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vladimir2022 [97]

3 years ago

9

A 75 W lightbulb is being run on a 110 V outlet. Determine the resistance of the lightbulb. Provide a detailed description of th

e process.

1 answer:

Sauron [17]3 years ago

8 0

Power is defined as

P = I*V

where I is the current and V is the voltage

Ohm's law gives us the relation betwen Voltage and current in a resistive component

V = I*R , Then

P = V² / R

We solve for R,

R = (110 V)²/ 75W = 161.33 ohms

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An object of mass m = 5.0 kg hangs from a cord around a light pulley: The length of the cord between the oscillator and the pull

puteri [66]

Answer:

$\mu=0.0049Kg/m$

Explanation:

When a standing wave is formed with six loops means the normal mode of the wave is n=6, the frequency of the normal mode is given by the expression:

$f_n=\frac{nv}{2L}$

Where $L$ is the length of the string and $v$ the velocity of propagation. Use this expression to find the value of $v$ .

$f_6=\frac{6v}{2L}\\(150)=\frac{6v}{2(2)} \\150=\frac{3v}{2} \\3v=150(2)\\ v=\frac{300}{3} \\v=100m/s$

The velocity of propagation is given by the expression:

$v=\sqrt{\frac{T}{\mu }$

Where $\mu$ is the desirable variable of the problem, the linear mass density, and $T$ is the tension of the cord. The tension is equal to the weight of the mass hanging from the cord:

$T=W=mg=(5)(9.81)=49.05N$

With the value of the tension and the velocity you can find the mass density:

$v=\sqrt{\frac{T}{\mu}$

$v^2=\frac{T}{\mu}\\ \mu=\frac{T}{v^2} =\frac{49.05}{(100)^2} =\frac{49.05}{10000} =0.0049Kg/m$

6 0

2 years ago

Two bicycle tires are set rolling with the same initial speed of 4.0 m/s along a long, straight road, and the distance each trav

umka2103 [35]

Answer:

The coefficient of rolling friction will be "0.011".

Explanation:

The given values are:

Initial speed,

$v_i = 4.0 \ m/s$

then,

$v_f=\frac{4.0}{2}$

$=2.0 \ m/s$

Distance,

s = 18.2 m

The acceleration of a bicycle will be:

⇒ $a=\frac{v_f^2-v_i^2}{2s}$

On substituting the given values, we get

⇒ $=\frac{(2.0)^2-(4.0)^2}{2\times 18.2}$

⇒ $=\frac{4-8}{37}$

⇒ $=\frac{-4}{37}$

⇒ $=0.108 \ m/s^2$

As we know,

⇒ $f=ma$

and,

⇒ $\mu_rmg=ma$

⇒ $\mu_r=\frac{a}{g}$

On substituting the values, we get

⇒ $=\frac{0.108}{9.8}$

⇒ $=0.011$

7 0

3 years ago

Bullets spin when shot from a rifle or handgun. What causes this spinning?

tamaranim1 [39]

Answer:

The spark from the primer ignites the gunpowder. Gas converted from the burning powder rapidly expands in the cartridge. The expanding gas forces the bullet out of the cartridge and down the barrel with great speed. The rifling in the barrel causes the bullet to spin as it travels out of the barrel.

Explanation:

7 0

2 years ago

A rugby player sits on a scrum machine that weighs 200 Newtons. Given that the coefficient of static friction is 0.67, the coeff

Trava [24]

a. 850 N is the minimum force needed to get the machine/player system moving, which means this is the maximum magnitude of static friction between the system and the surface they stand on.

By Newton's second law, at the moment right before the system starts to move,

• net horizontal force

∑ F[h] = F[push] - F[s. friction] = 0

• net vertical force

∑ F[v] = F[normal] - F[weight] = 0

and we have

F[s. friction] = µ[s] F[normal]

It follows that

F[weight] = F[normal] = (850 N) / (0.67) = 1268.66 N

where F[weight] is the combined weight of the player and machine. We're given the machine's weight is 200 N, so the player weighs 1068.66 N and hence has a mass of

(1068.66 N) / g ≈ 110 kg

b. To keep the system moving at a constant speed, the second-law equations from part (a) change only slightly to

∑ F[h] = F[push] - F[k. friction] = 0

∑ F[v] = F[normal] - F[weight] = 0

so that

F[k. friction] = µ[k] F[normal] = 0.56 (1268.66 N) = 710.45 N

and so the minimum force needed to keep the system moving is

F[push] = 710.45 N ≈ 710 N

4 0

1 year ago

The ocean’s tides are not only affected by the shape of the coastlines, or slope of the ocean floor, but also by the gravitation

dexar [7]

Answer:

a) According to Newton's law of gravitation, as the distance between the Moon and the Earth decreases, the gravitational attraction increases and vice versa

The gravitational force of the Moon on the Earth causes the Earth to be slightly bulged on the side directly facing the Moon

The gravitational force also pulls the water bodies on the Earth's surface towards the Moon in the same manner and the effect is more pronounced due to the ability of the liquid water to assume a shape based on the magnitude of the gravitational field attracting it

Therefore, the region where the Moon is closest to the Earth we have a high tide as the water level rises and the region which is perpendicular to where the Moon is located has a low tide

b) The two special types of tides are

1) The neap tide

2) The spring tide

Neap tide

Neap tide occurs when the Sun and Moon are 90° apart from each other when they are viewed by an observer from Earth

The gravitational pull of the Sun cancels (partially) the effect of the gravitational pull and tidal force of the Moon, resulting in minimum tidal range

Spring Tide

Spring tide occurs when the Earth, the Moon, and the Sun are simultaneously inline, such that the Sun reinforces the gravitational pull and tidal force of the Moon, resulting in a maximum tidal range

Explanation:

4 0

3 years ago

Read 2 more answers