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Amanda [17]
3 years ago
5

HELP ASAP WHAT ARE PROPERTIES OF WATER

Physics
1 answer:
iren [92.7K]3 years ago
3 0
Pls mark as brainiest!!

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Which object has the greatest amount of kinetic energy?
EleoNora [17]
I think that the answer is A
4 0
4 years ago
In a laboratory, you determine that the density of a certain solid is 5.23×10−6kg/mm3. Convert this density into kilograms per c
kiruha [24]

Answer:

5.23\cdot 10^3 kg/m^3

Explanation:

The density of the solid is

d = 5.23\cdot 10^{-6}kg/mm^3

we want to convert it into kg/m^3. We must note that:

1 m^3 = 1 m \cdot 1 m \cdot \1m =1000 mm\cdot 1000 mm \cdot 1000 mm=1\cdot 10^9 mm^3

Therefore, the conversion can be done as follows:

d=5.23\cdot 10^{-6} \frac{kg}{mm^3} \cdot (1\cdot 10^9 \frac{mm^3}{m^3}) =5.23\cdot 10^3 kg/m^3

8 0
3 years ago
A charged particle moving through a magnetic field at right angles to the field with a speed of 24.7 m/s experiences a magnetic
Sunny_sXe [5.5K]

Answer:

Explanation:

let the charge is q. velocity, v = 24.7 m/s

magnetic force, F = 2.38 x 10^-4 N

Let the magnetic field is B.

Velocity, v' = 5.64 m/s

angle, θ = 21.2°

The force experienced by a charged particle placed in a magnetic field is given by

F = q x v x B x Sinθ

in first case

2.38 x 10^-4 = q x 24.7 x B x Sin 90 .... (1)

in second case

F = q x 5.64 x B x Sin 21.2°      .... (2)

Divide equation (2) by equation (1), we get

\frac{F}{2.38\times 10^{-4}}=\frac{5.64\times Sin 21.2}{24.7\times Sin 90}

F = 1.97 x 10^-5 N

7 0
4 years ago
N=-D(n2-n1)/(x2-x1) D is diffastion.What are the dimensions of D.
MrMuchimi

We don't have much to go on.

The dimensions of D depend on the dimensions of N, n, and x, and we don't know what any of those stand for.

It might help if we had ever heard of 'diffastion', but we're striking out there too.

8 0
4 years ago
Letting D D represent the maximum displacement, the extremes of the block's motion are at position A, where x = − D x=−D, and at
Ksju [112]

Answer:

The answer is at x = 0, which represents position B

Explanation:

The full question is:

"A block is attached to a horizontal spring and set in a

simple harmonic motion, as shown from above in the figure. When the spring is relaxed, the block is a position B, where the displacement x from the equilibrium position is 0. Letting D represent the maximum displacement, the extremes of the block's motion are at position A, where x= -D, and at position C, where x= D.

At what point in the motion is the speed of the block at its maximum?"

And you can see the figure on the attached file.

Simple Harmonic motion equations

We can start from the equation that describes the position that is

x(t)=D \sin\left(\omega t)

Here D stands for the amplitude which is the maximum displacement, and \omega is the angular velocity, thus we can find the derivative to find the velocity equation, so we get

v(t)=D \omega \cos (\omega t)

And we can find the derivative again to find the acceleration.

a(t) = -D\omega^2 \sin (\omega t)

Maximum speed

We reach the maximum speed when the acceleration equation is equal to 0,

a(t) =0\\-D\omega^2 \sin (\omega t)=0

Thus it happens when

\sin (\omega t)=0

So if we replace that on the position equation we get

x(t)=D \sin(\omega t) \\x(t)=D(0)\\x(t)=0

Thus the position where the speed of the block is at at its maximum is when it is going back to the origin, that is x = 0, so point b.

7 0
3 years ago
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