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4 years ago

Different color ca

Physics

1 answer:

777dan777 [17]4 years ago

4 0

Answer: The color of candy cane

Explanation:

An independent variable is the Variable that the scientist changed in order to change the dependent variable. The scientist changes the color of the candy cane to test the rate at which they dissolve.

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A cart is moving to the right with a constant speed of 20 m s . A box of mass 80 kg moves with the cart without slipping. The co

nignag [31]

Answer:

Explanation:

Given

mass of box $m=80 kg$

coefficient of kinetic friction $\mu _k=0.15$

coefficient of Static friction $\mu _s=0.30$

cart is moving with constant velocity therefore Net Force is zero

Since there is no net acceleration therefore friction force will be zero

mathematically

$f_r=ma$

where $f_r=frictional\ Force$

$a=acceleration$

$a=0$

$f_r=0$

4 0

3 years ago

A low C (f = 65Hz) is sounded on a piano. If the length of the piano wire is 2.0 m and its mass density is 5.0 g/m2, determine t

LuckyWell [14K]

Answer:

Tension of the wire(T) = 169 N

Explanation:

Given:

f = 65Hz

Length of the piano wire (L) = 2 m

Mass density = 5.0 g/m² = 0.005 kg/m²

Find:

Tension of the wire(T)

Computation:

f = v / λ

65 = v / 2L

65 = v /(2)(2)

v = 260 m/s

T = v² (m/l)

T = (260)²(0.005/2)

T = 169 N

Tension of the wire(T) = 169 N

6 0

4 years ago

If a substance besides water is changed from a gas to a liquid and then to a solid, what happens to the particles?

NARA [144]

The correct answer is they get closer together

Going from gas into a liquid makes the molecules move closer together, which happens again when you turn it into a solid.

3 0

3 years ago

Read 2 more answers

PLEASE HELP ME!!!

s344n2d4d5 [400]

Answer:

I think is B. sound, nuclear, electrical

but if I wrong then I am so sorry

3 0

3 years ago

As the radius of the Gaussian surface increases, the magnitude of the electric field at the surface does which of the followingI

marshall27 [118]

Answer:

Decreases

Explanation:

Gauss's law states that the flux of the electric field through a closed surface is equal to the quotient between the charge inside that surface divided by the permittivity of free space:

$\int\limits {E\cdot dS} \,=\frac{Q_{enc}}{\epsilon_0}\\E=\frac{Q_{enc}}{A \epsilon_0}$

Therefore, the magnitude of the electric field is inversely proportional to the area, the area is directly proportional to the radius. So, as the radius of the Gaussian surface increases, the magnitude of the electric field at the surface decreases.

5 0

3 years ago