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3 years ago

I use 1 5/7 cans of paint to paint 7/8 of my room. How many cans of paint will I need to paint the entire room?

Physics

1 answer:

sergejj [24]3 years ago

8 0

Answer:

y = 2.45 x

Explanation:

given,

Let 'x' be the number of cans.

and 'y' be the total area of the room

now,

From the given statement in the question

$\dfrac{15}{7}\times x = \dfrac{7}{8}\times y$

$y = \dfrac{15}{7}\times \dfrac{8}{7} \times x$

y = 2.45 x

From the Above equation we can state that

To pain the entire room 2.45 cans of paint is required.

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Friction between solids can be most accurately defined as the force that opposes the sliding motion of two _____that are touchin

OlgaM077 [116]

Answer:

surfaces

Explanation:

3 0

3 years ago

A pendulum consists of a 2.0 kg stone swinging on a4.0 m string of negligible mass. The stone has a speed of 8.0 m/swhen it pass

arlik [135]

Answer:

a) $v_{60^{o}} =4.98 m/s$

b) $\theta_{max}=79.34^{o}$

Explanation:

This problem can be solved by doing an energy analysis on the given situation. So the very first thing we can do in order to solve this is to draw a diagram of the situation. (see attached picture)

So, in an energy analysis, basically you will always have the same amount of energy in any position of the pendulum. (This is in ideal conditions) So in this case:

$K_{lowest}+U_{lowest}=K_{60^{o}}+U_{60^{0}}$

where K is the kinetic energy and U is the potential energy.

We know the potential energy at the lowest of its trajectory will be zero because it will have a relative height of zero. So the equation simplifies to:

$K_{lowest}=K_{60^{o}}+U_{60^{0}}$

So now, we can substitute the respective equations for kinetic and potential energy so we get:

$\frac{1}{2}mv_{lowest}^{2}=\frac{1}{2}mv_{60^{o}}^{2}+mgh_{60^{o}}$

we can divide both sides of the equation into the mass of the pendulum so we get:

$\frac{1}{2}v_{lowest}^{2}=\frac{1}{2}v_{60^{o}}^{2}+gh_{60^{o}}$

and we can multiply both sides of the equation by 2 to get:

$v_{lowest}^{2}=v_{60^{o}}^{2}+2gh_{60^{o}}$

so we can solve this for $v_{60^{o}}$ . So we get:

$v_{60^{o}}=\sqrt{v_{lowest}^{2}-2gh_{60^{0}}}$

so we just need to find the height of the stone when the pendulum is at a 60 degree angle from the vertical. We can do this with the cos function. First, we find the vertical distance from the axis of the pendulum to the height of the stone when the angle is 60°. We will call this distance y. So:

$cos \theta = \frac{y}{4m}$

so we solve for y to get:

$y = 4cos \theta$

so we substitute the angle to get:

y=4cos 60°

y=2 m

so now we can find the height of the stone when the angle is 60°

$h_{60^{o}}=4m-2m$

$h_{60^{o}}=2m$

So now we can substitute the data in the velocity equation we got before:

$v_{60^{o}}=\sqrt{v_{lowest}^{2}-2gh_{60^{0}}}$

$v_{60^{o}} = \sqrt{(8 m/s)^{2}-2(9.81 m/s^{2})(2m)}$

$v_{60^{o}}=4.98 m/s$

b) For part b, we can do an energy analysis again to figure out what the height of the stone is at its maximum height, so we get.

$K_{lowest}+U_{lowest}=K_{max}+U_{max}$

In this case, we know that $U_{lowest}$ will be zero and $K_{max}$ will be zero as well since at the maximum point, the velocity will be zero.

So this simplifies our equation.

$K_{lowest} =U_{max}$

And now we substitute for the respective kinetic energy and potential energy equations.

$\frac{1}{2}mv_{lowest}^{2}=mgh_{max}$

again, we can divide both sides of the equation into the mass, so we get:

$\frac{1}{2}v_{lowest}^{2}=gh_{max}$

and solve for the height:

$h_{max}=\frac{v_{lowest}^{2}}{2g}$

and substitute:

$h_{max}=\frac{(8m/s)^{2}}{2(9.81 m/s^{2})}$

to get:

$h_{max}=3.26m$

This way we can find the distance between the axis and the maximum height to determine the angle of the pendulum about the vertical.

y=4-3.26 = 0.74m

next, we can use the cos function to find the max angle with the vertical.

$cos \theta_{max}= \frac{0.74}{4}$

$\theta_{max}=cos^{-1}(\frac{0.74}{4})$

so we get:

$\theta_{max}=79.34^{o}$

5 0

3 years ago

The formula v = √ 2.3 r models the maximum safe speed, v , in miles per hour, at which a car can travel on a curved road with ra

Archy [21]

Answer:

562 miles per hour.

Explanation:

As given in the question, the formula for the maximum speed on a curved road is

$v=\sqrt{2.3}$ $r$

Given value of $r=370$ feet

So the maximum safe speed will be

$v=\sqrt{2.3} \times 370 = 1.52\times 370 = 562.4$ miles per hour.

Rounding off to the nearest whole number we get the maximum safe speed at the curved road is 562 miles per hour.

6 0

4 years ago

Cam Newton can sprint 40 meters in 5.79 seconds! How fast can he run?<br> Show your work

Setler79 [48]

Speed=Distance/Time

Distance=40m,time=5.79seconds

S=40/5.79

=6.908m/s

6 0

3 years ago

Read 2 more answers

What type of foods do bodyguards eat give 5 examples

Rainbow [258]

They eat rice, chicken, beef, cereal they eat what we eat in a daily basis I guess

8 0

2 years ago