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3 years ago

Density is the ratio of an object’s mass to its volume. would you expect density to be a vector or a scalar quantity? explain.

1 answer:

4 0

Vectors are quantities described by a number and a direction, thus giving them a line value. these include acceleration, velocity and weight
Scalar quantities only have a magnitude and include things like volume, distance and temperature. Because density uses volume and mass, it is considered a scalar.

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Un cuerpo se mueve, partiendo del reposo, con una aceleración constante de 8 m/s². Calcular la velocidad que tiene al cabo de 5s

nataly862011 [7]

Answer:

v = 40 m / s

Explanation:

Let's use the expressions for accelerated motion

v = v₀ + a t

where vo is the initial velocity, at the acceleration and t is the time.

as the body starts from rest its initial velocity is zero

v = 0 + at

let's calculate

v = 8 5

v = 40 m / s

8 0

2 years ago

A rock, with a density of 3.55 g/cm^3 and a volume of 470 cm^3, is thrown in a lake. a) What is the weight of the rock out of th

SIZIF [17.4K]

Answer:

a) Weight of the rock out of the water = 16.37 N

b) Buoyancy force = 4.61 N

c) Mass of the water displaced = 0.47 kg

d) Weight of rock under water = 11.76 N

Explanation:

a) Mass of the rock out of the water = Volume x Density

Volume = 470 cm³

Density = 3.55 g/cm³

Mass = 470 x 3.55 = 1668.5 g = 1.6685 kg

Weight of the rock out of the water = 1.6685 x 9.81 = 16.37 N

b) Buoyancy force = Volume x Density of liquid x Acceleration due to gravity.

Volume = 470 cm³

Density of liquid = 1 g/cm³

$\texttt{Buoyancy force}= \frac{470\times 1\times 9.81}{1000} = 4.61 N$

c) Mass of the water displaced = Volume of body x Density of liquid

Mass of the water displaced = 470 x 1 = 470 g = 0.47 kg

d) Weight of rock under water = Weight of the rock out of the water - Buoyancy force

Weight of rock under water = 16.37 - 4.61 =11.76 N

3 0

3 years ago

If an object is moving to the right and there is a net force acting on it to the left, what will happen to the object?

Ksju [112]

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7 0

3 years ago

. Determine if approximate cylindrical symmetry holds for the following situations. State why or why not. (a) A 300-cm long copp

MA_775_DIABLO [31]

Answer:

a) Yes

b) No

Explanation:

In the first case, part a, yes we can say for certainty that cylinderical symmetry holds. Why so? You may ask. This is because from the question, we are told that the length of the rod is 300 cm. And this said length is longer than the distance to the point from the center of the rod, which is 5 cm.

In the second half of the question, I beg to disagree that cylindrical symmetry holds. Again, you may ask why, this is because the length of the rod in this case, is having the same order of magnitude as the distance to the center of the rod. Thus, it is not symmetrical.

6 0

2 years ago

Two sound waves (speed 343 m/s) have different wavelengths. The first has a wavelength of 5.72 m, and the second a wavelength of

lys-0071 [83]

Answer:

The beat frequency is 30 Hz

Explanation:

Given;

velocity of the two sound waves, v = 343 m/s

wavelength of the first wave, λ₁ = 5.72 m

wavelength of the second wave, λ₂ = 11.44 m

The frequency of the first wave is calculated as follows;

F₁ = v/λ₁

F₁ = 343 / 5.72

F₁ = 59.97 HZ

The frequency of the second wave is calculated as follows;

F₂ = v/λ₂

F₂ = 343 / 11.44

F₂ = 29.98 Hz

The beat frequency is calculated as;

Fb = F₁ - F₂

Fb = 59.97 HZ - 29.98 Hz

Fb = 30 Hz

6 0

3 years ago