How long does 100m long train take a cross a bridge of length 4.9 km at a speed of 50m/s

What is moment? Write down the law of moment.A long spanner is used to unscrew the tight nut.Why?

vfiekz [6]

Answer:

Moment is the product of force and its perpendicular distance from a point along its line of action.

The net moments on clockwise and anticlockwise is zero at a point.

The net force applied on a spanner handle is equal to the moment of the rotation

3 0

2 years ago

A 0.5 kg block of aluminum (caluminum=900J/kg⋅∘C) is heated to 200∘C. The block is then quickly placed in an insulated tub of co

Genrish500 [490]

Answer: When 1.0kg of aluminium block is used, the final temperature of the mixture will be T = 36.2∘C

If 1.0kg copper block is used, T of the mixture will be = 17.4∘C

If 100g (0.1kg) of ice at 0∘C is used, T will be = 64.9∘C

If 25g (0.025Kg) of ice is used, T will be= 147.1∘C

Explanation:

H = mcΘ

heat lost by block = heat gained by water

m₁c₁Θ₁ = m₂c₂Θ₂ where m₁ is mass of aluminium, m₂ is mass of water, c₁ is cAluminium, c₂ is cWater, Θ₁ is temperature change for aluminium, Θ₂ is temperature change for water.

0.5*900*(200-20) = m₁*4186*(20-0)

m₁ = 450*180/83270

m₁ = 0.973kg

when 1.0kg of aluminium block is used, the final temperature of the mixture will be T

heat lost by block = heat gained by water

1.0*900*(200-T) = 0.973*4186*(T-0)

180000 - 900T = 4073T

4973T = 180000

T = 180000/4973 = 36.2∘C

If 1.0kg copper block is used, T of the mixture will be

heat lost by block = heat gained by water

1.0*387*(200-T) = 0.973*4186*(T-0)

77400 - 387T = 4073T

4460T = 77400

T = 77400/4460 = 17.4∘C

If 100g (0.1kg) of ice at 0∘C is used, T will be

heat lost by block = heat gained by water + heat used in melting ice to form water at 0∘C

heat used in melting 0.1kg of ice, H = ml, where l= 33600J/Kg

0.5*900*(200-T) = 0.1*4186*(T-0) + 0.1*33600J/Kg

90000 - 450T = 418.6T + 33600

418.6T + 450T = 90000 - 33600

868.6T = 56400

T = 56400/868.6 = 64.9∘C

If 25g (0.025Kg) of ice is used, T will be

0.5*900*(200-T) = 0.025*4186*(T-0) + 0.025*33600J/Kg

90000 - 450T = 104.65T + 8400

104.65T + 450T = 90000 - 8400

554.65T = 81600

T = 81600/554.65 = 147.1∘C

7 0

3 years ago

I will fan and give 2 medals!!!!!!

Maksim231197 [3]

Corona

I. Rare gaseous envelope

II. Irregular pearly glow

III. Total solar eclipse

IV. Surrounds darkened moon

V. Caused by diffraction

Solar Prominence

I. loop shape

II. large bright gaseous

III. luminous hydrogen gas

IV. rises above chromospheres

V. anchored to surface (sun’s surface)

Solar Flare

I. Intense High-energy

II. Causes electromagnetic disruptions

III. In solar atmosphere

IV. Varies in brightness

V. Emits radiation

Sunspots

I. Surface dark spots

II. Strong magnetic fields

III. Solar magnetic storms

IV. Visible through telescope

V. Cooler than photospheres

I did this in class I got 100%

7 0

2 years ago

If an object is rolling without slipping, how does its linear speed compare to its rotational speed? If an object is rolling wit

salantis [7]

Answer:

$v = r\omega$

Explanation:

If the object is rolling without slipping, every unit of rotated angle equals to a distance perimeter rotated.

Suppose the object complete 1 revolution within time t. The angular distance is 2π rad. Its angular velocity is 2π/t

The distance it covered is its circumference, which is 2πr, and so the speed is 2πr/t

So the linear speed compared to angular speed is

$\frac{v}{\omega} = \frac{2\pi r/t}{2\pi /t} = r$

$v = r\omega$

6 0

3 years ago

An airplane weighing 11,000 N climbs to a

Gennadij [26K]

The power in horsepower is 40.1 hp

Explanation:

We start by calculating the work done by the airplane during the climb, which is equal to its change in gravitational potential energy:

$W=(mg)\Delta h$

where

mg = 11,000 N is the weight of the airplane

$\Delta h = 1.6 km = 1600 m$ is the change in height

Substituting,

$W=(11,000)(1600)=17.6\cdot 10^6 J$

Now we can calculate the power delivered, which is given by

$P=\frac{W}{t}$

where

$W=17.6\cdot 10^6 J$ is the work done

$t=9.8 min \cdot 60 = 588 s$ is the time taken

Substituting,

$P=\frac{17.6\cdot 10^6 J}{588}=2.99\cdot 10^4 W$

Finally, we can convert the power into horsepower (hp), keeping in mind that

$1 hp = 746 W$

Therefore,

$P=\frac{2.99\cdot 10^4}{746}=40.1 hp$

Learn more about power:

brainly.com/question/7956557

#LearnwithBrainly

8 0

3 years ago