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AlexFokin [52]
3 years ago
10

Dimension of density ​

Physics
1 answer:
Rom4ik [11]3 years ago
8 0

Answer:

See the explanation below.

Explanation:

We know that density is defined as the relationship between mass and volume.

Ro = m/V

where:

m = mass [kg]

V = volume [m³]

Therefore Ro is given in:

[kg/m^{3} ]

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marysya [2.9K]

Answer:

oh thank you for the free points :D

Explanation:

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3 years ago
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A soccer ball is rolling past Ramon. He kicks it in the same direction that it is rolling. What will MOST LIKELY happen? A) The
LUCKY_DIMON [66]
I think it’s A (MORE CHARACTERS CAUSE ITS TOO SHORT)
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3 years ago
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How much work must be done to bring three electrons from a great distance apart to 3.0×10−10 m from one another (at the corners
Natasha_Volkova [10]

Answer:

Potential\ Energy=Work \ Done=2.301*10^{-18} J

Work  done to bring three electrons from a great distance apart to 3.0×10−10 m from one another (at the corners of an equilateral triangle) is 2.301*10^{-18} Joules

Explanation:

The potential energy is given by:

U=Q*V

where:

Q is the charge

V is the potential difference

Potential Difference=V=\frac{kq}{r}

So,

Potential\ Energy=\frac{Qkq}{r} \\Q=q\\Potential\ Energy=\frac{kq^2}{r}

Where:

k is Coulomb Constant=8.99*10^9 Nm^2/C^2

q is the charge on electron=-1.6*10^-19 C

r is the distance=3.0*10^{-10}m

For 3 Electrons Potential Energy or work Done is:

Potential\ Energy=3*\frac{kq^2}{r}

Potential\ Energy=3*\frac{(8.99*10^9)(-1.6*10^{-19})^2}{3*10^{-10}}\\Potential\ Energy=2.301*10^{-18} J

Work  done to bring three electrons from a great distance apart to 3.0×10−10 m from one another (at the corners of an equilateral triangle) is 2.301*10^{-18} Joules

7 0
3 years ago
A student weighing 700 N climbs at constant speed to the top of an 8 m vertical rope in 10 s. The average power expended by the
Goshia [24]

To solve this problem we will use the concepts related to power, defined as the amount of energy applied over a period of time.

The energy in this case is the accumulated in the form of potential energy, over a period of time. Thus we will have that the mathematical expression of the power can be expressed as

P = \frac{E}{t}

Here,

E = Energy

t = time

As the energy is equal to the potential Energy we have tat

P = \frac{mgh}{t}

The weight (mg) of the man is 700N, the height (h) is 8m and the time is 10s, then:

P = \frac{700*8}{10}

P = 560W

Therefore the correct answer is A.

8 0
4 years ago
The time delay between transmission and the arrival of the reflected wave of a signal using ultrasound traveling through a piece
Brilliant_brown [7]

Answer:

10.01 cm

Explanation:

Given that,

The time delay between transmission and the arrival of the reflected wave of a signal using ultrasound traveling through a piece of fat tissue was 0.13 ms.

The average propagation speed for sound in body tissue is 1540 m/s.

We need to find the depth when the reflection occur. We know that, the distance is double when transmitting and arriving. So,

v=\dfrac{2d}{t}\\\\d=\dfrac{vt}{2}\\\\d=\dfrac{1540\times 0.13\times 10^{-3}}{2}\\\\d= $$0.1001\ m

or

d = 10.01 cm

So, the reflection will occur at 10.01 cm.

8 0
3 years ago
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