Which of the following iS an example of a falsifiable hypothesis that could lead to scientific knowledge?

Find the density of an object that has a mass of 5 kg and a volume of 50 cm3

serg [7]

Density = mass/volume. 5000/50= 100 grams per cubic centimeter. No substance on Earth has anywhere near that density. Tops is about 20 for osmium and gold.

3 0

3 years ago

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A gas is compressed from 600cm3 to 200cm3 at a constant pressure of 450kPa . At the same time, 100J of heat energy is transferre

Paraphin [41]

Answer: 80J

Explanation:

According to the first principle of thermodynamics:

<em>"Energy is not created, nor destroyed, but it is conserved." </em>

Then this priciple (also called Law) relates the work and the transferred heat exchanged in a system through the internal energy $U$ , which is neither created nor destroyed, it is only transformed. So, in this especific case of the compressed gas:

$\Delta U=Q+W$ (1)

Where:

$\Delta U$ is the variation in the internal (thermal) energy of the system (the value we want to find)

$Q=-100J$ is the heat transferred out of the gas (that is why it is negative)

$W$ is the work is done on the gas (as the gas is compressed, the work done on the gas must be considered positive )

On the other hand, the work done on the gas is given by:

$W=-P \Delta V$ (2)

Where:

$P=450kPa=450(10)^{3}Pa$ is the constant pressure of the gas

$\Delta V=V_{f}-V_{i}$ is the variation in volume of the gas

In this case the initial volume is $V_{i}=600{cm}^{3}=600(10)^{-6}m^{3}$ and the final volume is $V_{f}=200{cm}^{3}=200(10)^{-6}m^{3}$ .

This means:

$\Delta V=200(10)^{-6}m^{3}-600(10)^{-6}m^{3}=-400(10)^{-6}m^{3}$ (3)

Substituting (3) in (2):

$W=-450(10)^{3}Pa(-400(10)^{-6}m^{3})$ (4)

$W=180J$ (5)

Substituting (5) in (1):

$\Delta U=-100J+180J$ (6)

Finally:

$\Delta U=80J$ This is the change in thermal energy in the compression process.

8 0

3 years ago

A projectile of mass 2.0 kg is fired in the air at an angle of 40.0 ° to the horizon at a speed of 50.0 m/s. At the highest poin

tekilochka [14]

Answer:

a) The fragment speeds of 0.3 kg is 33.3 m / s on the y axis

0.7 kg is 109.4 ms on the x axis

b) Y = 109.3 m

Explanation:

This is a moment and projectile launch exercise.

a) Let's start by finding the initial velocity of the projectile

sin 40 = voy / v₀

$v_{oy}$ = v₀ sin 40

$v_{oy}$ = 50.0 sin40

$v_{oy}$ = 32.14 m / s

cos 40 = v₀ₓ / V₀

v₀ₓ = v₀ cos 40

v₀ₓ = 50.0 cos 40

v₀ₓ = 38.3 m / s

Let us define the system as the projectile formed t all fragments, for this system the moment is conserved in each axis

Let's write the amounts

Initial mass of the projectile M = 2.0 kg

Fragment mass 1 m₁ = 1.0 kg and its velocity is vₓ = 0 and $v_{y}$ = -10.0 m / s

Fragment mass 2 m₂ = 0.7 kg moves in the x direction

Fragment mass 3 m₃ = 0.3 kg moves up (y axis)

Moment before the break

X axis

p₀ₓ = m v₀ₓ

Y Axis y

$p_{oy}$ = 0

After the break

X axis

$p_{fx}$ = m₂ v₂

Axis y

$p_{fy}$ = m₁ v₁ + m₃ v₃

Let's write the conservation of the moment and calculate

Y Axis

0 = m₁ v₁ + m₃ v₃

Let's clear the speed of fragment 3

v₃ = - m₁ v₁ / m₃

v₃ = - (-10) 1 / 0.3

v₃ = 33.3 m / s

X axis

M v₀ₓ = m₂ v₂

v₂ = v₀ₓ M / m₂

v₂ = 38.3 2 / 0.7

v₂ = 109.4 m / s

The fragment speeds of 0.3 kg is 33.3 m / s on the y axis

0.7 kg is 109.4 ms on the x axis

b) The speed of the fragment is 33.3 m / s and has a starting height of where the fragmentation occurred, let's calculate with kinematics

$v_{fy}$ ² = $v_{oy}$ ² - 2 gy

0 = $v_{oy}$ ²-2gy

y = $v_{oy}$ ² / 2g

y = 32.14² / 2 9.8

y = 52.7 m

This is the height where the break occurs, which is the initial height for body movement of 0.3 kg

$v_{f}$ ² = $v_{y}$ ² - 2 g y₂

0 = $v_{y}$ ² - 2 g y₂

y₂ = $v_{y}$ ² / 2g

y₂ = 33.3²/2 9.8

y₂ = 56.58 m

Total body height is

Y = y + y₂

Y = 52.7 + 56.58

Y = 109.3 m

8 0

2 years ago

A force of 3600 N is exerted on a piston that has an area of 0.030 m2. What force is exerted on a second piston that has an area

Airida [17]

For Pascal's law, the pressure is transmitted with equal intensity to every part of the fluid:
$p_1 = p_2$
which becomes
$\frac{F_1}{A_1}= \frac{F_2}{A_2}$
where
$F_1=3600 N$ is the force on the first piston
$A_1=0.030 m^2$ is the area of the first piston
$F_2$ is the force on the second piston
$A_2=0.015 m^2$ is the area of the second piston

If we rearrange the equation and we use these data, we can find the intensity of the force on the second piston:
$F_2=F_1 \frac{A_2}{A_1}=(3600 N) \frac{0.015 m^2}{0.030 m^2}= 1800 N$

7 0

3 years ago

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A maroon plane has a takeoff speed of 88.3 m/s and requires 1635 m to reach that speed. Determine the acceleration of the plane

tensa zangetsu [6.8K]

18.5164213 if you divide them both you get that number so the volceity is the number shown above.

6 0

3 years ago