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4 years ago

When a can is crushed did it undergo a physical change

Physics

1 answer:

brilliants [131]4 years ago

4 0

Because even though the object got crush and misshape it still has the same identity. the identity never change

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The diagram below shows the movement of matter in a portion of the water cycle.

Alisiya [41]

Answer:

d) precipitation

Hope it helps you

And if you want to, pls mark it as the brainliest answer

7 0

3 years ago

3) If a ball launched at an angle of 10.0 degrees above horizontal from an initial height of 1.50 meters has a final horizontal

Irina-Kira [14]

Answer:

35.6 m

Explanation:

3 0

4 years ago

The weight of an object is the same on two different planets. The mass of planet A is only sixty percent that of planet B. Find

natka813 [3]

Answer:

0.775

Explanation:

The weight of an object on a planet is equal to the gravitational force exerted by the planet on the object:

$F=G\frac{Mm}{R^2}$

where

G is the gravitational constant

M is the mass of the planet

m is the mass of the object

R is the radius of the planet

For planet A, the weight of the object is

$F_A=G\frac{M_Am}{R_A^2}$

For planet B,

$F_B=G\frac{M_Bm}{R_B^2}$

We also know that the weight of the object on the two planets is the same, so

$F_A = F_B$

So we can write

$G\frac{M_Am}{R_A^2} = G\frac{M_Bm}{R_B^2}$

We also know that the mass of planet A is only sixty percent that of planet B, so

$M_A = 0.60 M_B$

Substituting,

$G\frac{0.60 M_Bm}{R_A^2} = G\frac{M_Bm}{R_B^2}$

Now we can elimanate G, MB and m from the equation, and we get

$\frac{0.60}{R_A^2}=\frac{1}{R_B^2}$

So the ratio between the radii of the two planets is

$\frac{R_A}{R_B}=\sqrt{0.60}=0.775$

6 0

3 years ago

A long solid conducting cylinder with radius a = 12 cm carries current I1 = 5 A going into the page. This current is distributed

jenyasd209 [6]

Answer:

Explanation:

We shall use Ampere's circuital law to find magnetic field at required point.

The point is outside the circumference of two given wires so whole current will be accounted for .

Ampere's circuital law

B = ∫ Bdl = μ₀ I

line integral will be over circular path of radius r = 41 cm .

Total current I = 5A -3A = 2A .

∫ Bdl = μ₀ I

2π r B = μ₀ I

2π x .41 B = 4π x 10⁻⁷ x 2

B = 2 x 10⁻⁷ x 2 / .41

= 9.75 x 10⁻⁷ T . It will be along - ve Y - direction.

7 0

3 years ago

In LEDs, the longer leg (wire) is always

galben [10]

Positive

In an LED, the longer lead is always connected to the positive terminal

5 0

3 years ago

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