All categories

3 years ago

How does mass affect density? How does volume affect density?

Physics

1 answer:

Musya8 [376]3 years ago

7 0

Answer:

Here's the Density Formula: D = M/V

Q: How does mass affect density?

A: <em>Mass is a factor in density, the density is proportional to the mass. So as the mass increases, so does the density, provided the volume remains constant.</em>

Q: How does volume affect density?

A:<em> If an object has a larger mass than its volume it has a high density, if an object has a smaller mass than its volume it has a lower density.</em>

Explanation:

<em><u>I really Hope this Helps!!</u></em>

You might be interested in

What is the length of the color attribute?

astraxan [27]

The length of the colour attribute is 6 characters.

<h3>What is a colour attribute?</h3>

Each colour has a distinct look based on three essential characteristics: hue, chroma (saturation), and value (lightness). It's critical to use all three of these properties when describing colour to appropriately identify it and distinguish it from others.

<h3>What is the use of colour attributes?</h3>

The HTML <font> color Attribute specifies the text color within the font element. Values of Attributes: colour name: It uses the colour name to set the text colour.

It should be noticed that colours are represented by the hex triplet.

Learn more about colour attributes here:

brainly.com/question/18071208

#SPJ4

6 0

1 year ago

How long will it take to go 150000m traveling at 50km/hr

Tom [10]

Answer:

3000 hurs

Explanation: just divide 150000 by 50 and get 3000

4 0

3 years ago

A 4 kg car with frictionless wheels is on a ramp that makes a 25° angle with the horizontal. The car is connected to a 1 kg mass

stellarik [79]

Refer to the diagram shown below.

W₁ = (4 kg)*(9.8 m/s²) = 39.2 N
W₂ = (1 kg)*(9.8 m/s²) = 9.8 N

The normal reaction on the 4-kg mass is
N = (39.2 N)*cos(25°) = 35.5273 N
The force acting down the inclined plane due to the weight is
F = (39.2 N)*sin(25°) = 16.5666 N

The net force that accelerates the 4-kg mass at a m/²s down the plane is
F - W₂ = (4 kg)*(a m/s²)
4a = 16.5666 - 9.8
a = 1.6917 m/s²

Answer: 1.69 m/s² (nearest hundredth)

7 0

3 years ago

A 50.0-g object connected to a spring with a force constant of 35.0 n/m oscillates with an amplitude of 4.00 cm on a frictionles

Dimas [21]

A) The total energy of the system is sum of kinetic energy and elastic potential energy:
$E=K+U= \frac{1}{2}mv^2 + \frac{1}{2}kx^2$
where
m is the mass
v is the speed
k is the spring constant
x is the elongation/compression of the spring

The total energy is conserved, so we can calculate its value at any point of the motion. If we take the point of maximum displacement:
$x=A=4.00 cm = 0.04 m$
then the velocity of the system is zero, so the total energy is just potential energy, and it is equal to
$E=U= \frac{1}{2}kA^2 = \frac{1}{2}(35.0 N/m)(0.04 m)^2=0.028 J$

b) When the position of the object is
$x=1.00 cm = 0.01 m$
the potential energy of the system is
$U= \frac{1}{2}kx^2 = \frac{1}{2}(35.0 N/m)(0.01 m)^2 = 1.75 \cdot 10^{-3} J$
and so the kinetic energy is
$K=E-U=0.028 J - 1.75 \cdot 10^{-3}J =0.026 J$
since the mass is $m=50.0 g=0.05 kg$ , and the kinetic energy is given by
$K= \frac{1}{2}mv^2$
we can re-arrange the formula to find the speed of the object:
$v= \sqrt{ \frac{2K}{m} }= \sqrt{ \frac{2 \cdot 0.026 J}{0.05 kg} }=1.02 m/s$

c) The potential energy when the object is at
$x=3.00 cm=0.03 m$
is
$U= \frac{1}{2}kx^2 = \frac{1}{2}(35.0 N/m)(0.03 m)^2 =0.016 J$
Therefore the kinetic energy is
$K=E-U=0.028 J-0.016 J = 0.012 J$

d) We already found the potential energy at point c, and it is given by
$U= \frac{1}{2}kx^2 = \frac{1}{2}(35.0 N/m)(0.03 m)^2 =0.016 J$

5 0

3 years ago

Someone please help me will give BRAILIEST!!!!!

ratelena [41]

The speed increases, because as the angle increases (the wing slants up more steeply), the air has to go farther to get over the wing.

4 0

3 years ago