Answer: something must drop it over
Explanation:
 
        
                    
             
        
        
        
Given:
m₁ = 1540 g, mass of iron horseshoe
T₁ = 1445 °C, initial temperature of horseshoe
c₁ = 0.4494 J/(g-°C), specific heat
m₂ = 4280 g, mass of water
T₂ = 23.1 C, initial temperature of water
c₂ = 4.18 J/(g-°C), specific heat of water
L = 947,000 J heat absorbed by the water.
Let the final temperature be T °C.
For energy balance,
m₁c₁(T₁ - T) = m₂c₂(T - T₂) + L
(1540 g)*(0.4494 J/(g-C))*(1445-T C) = (4280 g)*(4.18 J/(g-C))*(T-23.1 C) + 947000 J
692.076(1445 - T) = 17890(T - 23.1) + 947000
10⁶ - 692.076T = 17890T - 413259 + 947000
466259 = 18582.076T
T = 25.09 °C
Answer: 25.1 °C
        
             
        
        
        
Answer:
The hill should be not less than 0.625 m high
Explanation:
This problem can be solved by using the principle of conservation of mechanical energy. In the absence of friction, the total mechanical energy is conserved. That means that
 is constant, being U the potential energy and K the kinetic energy
 is constant, being U the potential energy and K the kinetic energy


When the car is in the top of the hill, its speed is 0, but its height h should be enough to produce the needed speed v down the hill.
The Kinetic energy is then, zero. When the car gets enough speed we assume it is achieved at ground level, so the potential energy runs out to zero but the Kinetic is at max. So the initial potential energy is transformed into kinetic energy.

We can solve for h:

The hill should be not less than 0.625 m high
 
        
             
        
        
        
Answer:
Ws=8.75 Watts
Explanation:
As per fig. of prob 02.061, it is clear that R5 and R4 are in parallel, its equivalent Resistance will be:



Now, this equivalent Req45 is in series with R3, therefore:

This Req345 is in parallel with R2, i.e

Now this gets in series with R1:

Now, the power delivered Ws is:
