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3 years ago

If two objects are the

Physics

1 answer:

evablogger [386]3 years ago

8 0

Answer:

I would say the smaller object.

Explanation:

In any case, thermal energy can be spread throughout the object easier, and quicker, rather than the bigger object.

For example: You are boiling two pots of water, let's suppose Pot A has twice as much as Pot B:

|--------------------- | | |

| | |----------------------- |

|_____________| |______________|

Pot A Pot B

If you boil them at both the same time, temperature, and continue that for sometime, your results will be that Pot B will have risen in temperature alot faster than Pot A

The comparison:

Pot A and Pot B's Amount of water - Supposed to be the different size of the objects

Stove/Boiling: Would be if thermal heat were to enter the objects at the same temperatures.

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The aorta is approximately 25 mm in diameter. The mean pressure there is about 100 mmHg and the blood flows through the aorta at

Ksenya-84 [330]

Answer:

Explanation:

25 mm diameter

r₁ = 12.5 x 10⁻³ m radius.

cross sectional area = a₁

Pressure P₁ = 100 x 10⁻³ x 13.6 x 9.8 Pa

a )

velocity of blood v₁ = .6 m /s

Cross sectional area at blockade = 3/4 a₁

Velocity at blockade area = v₂

As liquid is in-compressible

a₁v₁ = a₂v₂

a₁ x .6 m /s = 3/4 a₁ v₂

v₂ = .8m/s

b )

Applying Bernauli's theorem formula

P₁ + 1/2 ρv₁² = P₂ + 1/2 ρv₂²

100 x 10⁻³ x 13.6 x10³x 9.8 + 1/2 X 1060 x .6² = P₂ + 1/2x 1060 x .8²

13328 +190.8 = P₂ + 339.2

P₂ = 13179.6 Pa

= 13179 / 13.6 x 10³ x 9.8 m of Hg

P₂ = .09888 m of Hg

98.88 mm of Hg

8 0

4 years ago

The instrument which is commonly used to measure the intensity of radioactivity is called a

tatyana61 [14]

The instrument is a Geiger counter and is used to measure radioactive level around people's bodies.

4 0

3 years ago

Read 2 more answers

7. Un bloque de 700 N se encuentra sobre una viga uniforme de 200 N y 6 m de longitud. El bloque está a una distancia de 1 m del

GrogVix [38]

Answer:

x = 0.176 m

Explanation:

For this exercise we will take the condition of rotational equilibrium, where the reference system is located on the far left and the wire on the far right. We assume that counterclockwise turns are positive.

Let's use trigonometry to decompose the tension

sin 60 = $T_{y}$ / T

T_{y} = T sin 60

cos 60 = Tₓ / T

Tₓ = T cos 60

we apply the equation

∑ τ = 0

-W L / 2 - w x + T_{y} L = 0

the length of the bar is L = 6m

-Mg 6/2 - m g x + T sin 60 6 = 0

x = (6 T sin 60 - 3 M g) / mg

let's calculate

let's use the maximum tension that resists the cable T = 900 N

x = (6 900 sin 60 - 3 200 9.8) / (700 9.8)

x = (4676 - 5880) / 6860

x = - 0.176 m

Therefore the block can be up to 0.176m to keep the system in balance.

5 0

4 years ago

Single Replacement: HCI + Zn

Kay [80]

During a single replacement, the Zn and H will switch places. That means the product formed will be ZnCl and H2.

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3 years ago

An athlete at the gym holds a 3.0 kg steel ball in his hand. His arm is 60 cm long and has a mass of 3.8 kg, with the center of

Serggg [28]

Answer:

(a) τ = 26.58 Nm

(b) τ = 18.79 Nm

Explanation:

(a)

First we find the torque due to the ball in hand:

τ₁ = F₁d₁

where,

τ₁ = Torque due to ball in hand = ?

F₁ = Force due to ball in hand = m₁g = (3 kg)(9.8 m/s²) = 29.4 N

d₁ = perpendicular distance between ball and shoulder = 60 cm = 0.6 m

τ₁ = (29.4 N)(0.6 m)

τ₁ = 17.64 Nm

Now, we calculate the torque due to the his arm:

τ₁ = F₁d₁

where,

τ₂ = Torque due to arm = ?

F₂ = Force due to arm = m₂g = (3.8 kg)(9.8 m/s²) = 37.24 N

d₂ = perpendicular distance between center of mass and shoulder = 40% of 60 cm = (0.4)(60 cm) = 24 cm = 0.24 m

τ₂ = (37.24 N)(0.24 m)

τ₂ = 8.94 Nm

Since, both torques have same direction. Therefore, total torque will be:

τ = τ₁ + τ₂

τ = 17.64 Nm + 8.94 Nm

τ = 26.58 Nm

(b)

Now, the arm is at 45° below horizontal line.

First we find the torque due to the ball in hand:

τ₁ = F₁d₁

where,

τ₁ = Torque due to ball in hand = ?

F₁ = Force due to ball in hand = m₁g = (3 kg)(9.8 m/s²) = 29.4 N

42.42 cm = 0.4242 m

τ₁ = (29.4 N)(0.4242 m)

τ₁ = 12.47 Nm

Now, we calculate the torque due to the his arm:

τ₁ = F₁d₁

where,

τ₂ = Torque due to arm = ?

F₂ = Force due to arm = m₂g = (3.8 kg)(9.8 m/s²) = 37.24 N

d₂ = perpendicular distance between center of mass and shoulder = 40% of (60 cm)(Cos 45°) = (0.4)(42.42 cm) = 16.96 cm = 0.1696 m

τ₂ = (37.24 N)(0.1696 m)

τ₂ = 6.32 Nm

Since, both torques have same direction. Therefore, total torque will be:

τ = τ₁ + τ₂

τ = 12.47 Nm + 6.32 Nm

τ = 18.79 Nm

3 0

3 years ago