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Anika [276]
3 years ago
9

The wooden block is removed from the water bath and placed in an unknown liquid. In this liquid, only one-third of the wooden bl

ock is submerged. Is the unknown liquid more or less dense than water?
Physics
1 answer:
anygoal [31]3 years ago
4 0

Answer:

1.5 g/cm^3

Explanation:

In the question i am assuming that wooden block of weight W floats with exactly one-half of its volume below the waterline.

Let ρ_{liquid} be the density of the unknown liquid , V= volume submerged

g= acceleration due to gravity

then

weight of block = \rho_{water}\times V\times g/2

when the block is in water

Also,

weight of block = \rho_{liquid}\times V\times g/3

dividing eq 1 by eq 2 we get

\rho_{liquid} = 3/2 \rho_{water}

⇒\rho_{liquid} = 3/2

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A 0.520-kg object attached to a spring with a force constant of 8.00 N/m vibrates in simple harmonic motion with an amplitude of
Veseljchak [2.6K]

Answer:

Explanation:

Given that,

Mass attached to spring

M = 0.52kg

Force constant K = 8N/m

Amplitude A = 11.6 cm

a. Maximum speed?

Angular velocity is calculated using

w = √k/m

w = √8/0.52

w = √15.385

w = 3.922rad/s

Then, the relation ship between angular velocity and linear velocity is given as

v = - w•A

v = - 3.922 × 11.6

v = - 45.5 cm/s

Then, the maximum velocity is

vmax = |v|= 45.5cm/s

b. Acceleration a?

Acceleration can be determine using the formula

a = -w²• A

a = -3.922² × 11.6

a = -178.46 cm/s²

Magnitude of the acceleration is 178.46cm/s²

c. Speed when the object is at 9.6cm from equilibrium position?

Generally,

The position of the object at equilibrium is

x(t) = A•Cos(wt)

x(t) = 11.6 Cos (3.922t)

Then, when x(t) = 9.6cm

9.6 = 11.6 Cos(3.92t)

Cos(3.922t) = 9.6/11.6

Cos(3.922t) = 0.8276

3.922t = ArcCos(0.8276)

Note: the angle is in radiant

3.922t = 0.596

t = 0.596/3.922

t = 0.152 second

Then, v(t) at that time is

v(t) = x'(t) = -11.6×3.92Sin(3.922t)

v(t) = -45.5Sin(3.922t)

Now, when t =0.152

v(t) = -45.5 Sin(3.922×0.152)

v(t) = -45.5Sin(0.596)

v(t) = -25.5 cm/s

Then, it's magnitude is 25.5cm/s

d. Acceleration at same position

t = 0.152s

a(t) = v'(t) = - 45.5×3.922Cos(3.922t)

a(t) = -178.46Cos(3.92t)

a(t) = -178.46 Cos(3.92×0.152)

a(t) = -178.46 Cos(0.596)

a(t) = -147.68 cm/s²

Magnitude of the acceleration is 147.68 cm/s²

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3 years ago
The "atomic weight" of an atom reflects the average number of
Illusion [34]

E) Protons, neutrons, and electrons

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3 years ago
Compare tidal range to spring tide
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Spring tides have higher high tides and lower low tides whereas neap tides have lower high tides and higher low tides. Hence, the range is much larger in a spring tide than in a low tide.

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3 years ago
A charged particle of mass 0.0050 kg is subjected to a 5.0 T magnetic field which acts at a right angle to its motion. If the pa
Irina18 [472]

Answer:

0.01 C

Explanation:

Applying,

F = qvBsinФ................ Equation 1

Where F = Force on the charged particle, q = charge on the particle, v = velocity, B = magnetic field, Ф =  angle

Since the charged particle noves in a circle,

F = mv²/r................. Equation 2

Where m = mass of the particle, v = velocity of the particle, r = radius of the  circle

Substitute equation 2 into equation 1

mv²/r = qvBsinФ

make q the subject of the equation

q = mv/(rBsinФ)............. Equation 3

Given: m = 0.005 kg, v = 2 m/s, r = 0.2 m, B = 5 T, Ф = 90° (Act at right angle)

Substitute these values into equation 3

q = (0.005×2)/(0.2×5×sin90°)

q = 0.01/(1)

q = 0.01 C

5 0
3 years ago
A- what constant acceleration, in SI units, must a car have to gofrom zero to 60 mph in 10 s?
umka2103 [35]

Answer:

(a) 0.017m/s^2

(b) 17/100,000

(c) 0.17m, 0.558ft

Explanation:

(a) speed = 60mph = 60m/1h × 1h/3600s = 0.017m/s, time = 10s

Acceleration (a) = speed ÷ time = 0.017m/s ÷ 10s = 0.0017m/s^2

(b) g = 9.8m/s^2, a = 0.0017m/s^2

a/g = 0.0017/9.8 = 0.00017 = 17/100,000

(c) Distance = speed × time = 0.017m/s × 10s = 0.17m

Distance in foot = 0.17 × 3.2808ft = 0.558ft

3 0
3 years ago
Read 2 more answers
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