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Eddi Din [679]
3 years ago
8

A 5 kg mass is moving at 3 ms-1. Calculate its kinetic energy.

Physics
1 answer:
cupoosta [38]3 years ago
8 0

Answer:

1/2 MV squared

1/2*5*3 squared

=22.5

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You accelerate from 2m/s to 6m/s while traveling a distance of 2m; what was your acceleration?
nata0808 [166]

Answer:

The acceleration is 8 m/s²

Explanation:

The given parameters are;

The initial velocity, u = 2 m/s

The final velocity, v = 6 m/s

The distance the acceleration took place, s = 2 m

The acceleration, a, can be found from the following kinematic equation;

v² = u² + 2·a·s

By substituting the values, we have;

6² = 2² + 2 × a × 2

6² - 2² = 2 × a × 2

32 = 4·a

a = 32/4 = 8 m/s²

The acceleration, a, of the given motion = 8 m/s².

7 0
3 years ago
Analyze how buffers allow you to eat acidic and basic foods without changing your blood pH.
blondinia [14]
Buffers neutralize the acid and the bases
7 0
4 years ago
What is the order of magnitude of the distance of Sun to nearest star in meters?
neonofarm [45]

Answer:

Approximating the Milky Way as a disk and using the density in the solar neighborhood, there are about 100 billion stars in the Milky Way.

Explanation:

Since we are making an order of magnitude estimate, we will make a series of simplifying assumptions to get an answer that is roughly right.

Let's model the Milky Way galaxy as a disk.

The volume of a disk is:

V

=

π

⋅

r

2

⋅

h

Plugging in our numbers (and assuming that

π

≈

3

)

V

=

π

⋅

(

10

21

m

)

2

⋅

(

10

19

m

)

V

=

3

×

10

61

m

3

Is the approximate volume of the Milky Way.

Now, all we need to do is find how many stars per cubic meter (

ρ

) are in the Milky Way and we can find the total number of stars.

Let's look at the neighborhood around the Sun. We know that in a sphere with a radius of

4

×

10

16

m there is exactly one star (the Sun), after that you hit other stars. We can use that to estimate a rough density for the Milky Way.

ρ

=

n

V

Using the volume of a sphere

V

=

4

3

π

r

3

ρ

=

1

4

3

π

(

4

×

10

16

m

)

3

ρ

=

1

256

10

−

48

stars /

m

3

Going back to the density equation:

ρ

=

n

V

n

=

ρ

V

Plugging in the density of the solar neighborhood and the volume of the Milky Way:

n

=

(

1

256

10

−

48

m

−

3

)

⋅

(

3

×

10

61

m

3

)

n

=

3

256

10

13

n

=

1

×

10

11

stars (or 100 billion stars)

Is this reasonable? Other estimates say that there are are 100-400 billion stars in the Milky Way. This is exactly what we found.

4 0
3 years ago
Read 2 more answers
A _______ is a repeating disturbance or vibration that transfers or moves energy from place to place without transporting mass.
cupoosta [38]

the answer is a wave


7 0
3 years ago
Read 2 more answers
Write a short description of how the motion of the racers might change from the start of the race to the finish line
Ad libitum [116K]
The motion of the racers might change from the start because the pressure goes up so all the racer wants is to speed up and win, so when the racer first starts he or she is calm because he's not driving yet and when he or she is on his/hers way to he finish line he/she just wants to win and gets under pressure so he speeds up even more and drifts. Your welcome
6 0
3 years ago
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