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Elena-2011 [213]
3 years ago
13

Which of these temperatures is likely when a container of water at 20° C is mixed with water at 28° C?

Physics
1 answer:
Sliva [168]3 years ago
8 0
Ok when two differing temperatures are mixed together with no special circumstances indicated they will form a equilibrium temperature between the two differing temperature by effectively "sharing" their energy. So the answer should be B.23*C
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A 12.0-g bullet is fired horizontally into a 109-g wooden block that is initially at rest on a frictionless horizontal surface a
kykrilka [37]

Answer:

v₀ = 280.6 m / s

Explanation:

we have the shock between the bullet and the block that we can work with at the moment and another part where the assembly (bullet + block) compresses a spring, which we can work with mechanical energy,

We write the mechanical energy when the shock has passed the bodies

   Em₀ = K = ½ (m + M) v²

We write the mechanical energy when the spring is in maximum compression

Em_{f} = K_{e} \\= \frac{1}{2} kx^2\\    Em_0 = Em_{f}

½ (m + M) v² = ½ k x²

Let's calculate the system speed

   v = √ [k x² / (m + M)]

   v = √[152 ×0.78² / (0.012 +0.109) ]

   v = 27.65 m / s

This is the speed of the bullet + Block system

Now let's use the moment to solve the shock

Before the crash

   p₀ = m v₀

After the crash

p_{f} = (m + M) v

The system is formed by the bullet and block assembly, so the forces during the crash are internal and the moment is preserved

 p_0 =  p_{f}

  m v₀ = (m + M) v

  v₀ = v (m + M) / m

let's calculate

v₀ = 27.83 (0.012 +0.109) /0.012

  v₀ = 280.6 m / s

4 0
3 years ago
Please Help ASAP!!!
const2013 [10]

Answer:

The Forces of Flight

At any given time, there are four forces acting upon an aircraft.  

These forces are lift, weight (or gravity), drag and thrust. Lift is  

the key aerodynamic force that keeps objects in the air. It is the  

force that opposes weight; thus, lift helps to keep an aircraft in  

the air. Weight is the force that works vertically by pulling all  

objects, including aircraft, toward the center of the Earth. In order  

to fly an aircraft, something (lift) needs to press it in the opposite  

direction of gravity. The weight of an object controls how strong  

the pressure (lift) will need to be. Lift is that pressure. Drag is a  

mechanical force generated by the interaction and contract of a  

solid body, such as an airplane, with a fluid (liquid or gas). Finally,  

the thrust is the force that is generated by the engines of an  

aircraft in order for the aircraft to move forward.

Explanation:

5 0
2 years ago
An 85 kg man and his 35 kg daughter are sitting on opposite ends of a 3.00 m see-saw. The see-saw is anchored in the center. If
wolverine [178]

Answer:

0.54m

Explanation:

Step one:

given data

length of seesaw= 3m

mass of man m1= 85kg

weight = mg

W1= 85*10= 850N

mass of daughter m2= 35kg

W2= 35*10= 350N

distance from the center= (1.5-0.2)= 1.3m

Step two:

we know that the sum of clockwise moment equals the anticlockwise moment

let the distance the must sit to balance the system be x

taking moment about the center of the system

350*1.3=850*x

455=850x

divide both sides by  850

x=455/850

x=0.54

Hence the man must sit 0.54m from the right to balance the system

3 0
3 years ago
A 600kg car is moving at 5 m/s to the right and elastically collided with a stationary 900 kg car. What is the velocity of the 9
11111nata11111 [884]

Answer:

\mathrm{v}_{2} \text { velocity after the collision is } 3.3 \mathrm{m} / \mathrm{s}

Explanation:

It says “Momentum before the collision is equal to momentum after the collision.” Elastic Collision formula is applied to calculate the mass or velocity of the elastic bodies.

m_{1} v_{1}=m_{2} v_{2}

\mathrm{m}_{1} \text { and } \mathrm{m}_{2} \text { are masses of the object }

\mathrm{v}_{1} \text { velocity before the collision }

\mathrm{v}_{2} \text { velocity after the collision }

\mathrm{m}_{1}=600 \mathrm{kg}

\mathrm{m}_{2}=900 \mathrm{kg}

\text { Velocity before the collision } v_{1}=5 \mathrm{m} / \mathrm{s}

600 \times 5=900 \times v_{2}

3000=900 \times v_{2}

\mathrm{v}_{2}=\frac{3000}{900}

\mathrm{v}_{2}=3.3 \mathrm{m} / \mathrm{s}

\mathrm{v}_{2} \text { velocity after the collision is } 3.3 \mathrm{m} / \mathrm{s}

5 0
3 years ago
Help would be greatly appreciated:) thank you! a pendulum clock is brought to mars. How does the bob move on Mars as compared to
Alborosie
It runs slower <span>as gravity is lower so acceleration due to gravity is smaller</span>
8 0
3 years ago
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