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Roman55 [17]
3 years ago
11

Two astronauts are taking a spacewalk outside the International Space

Physics
1 answer:
Tems11 [23]3 years ago
5 0

Answer:

im not sure

Explanation:

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Which of the following is an example of a lever? knife ramp pencil sharpener wheelbarrow
Kay [80]
I believe it is a knife
5 0
3 years ago
Read 2 more answers
In which of the two situations described is more energy transferred?
Furkat [3]

Answer:

More energy is transferred in situation A

Explanation:

Each of the situations are analyzed as follows;

Situation A

The temperature of the cup of hot chocolate = 40 °C

The temperature of the interior of the freezer in which the chocolate is placed = -20 °C

We note that at 0°C, the water in the chocolate freezes

The energy transferred by the chocolate to the freezer before freezing is given approximately as follows;

E₁ = m×c₁×ΔT₁

Where;

m = The mass of the chocolate

c₁ = The specific heat capacity of water = 4.184 kJ/(kg·K)

ΔT₁ = The change in temperature from 40 °C to 0°C

Therefore, we have;

E₁ = m×4.184×(40 - 0) = 167.360·m kJ

The heat the coffee gives to turn to ice is given as follows;

E₂ = m·H_f

Where;

H_f = The latent heat of fusion = 334 kJ/kg

∴ E₂ = m × 334 kJ/kg = 334·m kJ

The heat required to cool the frozen ice to -20 °C is given as follows;

E₃ = m·c₂·ΔT₂

Where;

c₂ = The specific heat capacity of ice = 2.108 kJ/(kg·K)

Therefore, we have;

E₃ = m × 2.108 ×(0 - (-20)) = 42.16

E₃ = 42.16·m kJ/(kg·K)

The total heat transferred = (167.360 + 334 + 42.16)·m kJ/(kg·K) = 543.52·m kJ/(kg·K)

Situation B

The temperature of the cup of hot chocolate = 90 °C

The temperature of the room in which the chocolate is placed = 25 °C

The heat transferred by the hot cup of coffee, E, is given as follows;

E = m×4.184×(90 - 25) = 271.96

∴ E = 271.96 kJ/(kg·K)

Therefore, the total heat transferred in situation A is approximately twice the heat transferred in situation B and is therefore more than the heat transferred in situation B

Energy transferred in situation A = 543.52 kJ/(kg·K)

Energy transferred in situation B = 271.96 kJ/(kg·K)

Energy transferred in situation A ≈ 2 × Energy transferred in situation B

∴ Energy transferred in situation A > Energy transferred in situation B.

3 0
3 years ago
(b) A car of mass 3000 kg travels at a constant velocity of 5.0 m/s.
Tamiku [17]

Answer:

16.7 s

Explanation:

T= <u>Vf - Vo</u>          a= <u>F</u>

         a                    m

4,500 / 3000 = 1.5 (a)

30 - 5 / 1.5(a) = 16.7 s      

4 0
3 years ago
Your friend says that if Newton’s third law is correct, no object would ever start moving. Here is his argument: You pull a sled
m_a_m_a [10]

Answer:

Explanation:

You pull a sled exerting a 50 N force on it , sled also exerts a force on you . These forces are action and reaction force , as per third law of Newton . These two forces are equal and  opposite . But they do not act on the same object so they do not cancel each other . They act on different objects , one on the sledge and the other on you . Due to force on sledge , sledge moves in the direction of force or towards you . You will start moving in opposite direction if frictional force of ground is nil or less .

6 0
3 years ago
A vector has an x-component of length 10 and a y-component of length 3. What is the angle of the vector? (Hint: Use the inverse
Sonja [21]
The vector, the x-component and the y-component form a rectangle triangle where the vector is the hypothenuse and the x and y components are the two sides.
Calling \alpha the angle between the vector and the horizontal direction (x), the two sides are related to \alpha by
\tan \alpha  =  \frac{v_y}{v_x}
where vy and vx are the two components on the y- and x-axis. Using vx=10 and vy=3 we find
\tan \alpha  =  \frac{3}{10} =0.3
And so the angle is
\alpha = \arctan (0.3)=16.7^{\circ}
5 0
3 years ago
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