Why do water and table salt have very different properties?

,how do charged objects react???quiet urgent

son4ous [18]

Any charged object can exert the force upon other objects ... i think tell me if im right

8 0

2 years ago

Read 2 more answers

A 4 kg bowling bowl is sitting on a table 1 meter off the ground. How much potential energy does it have?

il63 [147K]

Answer:

$\huge\boxed{\sf P.E. = 39.2\ Joules}$

Explanation:

Given Data:

Mass = m = 4 kg

Acceleration due to gravity = g = 9.8 m/s²

Height = h = 1 m

Required:

Potential Energy = P.E. = ?

Formula:

P.E. = mgh

Solution:

P.E. = (4)(9.8)(1)

P.E. = 39.2 Joules

$\rule[225]{225}{2}$

Hope this helped!

5 0

2 years ago

Please help me I need this for a test

Marysya12 [62]

Answer:

Explanation:

Lol thats big tufffffffffvffffffftg

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8 0

3 years ago

A rifle bullet with mass 8.00g strikes and embeds itself in a block with mass 0.992kg that rests on a frictionless, horizontal s

mafiozo [28]

Answer:

block velocity v = 0.09186 = 9.18 10⁻² m/s and speed bollet v₀ = 11.5 m / s

Explanation:

We will solve this problem using the concepts of the moment, let's try a system formed by the two bodies, the bullet and the block; In this system all scaffolds during the crash are internal, consequently, the moment is preserved.

Let's write the moment in two moments before the crash and after the crash, let's call the mass of the bullet (m) and the mass of the Block (M)

Before the crash

p₀ = m v₀ + 0

After the crash

$p_{f}$ = (m + M) v

p₀ = $p_{f}$

m v₀ = (m + M) v (1)

Now let's lock after the two bodies are joined, in this case the mechanical energy is conserved, write it in two moments after the crash and when you have the maximum compression of the spring

Initial

Em₀ = K = ½ m v2

Final

E $m_{f}$ = Ke = ½ k x2

Emo = E $m_{f}$

½ m v² = ½ k x²

v² = k/m x²

Let's look for the spring constant (k), with Hook's law

F = -k x

k = -F / x

k = - 0.75 / -0.25

k = 3 N / m

Let's calculate the speed

v = √(k/m) x

v = √ (3/8.00) 0.15

v = 0.09186 = 9.18 10⁻² m/s

This is the spped of the block plus bullet rsystem right after the crash

We substitute calculate in equation (1)

m v₀ = (m + M) v

v₀ = v (m + M) / m

v₀ = 0.09186 (0.008 + 0.992) /0.008

v₀ = 11.5 m / s

6 0

2 years ago

2. Below what depth would a submarine have to submerge so that it would not be swayed by surface waves with a wavelength of 24 m

Travka [436]

Answer: Below 12m of depth, the submarine has to submerge so that it would not be swayed by surface waves

Explanation:

To avoid the surface waves, a submarine has to submerge below the wave base. It is the position below which the motion of the waves is negligible.

This wave base is equal to half of the wavelength. The equation becomes:

Wave base = $\frac{\text{Wavelength}}{2}$

We are given:

Wavelength = 24 m

Putting values in above equation, we get:

Wave base = $\frac{24m}{2}=12m$

Hence, below 12m of depth, the submarine has to submerge so that it would not be swayed by surface waves

4 0

2 years ago