<span>A sheet of copper could cause the object to lose the most amount of heat. Copper is an essential element and a good conductor of heat. Heat can transfer from one end of a piece of copper to the other end.</span>
Answer:
3.1 m/s
Explanation:
First, find the time it takes for the cat to land. Take down to be positive.
Given:
Δy = 0.61 m
v₀ = 0 m/s
a = 9.81 m/s²
Find: t
Δy = v₀ t + ½ at²
(0.61 m) = (0 m/s) t + ½ (9.81 m/s²) t²
t = 0.353 s
Now find the horizontal velocity needed to travel 1.1 m in that time.
Given:
Δx = 1.1 m
a = 0 m/s²
t = 0.353 s
Find: v₀
Δx = v₀ t + ½ at²
(1.1 m) = v₀ (0.353 s) + ½ (0 m/s²) (0.353 s)²
v₀ = 3.1 m/s
<span>Well, It is the aphelion point, When the Earth is farthest away from the Sun, when the Northern Hemisphere is warm. the Earth is closest to the Sun, or at the perihelion, 2 weeks after the June Solstice, when the Northern Hemisphere is enjoying warm summer months. Well this kind of weather is very nice.</span>
Answer:Water Only
Explanation:
Given
vessel is insulated therefore no heat can be added or removed i.e. heat exchange is zero
If hot water at is mixed with cold water at then at equilibrium vessel contains only water and final temperature of water will be between and
Heat released by hot water is equal to heat gain by cold water .
Answer:
the question is incomplete, the complete question is
"A circular coil of radius r = 5 cm and resistance R = 0.2 ? is placed in a uniform magnetic field perpendicular to the plane of the coil. The magnitude of the field changes with time according to B = 0.5 e^-t T. What is the magnitude of the current induced in the coil at the time t = 2 s?"
2.6mA
Explanation:
we need to determine the emf induced in the coil and y applying ohm's law we determine the current induced.
using the formula be low,
where B is the magnitude of the field and A is the area of the circular coil.
First, let determine the area using where r is the radius of 5cm or 0.05m
since we no that the angle is at
we determine the magnitude of the magnetic filed
the Magnitude of the voltage is 0.000532V
Next we determine the current using ohm's law