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jok3333 [9.3K]
3 years ago
6

Which is a characteristic of a physical change? (1 point)

Physics
1 answer:
krek1111 [17]3 years ago
6 0
The answer would be a.

a chemical change is a change to the chemical makeup of a substance so if the bonds are unchanged it would be a characteristic of a physical change
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What is the magnitude of the velocity of a 25 kg mass that is moving with a momentum of 100 kg*m/s?
Gekata [30.6K]

Answer:

v= 4 m/s

Explanation:

Momenutm is, by definition, the product of mass and velocity.

p = mv

Let's replace what we know and solve for whatever's left

100 kg\cdot m/s = 25kg \cdot v \rightarrow v= 4 m/s

7 0
3 years ago
Match the choices to the appropriate blank . Use each choice only once.
pav-90 [236]

Answer:

1. about 1.5 AU

2. about 5 AU

3. about 8 light-years

4. about 100,000 light-years

5. less than 0.01 AU

Explanation:

a. Mars is about 1.5 AU from the Sun.

b. Jupiter is about 5 AU from the Sun.

c. The star Sirius is about 8 light-years from the Sun.

d. The diameter of the Milky Way Galaxy is about 100,000 light-years.

e. The distance from Earth to the Moon is less than 0.01 AU.

Note: AU is an acronym for Astronomical Unit and it is a standard unit by astronomers to illustrate the distance between the planetary bodies found in the solar system.

5 0
3 years ago
What is your interpretation of 'nothing'? (Talking astrophysics) More answers the better!
Sladkaya [172]
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5 0
3 years ago
1. If a car travels 400m in 20 seconds how fast is it going?
Marrrta [24]

Answer:

1) 20 m/s

2) 5 m/s

3) 2 m/s

4) 395,000m/9000s

5) 16 km/0.25h

Explanation:

i dunno

5 0
3 years 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
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