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Oliga [24]
4 years ago
4

What relationship is discussed in the zeroth law of thermodynamics

Physics
1 answer:
klemol [59]4 years ago
6 0
The answer is Thermal Equilibrium.                                                                      
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When a trebuchet counterweight is hoisted by soldiers, what form of energy is
vladimir1956 [14]
10. A safe place to use the trebuchet would be away from other buildings and people. A good example of a place would be a large field with no nearby structures. 
14. Many factors need to be kept consistent throughout the experiment. One example of a variable that would need to be consistent is the weight and size of the projectile. 
15. It is important to do many trials so that you can make sure that the results of each trial are nearly the same. If they are all vastly different, then it means that something has gone wrong. 

Sorry I was only able to answer a few questions, but I hope these few answers help! :) 
4 0
3 years ago
Read 2 more answers
A 65.0 kg ice skater standing on frictionless ice throws a 0.15 kg snowball horizontally at a speed of 32.0 m/s. What is the vel
spin [16.1K]

Answer:

(d) 0.07 m/s

Explanation:

Given Data

Snowball mass m₁=0.15 kg

Ice skater mass m₂=65.0 kg

Snowball velocity v₁=32.0 m/s

To find

Velocity of Skater v₂=?

Solution

From law of conservation of momentum

m_{1}v_{1}=m_{2}v_{2}\\  v_{2}=\frac{m_{1}v_{1}}{m_{2}}\\ v_{2}=\frac{(0.15kg)(32.0m/s)}{65.0kg}\\ v_{2}=0.0738m/s\\or\\v_{2}=0.07 m/s

So Option d is correct one

5 0
4 years ago
Suppose a certain car supplies a constant deceleration of A meter per second per second. If it is traveling at 90km/hr. When. th
aksik [14]

Answer:

i)-6.25m/s

ii)18 metres

iii)26.5 m/s or 95.4 km/hr

Explanation:

Firstly convert 90km/hr to m/s

90 × 1000/3600 = 25m/s

(i) Apply v^2 = u^2 + 2As...where v(0m/s) is the final speed and u(25m/s) is initial speed and also s is the distance moved through(50 metres)

0 = (25)^2 + 2A(50)

0 = 625 + 100A....then moved the other value to one

-625 = 100A

Hence A = -6.25m/s^2(where the negative just tells us that its deceleration)

(ii) Firstly convert 54km/hr to m/s

In which this is 54 × 1000/3600 = 15m/s

then apply the same formula as that in (i)

0 = (15)^2 + 2(-6.25)s

-225 = -12.5s

Hence the stopping distance = 18metres

(iii) Apply the same formula and always remember that the deceleration values is the same throughout this question

0 = u^2 + 2(-6.25)(56)

u^2 = 700

Hence the speed that the car was travelling at is the,square root of 700 = 26.5m/s

In km/hr....26.5 × 3600/1000 = 95.4 km/hr

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3 years ago
When electrons in an atom in an excited state fall to lower energy levels, energy is?
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Radiated away as electromagnetic radiation.
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I have a pure breed homozygous long claw monster (CC), Claude and
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