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Svetradugi [14.3K]
3 years ago
9

___________ is a tiny, hair-like fiber that aids in movement.

Physics
1 answer:
Vinvika [58]3 years ago
5 0
Cilium are tiny, hair like fiber that aids in movement.

hope the answer helped you.
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Standing and holding a barbell that has a mass of 100kg at a height of 2m involves being done on the barbell to maintain
Basile [38]

Answer:

1960Joules

Explanation:

Since we are not told what too find, we can as well find the Gravitational Potential Energy.

GPE = mass * acceleration due to gravity * height

GPE = 100*9.8 * 2

GPE = 980*2

GPE = 1960Joules

Hence the gravitational potential Energy is 1960Joules

6 0
3 years ago
A 82-kg fisherman in a 112-kg boat throws a package of mass m = 15 kg horizontally toward the right with a speed of vi = 4.8 m/s
abruzzese [7]

Answer:

0.37 m/s to the left

Explanation:

Momentum is conserved.  Initial momentum = final momentum.

m₁ u₁ + m₂ u₂ = m₁ v₁ + m₂ v₂

Initially, both the fisherman/boat and the package are at rest.

0 = m₁ v₁ + m₂ v₂

Plugging in values and solving:

0 = (82 kg + 112 kg) v + (15 kg) (4.8 m/s)

v = -0.37 m/s

The boat's velocity is 0.37 m/s to the left.

8 0
2 years ago
Analyzing the Light Bulb: You should have noticed that the light bulb doesn't have a single well-defined "resistance," since the
VLD [36.1K]

Answer:

Resistance increases with increase in temperature which depends on power supplied which also depends on voltage.

Thermal expansion will make resistance larger.

Explanation:

Light bulb is a good example of a filament lamp. If we plot the graph of voltage against current we will notice that resistance is constant at constant temperature.

The filament heats up when an electric current passes through it, and produces light as a result.

The resistance of a lamp increases as the temperature of its filament increases. The current flowing through a filament lamp is not directly proportional to the voltage across it.

tensile stress begins to appear in resistor as the temperature rises. Thus, the resistance value increases as the temperature rises. Resistance value can only decrease as the temperature rises in case of thin film resistor with aluminium substrate.

In case of a filament bulb, the resistance will increase as increase in length of the wire. The thermal expansion in this regard is linear expansivity in which resistance is proportional to length of the wire.

Resistance therefore get larger.

6 0
3 years ago
A 6.0 m wire with a mass of 50 g, is under tension. A transverse wave, for which the frequency is 810 Hz, the wavelength is 0.40
MrRissso [65]

Answer:

a) t = 0.0185 s = 18.5 ms

b) T = 874.8 N

Explanation:

a)

First we find the seed of wave:

v = fλ

where,

v = speed of wave

f = frequency = 810 Hz

λ = wavelength = 0.4 m

Therefore,

v = (810 Hz)(0.4 m)

v = 324 m/s

Now,

v = L/t

where,

L = length of wire = 6 m

t = time taken by wave to travel length of wire

Therefore,

324 m/s = 6 m/t

t = (6 m)/(324 m/s)

<u>t = 0.0185 s = 18.5 ms</u>

<u></u>

b)

From the formula of fundamental frquency, we know that:

Fundamental Frequency = v/2L = (1/2L)(√T/μ)

v = √(T/μ)

where,

T = tension in string

μ = linear mass density of wire = m/L = 0.05 kg/6 m = 8.33 x 10⁻³ k gm⁻¹

Therefore,

324 m/s = √(T/8.33 x 10⁻³ k gm⁻¹)

(324 m/s)² = T/8.33 x 10⁻³ k gm⁻¹

<u>T = 874.8 N</u>

8 0
2 years ago
A 120 g, 8.0-cm-diameter gyroscope is spun at 1000 rpm and allowed to precess. What is the precession period?
dolphi86 [110]

To solve this problem we will derive the expression of the precession period from the moment of inertia of the given object. We will convert the units that are not in SI, and finally we will find the precession period with the variables found. Let's start defining the moment of inertia.

I = MR^2

Here,

M = Mass

R = Radius of the hoop

The precession frequency is given as

\Omega = \frac{Mgd}{I\omega}

Here,

M = Mass

g= Acceleration due to gravity

d = Distance of center of mass from pivot

I = Moment of inertia

\omega= Angular velocity

Replacing the value for moment of inertia

\Omega= \frac{MgR}{MR^2 \omega}

\Omega = \frac{g}{R\omega}

The value for our angular velocity is not in SI, then

\omega = 1000rpm (\frac{2\pi rad}{1 rev})(\frac{1min}{60s})

\omega = 104.7rad/s

Replacing our values we have that

\Omega = \frac{9.8m/s^2}{(8*10^{-2}m)(104.7rad)}

\Omega = 1.17rad/s

The precession frequency is

\Omega = \frac{2\pi rad}{T}

T = \frac{2\pi rad}{\Omega}

T = \frac{2\pi}{1.17}

T = 5.4 s

Therefore the precession period is 5.4s

7 0
2 years ago
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