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3 years ago

12

What is the mechanical advantage of a wedge that is 2 inches at its widest part and has a sloped side with a length of 10 inches

?
a. 0.2b. 5c. 8d. 20

1 answer:

Alex_Xolod [135]3 years ago

6 0

Answer:

B) 5

Explanation:

Mechanical advantage = D/W

= 10/2

= 5.

It has no unit because the inches in the length and width will cancel out.

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The roller coaster car has a mass of 700 kg, including its passenger. If it is released from rest at the top of the hill A, dete

sweet [91]

Answer:

$h = 18.75 m$

Now when it will reach at point B then its normal force is just equal to ZERO

$N_B = 0$

$F_n = 1.72 \times 10^4$

Explanation:

Since we need to cross both the loops so least speed at the bottom must be

$v = \sqrt{5 R g}$

also by energy conservation this is gained by initial potential energy

$mgh = \frac{1}{2}mv^2$

$v = \sqrt{2gh}$

so we will have

$\sqrt{2gh} = \sqrt{5Rg}$

now we have

$h = \frac{5R}{2}$

here we have

R = 7.5 m

so we have

$h = \frac{5(7.5)}{2}$

$h = 18.75 m$

Now when it will reach at point B then its normal force is just equal to ZERO

$N_B = 0$

now when it reach point C then the speed will be

$mgh - mg(2R_c) = \frac{1}{2]mv_c^2$

$v_c^2 = 2g(h - 2R_c)$

$v_c = 13.1 m/s$

now normal force at point C is given as

$F_n = \frac{mv_c^2}{R_c} - mg$

$F_n = \frac{700\times 13.1^2}{5} - (700 \times 9.8)$

$F_n = 1.72 \times 10^4$

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3 years ago

In your own words explain how energy is involved in physical changes and in chemical changes

olasank [31]

Answer:

All adjustments in issue include changes in energy. Energy is either delivered or consumed. The energy is often in the form of heat, however it might be in the form of sound or light.

Explanation:

this is in my own words btw:)

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2 years ago

How do i solve the ones in red? A student fires a cannonball horizontally with a speed of 24.0m/s from a height of 51.0m. Neglec

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For the initial vertical speed will are considering the vertical component. Therefore,

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$\text{ Initial vertical velocity = 0 m/s}$

let's find how long the ball remained in the air.

$\begin{gathered} 0=51-\frac{1}{2}(9.8)t^2 \\ 4.9t^2=51 \\ t^2=\frac{51}{4.9} \\ t^2=10.4081632653 \\ t=\sqrt[]{10.4081632653} \\ t=3.22 \\ t=3.22\text{ s} \end{gathered}$

Finally, let's find the how far from the base of the building the ball landed(horizontal distance)

$\begin{gathered} S_x=u_xt+\frac{1}{2}a_xt^2 \\ S_x=24\times3.23 \\ S_x=\text{horizontal distance=77.28 m} \\ \text{note} \\ a_x=0m/s^2 \end{gathered}$

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1 year ago

By what angle should the second polarized sheet be rotated relative to the first to reduce the transmitted intensity to one-half

algol [13]

Answer:

θ = 45º

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½ I₀ = I₀ cos² θ

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θ = 45º

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docker41 [41]

Here are the answers to the following questions:

1. spiral

2. spiral

3. rapidly spinning protogalactic clouds

I am hoping that these answers have satisfied your queries and it will be able to help you in your endeavors, and if you would like, feel free to ask another question.

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3 years ago

Read 2 more answers