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4 years ago

a teacher pushed a 10kg desk across a floor for a distance of 5m. she exerted a horizontal force of 20n. how much work was done?

1 answer:

7 0

Work Done = Force x distance
Since she exerted a horizontal force of 20N over a distance of 5m, the work done is 20N x 5m which is equals to 100 joules

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suppose you want to determine the surface area of this sugar cube. it has edges that are each 2 cm long. if you cut the cube in

ivolga24 [154]

Answer:

Half: 6 cm^2 Whole: 12 cm^2

Explanation:

First, we know that the edges of the cube are 2 cm long. So there are 6 faces on a cube. We do 2x6=12 cm^2 as our total surface area. Then it asks for each half. So you would divide it by 2 and get 6 cm^2 as your half.

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3 years ago

Help, It has multiple answers for this question.

evablogger [386]

The answer is ultra violet radiation. From the air

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3 years ago

a hockey puck with a mass of 0.11 kg is at rest on the horizontal frictionless surface of the rink. a player applies a horizonta

stira [4]

The solution to this ques is available in the image.

Given,

Force= 1N

Mass= 0.11kg

Time= 5sec

Force= mass X accelaration

Accelaration= velocity/ time

Speed=distance/ time

Hence, the speed is 45 m/s and the distance is 225 m.

To know more about speed and distance problems the link is given below:

brainly.com/question/19610984?

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8 0

1 year ago

An ice sled powered by a rocket engine starts from rest on a large frozen lake and accelerates at +44 ft/s2. After some time t1,

mash [69]

Answer:

a) t₁ = 4.76 s, t₂ = 85.2 s

b) v = 209 ft/s

Explanation:

Constant acceleration equations:

x = x₀ + v₀ t + ½ at²

v = at + v₀

where x is final position,

x₀ is initial position,

v₀ is initial velocity,

a is acceleration,

and t is time.

When the engine is on and the sled is accelerating:

x₀ = 0 ft

v₀ = 0 ft/s

a = 44 ft/s²

t = t₁

So:

x = 22 t₁²

v = 44 t₁

When the engine is off and the sled is coasting:

x = 18350 ft

x₀ = 22 t₁²

v₀ = 44 t₁

a = 0 ft/s²

t = t₂

So:

18350 = 22 t₁² + (44 t₁) t₂

Given that t₁ + t₂ = 90:

18350 = 22 t₁² + (44 t₁) (90 − t₁)

Now we can solve for t₁:

18350 = 22 t₁² + 3960 t₁ − 44 t₁²

18350 = 3960 t₁ − 22 t₁²

9175 = 1980 t₁ − 11 t₁²

11 t₁² − 1980 t₁ + 9175 = 0

Using quadratic formula:

t₁ = [ 1980 ± √(1980² - 4(11)(9175)) ] / 22

t₁ = 4.76, 175

Since t₁ can't be greater than 90, t₁ = 4.76 s.

Therefore, t₂ = 85.2 s.

And v = 44 t₁ = 209 ft/s.

3 0

3 years ago

Given that ethylene has a λmax of 175nm, butadiene has a λmax of 220nm, and 2-methyl-1,3-butadiene has a λmax or 215nm, what is

Vika [28.1K]

The λmax of 2,3,4-trimethylhexatriene is 280 nm.

Ethylene has a λmax of 175nm.

Butadiene has a λmax of 220nm.

2-methyl-1,3-butadiene has a λmax or 215nm.

1,3,5-hexatriene has a λmax of 258nm.

Woodward's rules, sometimes known as Woodward-Fieser rules (after Louis Fieser) and named after Robert Burns Woodward, are a number of sets of empirically developed principles that aim to forecast the wavelength of the absorption maximum (max) in an ultraviolet-visible spectrum of a certain molecule.

By using the Woodward Fieser rule,

R- (Alkyl Group) .... +5 nm = 5 × 2 = 10

RO- (Alkoxy Group) .. +6 = 6 × 2 = 12

Adding 22nm to the λmax of 1,3,5-hexatriene as it has 2 alkyl groups and 2 alkoxy groups to form 2,3,4-trimethylhexatriene.

The λmax of 2,3,4-trimethylhexatriene is 280 nm.

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5 0

2 years ago