Answer:
The correct solution will be "12.0 A".
Explanation:
The given values are:
By using the transformer formula, we get
⇒ 
⇒ 
On substituting the given values, we get
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⇒ 
Answer:
Water normally freezes at 0°C (32°F). Salt lowers the freezing temperature. (That is, it can remain a liquid at much lower temperatures.)
When sprinkled on ice, the salt lowers the freezing temperature of the water which effectively melts the ice when the salt dissolves into it. There is a limit to how low it can reduce the temperature, though. If the temperature drops below -9°C (15°F), it's too cold for the salt to dissolve into the ice.
When making ice cream, the salt lowers the temperature of the ice and water sufficiently enough to freeze the cream.
Answer:
3.0 cm
Explanation:
We can solve this problem by using the mirror equation:
where
f is the focal length of the mirror
p is the distance of the object from the mirror
q is the distance of the image from the mirror
In this problem we have:
f = 1.5 cm is the focal length of the mirror (positive for a concave mirror)
p = 3.0 cm is the distance of the object from the mirror
Therefore, the distance of the image is:
And the positive sign means that the image is real.
(The second part of the exercise is just the description of the image of the first exercise).
That’s really easy ask your teacher and also peace happy
Answer:
The angular acceleration α = 14.7 rad/s²
Explanation:
The torque on the rod τ = Iα where I = moment of inertia of rod = mL²/12 where m =mass of rod and L = length of rod = 4.00 m. α = angular acceleration of rod
Also, τ = Wr where W = weight of rod = mg and r = center of mass of rod = L/2.
So Iα = Wr
Substituting the value of the variables, we have
mL²α/12 = mgL/2
Simplifying by dividing through by mL, we have
mL²α/12mL = mgL/2mL
Lα/12 = g/2
multiplying both sides by 12, we have
Lα/12 × 12 = g/2 × 12
αL = 6g
α = 6g/L
α = 6 × 9.8 m/s² ÷ 4.00 m
α = 58.8 m/s² ÷ 4.00 m
α = 14.7 rad/s²
So, the angular acceleration α = 14.7 rad/s²
