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A spotlight on the ground shines on a wall 12 m away. If a man 2 m tall walks from the spotlight toward the building at a speed

mafiozo [28]

Answer:

0.6375 m/s

Explanation:

Let x be the distance of the man from the building

from the figure attached

initially the value of x=12

Given:

$\frac{dx}{dt}=-1.7m/s$

where the negative sign depicts that the distance of the man from the building is decreasing.

Now, Let The length of the shadow be = y

we have to calculate $\frac{dy}{dt}$ when x=4

from the similar triangles

we have,

$\frac{2}{12-x}=\frac{y}{12}$

or

$y=\frac{24}{12-x}$

Differentiating with respect to time 't' we get

$\frac{dy}{dt}=-\frac{24}{12-x}^2\frac{-dx}{dt}$

or

$\frac{dy}{dt}=\frac{24}{12-x}^2\frac{dx}{dt}$

Now for x = 4, and $\frac{dx}{dt}=-1.7m/s$ we have,

$\frac{dy}{dt}=\frac{24}{12-4}^2\times (-1.7)$

or

$\frac{dy}{dt}=-0.6375m/s$

here, the negative sign depicts the decrease in length and in the question it is asked the decreasing rate thus, the answer is 0.6375m/s

4 0

3 years ago

The rate at which a force displaces a mass a horizontal distance is measured in?

kodGreya [7K]

Answer:

c. joules

Explanation:

The rate at which a force displaces a mass a horizontal distance is measured in joules. In science, this phenomenon is known as work done.

Work done can be defined as the rate at which a force acting on an object or a body causes it to experience a displacement. The work done is a scalar quantity and is measured in joules (J).

Mathematically, work done is given by the formula;

Work done = force * distance

$W = F * d$

Where,

W is the work done

F represents the force acting on a body.

d represents the distance covered by the body.

For example, a bull pulling a plough through a farm, a girl pushing a shopping cart down the aisle of a supermarket etc.

6 0

3 years ago

What is the kinetic energy of a 10kg object that is moving with a speed of 60m/s.

Sergio039 [100]

The answer is 18000 J

I hope this helps!^^ , if you need the work to be shown please tell me, I hope you have a great day!^^

5 0

3 years ago

Dos cargas Q1=2pc y Q2=4pc estan separadas por una distancia de 6cm ¿con que fuerza se atraen?

noname [10]

Here we can use coulomb's law to find the force between two charges

As per coulombs law

]tex]F = \frac{kq_1q_2}{r^2}[/tex]

here we have

$k = 9 * 10^9$

$q_1 = 2pC$

$q_2 = 4pC$

$r = 6cm = 0.06 m$

now by using the above equation we have

$F = \frac{9*10^9 * 2*10^{-12} * 4*10^{-12}}{0.06^2}$

$F = 2 * 10^{-11} N$

so here the force between two charges is of above magnitude and this will be repulsive force between them as both charges are of same sign.

3 0

3 years ago

Without sea otters, sea urchins would overgraze on kelp beds, dramatically changing the marine community. true/false

Bumek [7]

True is the anwser to your question
Hope this helps

5 0

3 years ago

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