A 2.00 kg textbook is forced against one end of a horizontal spring of negligible mass and force constant 250 N/m, compressing t
1 answer:
Answer:
d = 1.875 m
Explanation:
Given that
m = 2 kg
k = 250 N/m
x= 0.3 m
Lets take the speed of book just before when it is leaving the spring
From energy conservation
250 x 0.3²= 2 x v²
v= 3.35
When it moves on the rough surface then de -acceleration a
a= μk g
a= 0.3 x 10 = 3 m/s²
The distance cover by book before getting to rest position is d
The final speed(V) of the block will be zero.
We know that
V² = v²- 2 a d
0 = 125 x 0.3² - 2 x 3 x d
d = 1.875 m
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C) 7.35*10⁶ N/C radially outward
Explanation:
- If we apply the Gauss'law, to a spherical gaussian surface with radius r=7 cm, due to the symmetry, the electric field must be normal to the surface, and equal at all points along it.
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- So, the +8 μC charge of the solid conducting sphere of radius 2cm, must be compensated by an equal and opposite charge on the inner surface of the conducting shell of total charge -4 μC.
- So, on the outer surface of the shell there must be a charge that be the difference between them:
- Replacing in (1) A = 4*π*ε₀, and Qenc = +4 μC, we can find the value of E, as follows:
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Answer : The final temperature of the mixture is
Explanation :
First we have to calculate the mass of water.
Mass = Density × Volume
Density of water = 1.00 g/mL
Mass = 1.00 g/mL × 180 cm³ = 180 g
In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.
where,
= specific heat of hot water (liquid) =
= specific heat of ice (solid)=
= mass of hot water = 180 g
= mass of ice = 20 g
= final temperature of mixture = ?
= initial temperature of hot water =
= initial temperature of ice =
Now put all the given values in the above formula, we get
Therefore, the final temperature of the mixture is