Answer:
285 Litre ( 1kg of water occupies 1 Litre )
which is identical to 285 dm^3 ie 285 cubes of length 10 cm
and as there are 10 of these linearly in 1 m then there are 10^3 = 1000 dm^3 in each m^3
giving a volume of 0.285 m^3
Of course if you KNEW that 1m^3 of water has a mass of 1000 kg then you would simply have said that the volume was 285/1000 m^3
Answer:
a)
85.05 N/m
b)
179.81 rad/s
Explanation:
a)
k = spring constant of the spring
m = mass of the block = 0.473 kg
x = stretch caused in the spring = 0.109 m
h = height dropped by the block = 0.109 m
Using conservation of energy
Spring potential energy gained by the spring = Potential energy lost by the block
(0.5) k x² = mgh
(0.5) k x² = mgx
(0.5) k x = mg
(0.5) k (0.109) = (0.473) (9.8)
k = 85.05 N/m
b)
angular frequency is given as
= 179.81 rad/s
Answer:
a. ac = 1844.66 m/s²
b. Fc = 265.63 N
Explanation:
a.
The centripetal acceleration of the ball is given as follows:
ac = v²/r
where,
ac = centripetal acceleration = ?
v = speed of ball = (87 mph)(1 h/ 3600 s)(1609.34 m / 1 mile) = 38.9 m/s
r = radius of path = 82 cm = 0.82 m
Therefore,
ac = (38.9 m/s)²/0.82 m
<u>ac = 1844.66 m/s²</u>
<u></u>
b.
The centripetal force is given as:
Fc = (m)(ac)
Fc = (0.144 kg)(1844.66 m/s²)
<u>Fc = 265.63 N</u>
Answer:
The density is
Explanation:
From the question we are told
The radius of the cylinder is
The length of the cylinder is
The density of the glucose is
The depth is
The net force on the string is
The net force on the string is mathematically represented as
Here is the force due to the weight of the cylinder
Which mathematically evaluated as
Where is the volume of liquid displaced which is equal to the volume of the cylinder which is mathematically represented as
=>
is the buoyant force acting on the cylinder due to the glucose
So substituting into formula for
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