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3 years ago

The mechanical advantage of a wheel and axle is the radius of the wheel divided by the radius of the axle. TRUE or FALSE.

Physics

1 answer:

viva [34]3 years ago

4 0

True! The mechanical advantage of the wheel and axle is equal to the ratio of the radius of the wheel over the radius of the axle.

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The lightest car in the world was built in London and had a mass of less than 10 kg. it's maximum speed was 25.0 km/h. Suppose t

Softa [21]

25km/h = 6.94 m/s
suvat
s=16
u=6.94
v=0
a=a
v^2=u^2+2as
(v^2-u^2)/2s = a =1.5ms^-2

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3 years ago

A model rocket is shot directly upward, rises to its maximum height and then returns to its launch position in 10.0 s. Assuming

worty [1.4K]

d. 49.0 m/s

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3 years ago

A golfer hits a 42 g ball, which comes down on a tree root and bounces straight up with an initial speed of 15.6 m/s. Determine

nevsk [136]

Answer:

12.2 m

Explanation:

Given:

v₀ = 15.6 m/s

v = 0 m/s

a = -10 m/s²

Find: Δy

v² = v₀² + 2aΔy

(0 m/s)² = (15.6 m/s)² + 2 (-10 m/s²) Δy

Δy = 12.2 m

5 0

3 years ago

Read 2 more answers

Martije has made a slight error in naming a compound monocarbon tetrabromide. What compound is she most likely naming, and what

Alik [6]

Answer: $CBr_4$ : carbon tetrabromide

Explanation:

$CBr_4$ is a covalent compound because in this compound the sharing of electrons takes place between carbon and bromine. Both the elements are non-metals. Hence, it will form covalent bond.

The naming of covalent compound is given by:

1. The less electronegative element is written first.

2. The more electronegative element is written second. Then a suffix is added with it. The suffix added is '-ide'.

3. If atoms of an element is greater than 1, then prefixes are added which are 'mono' for 1 atom, 'di' for 2 atoms, 'tri' for 3 atoms and so on.

Hence, the correct name for $CBr_4$ is carbon tetrabromide.

7 0

3 years ago

Read 2 more answers

A 2 eV electron encounters a barrier 5.0 eV high and width a. What is the probability b) 0.5 that it will tunnel through the bar

denis23 [38]

Answer:

The tunnel probability for 0.5 nm and 1.00 nm are $5.45\times10^{-4}$ and $7.74\times10^{-8}$ respectively.

Explanation:

Given that,

Energy E = 2 eV

Barrier V₀= 5.0 eV

Width = 1.00 nm

We need to calculate the value of $\beta$

Using formula of $\beta$

$\beta=\sqrt{\dfrac{2m}{\dfrac{h}{2\pi}}(v_{0}-E)}$

Put the value into the formula

$\beta = \sqrt{\dfrac{2\times9.1\times10^{-31}}{(1.055\times10^{-34})^2}(5.0-2)\times1.6\times10^{-19}}$

$\beta=8.86\times10^{9}$

(a). We need to calculate the tunnel probability for width 0.5 nm

Using formula of tunnel barrier

$T=\dfrac{16E(V_{0}-E)}{V_{0}^2}e^{-2\beta a}$

Put the value into the formula

$T=\dfrac{16\times 2(5.0-2.0)}{5.0^2}e^{-2\times8.86\times10^{9}\times0.5\times10^{-9}}$

$T=5.45\times10^{-4}$

(b). We need to calculate the tunnel probability for width 1.00 nm

$T=\dfrac{16\times 2(5.0-2.0)}{5.0^2}e^{-2\times8.86\times10^{9}\times1.00\times10^{-9}}$

$T=7.74\times10^{-8}$

Hence, The tunnel probability for 0.5 nm and 1.00 nm are $5.45\times10^{-4}$ and $7.74\times10^{-8}$ respectively.

6 0

3 years ago