = 15*60*4

Dw = 3,600 ft

- The displacement between man and woman at any instance is (s) which can be related by pythagoras theorem as follows:

s^2 = (dm + dw)^2 + 500^2

Where, dm : Distance travelled by man at any time Tm

dw : Distance travelled by woman at any time Tw

- Differentiate s with respect to t:

2s*ds/dt = 2*(dm + dw)*(Vm + Vw)

s*ds/dt = (dm + dw)*(Vm + Vw)

ds/dt = [ (dm + dw)*(Vm + Vw) ] / s

- Evaluate the rate of separation of man and woman ds/dt by evaluating at instance Tm = 20 mins and Tw = 15 mins. We have:

ds/dt = [ (Dm + Dw)*(Vm + Vw) ] / sqrt ( (Dm + Dw)^2 + 500^2 )

- Plug in the values:

ds/dt = [ (3600 + 3600)*(3 + 4) ] / [sqrt ( (3600 + 3600)^2 + 500^2 )]

ds/dt = 6.98 ft/s

7 0

2 years ago

An igneous rocks color is mainly determined by its

melomori [17]

An ignious rock's color is mainly determined by its silica content

4 0

3 years ago

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

A race car accelerates uniformly at 14.2 m/s2. If the race car starts from rest how fast will it

Kaylis [27]

Answer:

vf=94.4 m/s

Explanation:

acceleration is the final velocity minus initial velocity divided by time

a = (vf-vi)/t

Given:

a= 14.2 m/s^2

vi= 0 (at rest)

t = 6.6

Solve for vf

a = (vf-vi)/t

a*t=vf-vi

(14.2)*(6.6)=vf - 0

vf=94.4 m/s

6 0

2 years ago

What is the gravitational potential energy of a projectile with mass 5 kg at<br> a height of 10 m?

zimovet [89]

Answer:

Explanation:

The gravitational potential energy of a body can be found by using the formula

GPE = mgh

where

m is the mass

h is the height

g is the acceleration due to gravity which is 10 m/s²

From the question we have

GPE = 5 × 10 × 10

We have the final answer as

Hope this helps you

3 0

2 years ago