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3 years ago

At what speed does a 1400 compact car have the same kinetic energy as a 20000 truck going 25 ?

1 answer:

4 0

K=1400*V^2/2
K=20000*25^2/2. => 1400*V^2/2=20000*25^2/2 <=> 1400*V^2=20000*25^2
14*V^2=200*225
v^2=100*225/7
v=250/7^(1/2)

Answer: 250*7^(1/2)/7

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Please help me people

saveliy_v [14]

Answer :

(1) The density of asphalt is, $1200kg/m^3$

(2) (a) Length, width and thickness of sheet in meter is, 0.35 m, 1.1 m and 0.015 m respectively.

(b) The volume and mass of slab is, 0.005775 m³ and 15.59 kg respectively.

Explanation :

Part 1 :

As we are given:

Mass of block = 90 kg

Volume of block = $0.075m^3$

Formula used :

$\text{Density of block}=\frac{\text{Mass of block}}{\text{Volume of block}}$

Now put all the given values in this formula, we get:

$\text{Density of block}=\frac{90kg}{0.075m^3}=1200kg/m^3$

Thus, the density of asphalt is, $1200kg/m^3$

Part 2(a) :

As we are given that:

Length of aluminium sheet = 35 cm

Width of aluminium sheet = 11 dm

Thickness of aluminium sheet = 15 mm

Now we have to convert these dimensions into meters.

Conversions used:

1 cm = 0.01 m

1 dm = 0.1 m

1 mm = 0.001 m

Length of aluminium sheet = 35 cm = 35 × 0.01 = 0.35 m

Width of aluminium sheet = 11 dm = 11 × 0.1 = 1.1 m

Thickness of aluminium sheet = 15 mm = 15 × 0.001 = 0.015 m

Part 2(b) :

First we have to calculate the volume of aluminium sheet.

Volume of aluminum sheet (cuboid) = Length × Width × Thickness

Volume of aluminum sheet (cuboid) = 0.35 m × 1.1 m × 0.015 m

Volume of aluminum sheet (cuboid) = 0.005775 m³

Now we have to calculate the mass of aluminium sheet.

$\text{Density of aluminium}=\frac{\text{Mass of aluminium}}{\text{Volume of aluminium}}$

$2700kg/m^3=\frac{\text{Mass of aluminium}}{0.005775m^3}$

$\text{Mass of aluminium}=15.59kg$

Thus, the volume and mass of slab is, 0.005775 m³ and 15.59 kg respectively.

7 0

3 years ago

Read 2 more answers

Un the way to the moon, the Apollo astro-

kherson [118]

Answer:

Distance = 345719139.4[m]; acceleration = 3.33*10^{19} [m/s^2]

Explanation:

We can solve this problem by using Newton's universal gravitation law.

In the attached image we can find a schematic of the locations of the Earth and the moon and that the sum of the distances re plus rm will be equal to the distance given as initial data in the problem rt = 3.84 × 108 m

$r_{e} = distance earth to the astronaut [m].\\r_{m} = distance moon to the astronaut [m]\\r_{t} = total distance = 3.84*10^8[m]$

Now the key to solving this problem is to establish a point of equalisation of both forces, i.e. the point where the Earth pulls the astronaut with the same force as the moon pulls the astronaut.

Mathematically this equals:

$F_{e} = F_{m}\\F_{e} =G*\frac{m_{e} *m_{a}}{r_{e}^{2} } \\$

$F_{m} =G*\frac{m_{m}*m_{a} }{r_{m} ^{2} } \\where:\\G = gravity constant = 6.67*10^{-11}[\frac{N*m^{2} }{kg^{2} } ] \\m_{e}= earth's mass = 5.98*10^{24}[kg]\\ m_{a}= astronaut mass = 100[kg]\\m_{m}= moon's mass = 7.36*10^{22}[kg]$

When we match these equations the masses cancel out as the universal gravitational constant

$G*\frac{m_{e} *m_{a} }{r_{e}^{2} } = G*\frac{m_{m} *m_{a} }{r_{m}^{2} }\\\frac{m_{e} }{r_{e}^{2} } = \frac{m_{m} }{r_{m}^{2} }$

To solve this equation we have to replace the first equation of related with the distances.

$\frac{m_{e} }{r_{e}^{2} } = \frac{m_{m} }{r_{m}^{2} } \\\frac{5.98*10^{24} }{(3.84*10^{8}-r_{m} )^{2} } = \frac{7.36*10^{22} }{r_{m}^{2} }\\81.25*r_{m}^{2}=r_{m}^{2}-768*10^{6}* r_{m}+1.47*10^{17} \\80.25*r_{m}^{2}+768*10^{6}* r_{m}-1.47*10^{17} =0$

Now, we have a second-degree equation, the only way to solve it is by using the formula of the quadratic equation.

$r_{m1,2}=\frac{-b+- \sqrt{b^{2}-4*a*c } }{2*a}\\ where:\\a=80.25\\b=768*10^{6} \\c = -1.47*10^{17} \\replacing:\\r_{m1,2}=\frac{-768*10^{6}+- \sqrt{(768*10^{6})^{2}-4*80.25*(-1.47*10^{17}) } }{2*80.25}\\\\r_{m1}= 38280860.6[m] \\r_{m2}=-2.97*10^{17} [m]$

We work with positive value

rm = 38280860.6[m] = 38280.86[km]

Second part

The distance between the Earth and this point is calculated as follows:

re = 3.84 108 - 38280860.6 = 345719139.4[m]

Now the acceleration can be found as follows:

$a = G*\frac{m_{e} }{r_{e} ^{2} } \\a = 6.67*10^{11} *\frac{5.98*10^{24} }{(345.72*10^{6})^{2} } \\a=3.33*10^{19} [m/s^2]$

6 0

3 years ago

When a spring is compressed, the energy changes from kinetic to potential. Which best describes what is causing this change?

just olya [345]

Answer:

Well, Your answer will be is Work. Because, When a spring is compressed, the energy changes from kinetic to potential. This change is caused by work.

Good Luck!~

$^{By$ $^{Itsbrazts$