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Marrrta [24]
3 years ago
14

What matter does not take up the space in the container?

Physics
1 answer:
son4ous [18]3 years ago
5 0

Answer:

I think it is solids

Explanation:

gaseous materials take any space in any container they are stored in the same is for liquids.

But solids can not take up the space of a contractor because they have a definite shape .

You might be interested in
PHYSICS QUESTION!!!!
geniusboy [140]

<u>Answer:</u>

 A) Mass of dinosaur = 77,000 kg

     Weight of dinosaur = 755370 N

B)  Mass of dinosaur = 77,000 kg

      Weight of dinosaur = 377685 N

<u>Explanation:</u>

 A) Mass of dinosaur the brachiosaurus = 77,000 kg

     We have weight of body = mg = 77,000*9.81 = 755370 N

 B) Mass of a body is constant, it will not get affected by change in acceleration due to gravity value.

    So, Mass of dinosaur = 77,000 kg

     We have weight of body = mg' = m*g/2 = 77,000*9.81/2 = 377685 N

4 0
3 years ago
A block of mass m= 2.8 kg is attached to a spring of spring constant k= 500 N/m. the block is pulled to an initial position x =
padilas [110]

Answer:

Explanation:

The energy stored in a spring

= 1/2 k x²

where k is spring constant and x is extension in the spring.

= .5 x 500 x .05²

= .625 J

Work done by friction = energy dissipated

= - μmg x d , μ is coefficient of friction , m is mass , d is displacement

= - .35 x 2.8 x 9.8 x .05

= - .48 J

energy of the mass when it reaches equilibrium position

= .625 - .48

= .145 J

If v be its velocity at that time

1/2 m v ² = .145

.5 x 2.8 x v² = .145

v² = .10357

v = .32 m /s

32 cm /s

3 0
3 years ago
Use Hooke's Law to determine the work done by the variable force in the spring problem. A force of 450 newtons stretches a sprin
natima [27]

The work done in stretching the spring from 50 cm to 80 cm is 67.5 J.

<h3>Hooke's Law</h3>

Hooke's law states that the force applied to an elastic material is directly proportional to its extension, provided its elastic limit is not exceeded.

To calculate the amount of work done by Hooke's law, first, we need to find the force constant of the spring.

Formula:

  • F = ke................. Equation 1

Where:

  • F = Force applied
  • k = Spring constant
  • e = extension

make k the subject of the equation

  • k = F/e................ Equation 2

From the question,

Given:

  • F = 450 N
  • e = 30 cm = 0.3 m

Substitute these values into equation 2

  • k = 450/0.3
  • k = 1500 N/m.

Finally, To find the work done in stretching the spring from 50 cm to 80 cm, we use the formula below.

  • W = ke²/2........... Equation 3

Where:

  • W = Work done
  • k = spring constant
  • e = extension

Also, From the question,

Given:

  • e = (80-50) = 30 cm = 0.3 m
  • k = 1500 N/m

Substitute these values into equation 3

  • W = 1500(0.3²)/2
  • W = 67.5 J.

Hence, The work done in stretching the spring from 50 cm to 80 cm is 67.5 J.

Learn more about Hooke's law here: brainly.com/question/12253978

5 0
2 years ago
A medicine ball has a mass of 5 kg and is thrown with a speed of 2 m/s. what is its kinetic energy?
Fantom [35]
The kinetic energy of the ball is given by:
K= \frac{1}{2}mv^2
where m=5 kg is the mass of the ball and v=2 m/s is its speed. Substituting these numbers, we find the kinetic energy:
K= \frac{1}{2} (5kg)(2m/s)^2=10 J
7 0
3 years ago
A steel bar that is at 10 ° c is 5 meters long, a bar for heated to 120 ° c, how long is that bar? Α = 1.2.10- ° c
svet-max [94.6K]

Answer:

1) 5.0066 m

2A) β = 3×10⁻⁷ / °C

2B) 2500.045 cm²

3A) γ = 8.1×10⁻⁵ / °C

3B) 1618.144 cm³

Explanation:

1) Linear thermal expansion is:

ΔL = α L₀ ΔT

where ΔL is the change in length,

α is the linear thermal expansion coefficient,

L₀ is the original length,

and ΔT is the change in temperature.

Given L₀ = 5 m, ΔT = 110°C, and α = 1.2×10⁻⁵ / °C:

ΔL = (1.2×10⁻⁵ / °C) (5 m) (110°C)

ΔL = 0.0066 m

The length increases by , so the new length is:

L = L₀ + ΔL

L = 5 m + 0.0066 m

L = 5.0066 m

2A) The surface expansion coefficient is:

β = 2α

β = 2 (1.5×10⁻⁷ / °C)

β = 3×10⁻⁷ / °C

2B) The change in area is:

ΔA = β A₀ ΔT

ΔA = (3×10⁻⁷ / °C) (50 cm × 50 cm) (60°C)

ΔA =  0.045 cm²

So the new area is:

A = A + ΔA

A = 2500 cm² + 0.045 cm²

A = 2500.045 cm²

3A) The volumetric expansion coefficient is:

γ = 3α

γ = 3 (2.7×10⁻⁵ / °C)

γ = 8.1×10⁻⁵ / °C

3B) The change in volume is:

ΔV = γ V₀ ΔT

ΔV = (8.1×10⁻⁵ / °C) (1600 cm³) (140°C)

ΔV = 18.144 cm³

So the new area is:

V = V + ΔV

V = 1600 cm³ + 18.144 cm³

V = 1618.144 cm³

6 0
3 years ago
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