Weight of the body will be more in air water Milk or Mercury

Explain why the extrapolated temperature is used to determine the maximun temperature of the mixture rather than the highest rec

Oksi-84 [34.3K]

The extrapolated temperature is used to define the maximum temperature of the mixture relatively than the highest recorded temperature in which the conclusion will effect in a higher specific heat value. Heat is bound to escape from whatever apparatus is using, therefore it is needed to account for the loss of the heat that does not go into increasing the temperature of the mixture.

6 0

3 years ago

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How do you solve 0.004 dm + 0.12508 dm?

Effectus [21]

0.004 of something added to 0.12508 of the same thing
adds up to 0.12908 of it.

The thing could be a glass of water, a sheet of paper,
a pound of ground beef, a gallon of gas, or a snowball.
In this problem, it just happens to be a dm.

7 0

2 years ago

A ball is thrown vertically upward. What is its acceleration right before it hits the earth?

Bezzdna [24]

The entire time the ball is in the air, its acceleration is 9.8 m/s2 down provided this occurs on the surface of the Earth. Note that the acceleration can be either 9.8 m/s2 or -9.8 m/s2.
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7 0

3 years ago

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An elaborate pulley consists of four identical balls at the ends of spokes extending out from a rotating drum. A box is connecte

Klio2033 [76]

Answer:

its speed will be less than V

Explanation:

When the ball falls a distance d, its final kinetic energy plus rotational kinetic energy of the drum equals its initial potential energy.

K = U

With its speed V at the end of d, we have

1/2mV² + 1/2Iω² = mgd where I = rotational inertia of drum and balls, ω = angular speed of drum and balls and m = mass of box

1/2mV² + 1/2Iω² = mgd

1/2mV² = mgd - 1/2Iω²

V² = [2(mgd - 1/2Iω²)/m]

V = √[2(mgd - 1/2Iω²)/m]

When the four balls are moved inward closer to the drum, their rotational inertia increases and also its angular speed which thus causes an increase in rotational kinetic energy. But, since the box still falls the same distance of d, its final kinetic energy plus rotational kinetic energy of the drum plus balls still equals its initial potential energy

K = U

I' = new rotational inertia of drum and balls, ω' = new angular speed of drum and balls

With its new speed is now V' at the end of d,

1/2mV'² + 1/2I'ω'² = mgd

1/2mV'² = mgd - 1/2I'ω'²

V² = [2(mgd - 1/2I'ω'²)/m]

V' = √[2(mgd - 1/2I'ω'²)/m]

Since I' and ω' increase, the rotational kinetic energy of the drum and balls (1/2I'ω'²) increases. Thus, the difference (mgd - 1/2I'ω'²) < (mgd - 1/2Iω²) which implies that the kinetic energy of the box decreases. Hence, since its kinetic energy decreases, its speed V' also decreases.

So, V' < V

So, its speed will be less than V.

3 0

3 years ago

9. Captain America is chasing Red Skull. He plans to throw his shield to knock down Red Skull but needs to know how fast Red Sku

Sedaia [141]

Red Skull's relative velocity to Captain America, towards the left front of the

truck is approximately 33.23 m/s in a direction from the North of

approximately 9.18°.

Reasons:

Assumptions;

Taking the north direction as positive.

The activity takes place on the trucks.

The trucks are moving towards each other.

Solution:

Vector form of net speed of Red Skull, is given as follows;

v₁ = -( $\frac{\sqrt{2} }{2}$ × 3.5)·i + ( $\frac{\sqrt{2} }{2}$ × 3.5 + 12.5)·j

Vector form of the net speed of Captain America is given as follows;

v₂ = ( $\frac{\sqrt{2} }{2}$ × 4.0)·i - ( $\frac{\sqrt{2} }{2}$ × 4.0 + 15)·j

Relative velocity, v₁₂ = v₁ - v₂

∴ v₁₂ = (-( $\frac{\sqrt{2} }{2}$ × 3.5) - ( $\frac{\sqrt{2} }{2}$ × 4.0))·i + (( $\frac{\sqrt{2} }{2}$ × 3.5 + 12.5) + ( $\frac{\sqrt{2} }{2}$ × 4.0 + 15))·j

v₁₂ = $-\frac{ 15 \cdot \sqrt{2} }{4}$ ·i + $\frac{ 110 + 15 \cdot \sqrt{2} }{4}$ ·j

Red Skull's velocity relative to Captain America, v₁₂ = $-\frac{ 15 \cdot \sqrt{2} }{4}$ ·i + $\frac{ 110 + 15 \cdot \sqrt{2} }{4}$ ·j

v₁₂ ≈ -5.3·i + 32.8·j

Therefore;

Red Skull appears to be moving West at 5.3 m/s and North at 32.8 m/s

The direction is $arctan \left(\frac{32.8}{-5.3} \right) \approx -80.2^{\circ}$

Therefore;

Red Skull appear to be moving at 90° - 80.2° ≈ 9.18° towards the left front end of the truck moving North

The magnitude of the velocity, |v₁₂|, is given as follows;

$|v_{12}| = \sqrt{\left(-\frac{ 15 \cdot \sqrt{2} }{4}\right)^2 + \left(\frac{ 110 + 15 \cdot \sqrt{2} }{4}\right)^2} = \dfrac{ 5 \cdot \sqrt{130+33 \cdot\sqrt{2} } }{2} \approx 33.23$ ·

The magnitude of Red Skull's velocity relative to Captain America is,

therefore;

|v₁₂| ≈ 33.23 m/s

Learn more here:

brainly.com/question/24430414

6 0

2 years ago

Weight of the body will be more in air water Milk or Mercury​

Weight of the body will be more in air water Milk or Mercury