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Natasha2012 [34]
3 years ago
6

When a light bulb is connected to a 4.5 V battery, a current of 0.16 A passes through the bulb filament. What is the resistance

of the filament
Physics
1 answer:
larisa [96]3 years ago
5 0

Answer:

R = 28.125 ohms

Explanation:

Given that,

The voltage of a bulb, V = 4.5 V

Current, I = 0.16 A

We need to find the resistance of the filament. Using Ohm's law,

V = IR

Where

R is the resistance of the filament

So,

R=\dfrac{V}{I}\\\\R=\dfrac{4.5}{0.16}\\\\R=28.125\ \Omega

So, the resistance of the filament is equal to 28.125 ohms.

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A bicyclist starting from rest applies a force of F 195 N to ride his bicycle across flat ground for a distance of d- 290 m befo
laila [671]

Answer: a) 56,550 J b) 30.1 m/s c) 321 m

Explanation:

a) By definition, work, is the process that does an applied force, in order to produce a displacement in the same direction than the applied force, and can be written as follows;

W = F. d. (scalar product of two vectors)

In this case, as the force is parallel to the displacement, work is directly equal to the product of  the applied force times the displacement (in magnitude), so we can write the following:

W = F. D = 195 N. 290 m = 56,550 J

b) In absence of friction, the work done by the force is equal to the change in the kinetic energy, as it can be showed using the work-energy theorem.

So, in this case, we can put the following:

W = ΔK ⇒F. D = 1/2 mv²

Solving for v, we get:

v=√2.F.D/m = 30.1 m/s

c) Now, if we assume that there is no friction between the bike and the ground, all the kinetic energy must become gravitational potential energy, at some height h.

We can write the following equation

m.g. h = 1/2 mv²

Simplifying, and taking g = 9.8 m/s², we can find h:

h= 46.2 m

Now, we need to know the distance travelled up the incline, which is related with the height h, by the angle that the incline does with the horizontal, as follows:

sin 8.29° = h /d ⇒ d= h / sin 8.29° = 46.2 m / sin 8.29° = 321 m

6 0
4 years ago
A 0 .1 kg pinball is fired horizontally by a springwith force constant 40 N/m. If the spring is depressed 10 cm and the ball col
ANEK [815]

Answer:

Explanation:

We can find out the velocity of pinball after being fired from the equation of conservation of mechanical energy

1/2 m v² = 1/2 k x² , m is mass of ball , v is velocity after firing , k is spring constant , x is depression in spring.

.1 x v² = 40 x .1 x .1

v = 2 m /s

ball of mass .1 , collides with velocity 2 m /s ,

from the equation of elastic collision

v₂ = (m₂ - m₁) u₁ / (m₂ + m₁) + 2m₁m₂u₂ / (m₂ + m₁)

m₁ ,u₁ are mass and velocity of first ball , m₂ , u₂ are mass and velocity of second mass , v₂ is velocity of second mass after collision.

u₁ = 2 , u₂ = 0 , m₁ = .1  m₂ = .3

v₂ = (.3 - .1 ) x 2/ (.3 + .1 )

= (.2 / .4 )x 2

= 1 m /s

So .3 kg ball will move with velocity of 1 m /s in the same direction as .1 m mass collided with it.

.

4 0
3 years ago
The deck of a bridge is suspended 235 feet above a river. If a pebble falls off the side of the bridge, the height, in feet, of
Aleks04 [339]

Answer:

(a) 1, average velocity = -65.6 m/s

   2, average velocity = -64.8 m/s

   3, average velocity = -64.16 m/s

(b) The instantaneous velocity is -96 m/s

Explanation:

(a)

Average velocity is given  by;

y(t_2,t_1) = \frac{y(t_2) - y(t_1)}{t_2-t_1}

(1)

y(2.1,2) = \frac{(235-16*2.1^2) - (235-16*2^2)}{2.1-2}\\\\ y(2.1,2) = -65.6 \ m/s

(2)

y(2.05,2) = \frac{(235-16*2.05^2) - (235-16*2^2)}{2.05-2}\\\\ y(2.05,2) = -64.8 \ m/s

(3)

y(2.01,2) = \frac{(235-16*2.01^2) - (235-16*2^2)}{2.01-2}\\\\ y(2.01,2) = -64.16 \ m/s

b. y = 235 - 16t²

The instantaneous velocity is given by;

v = dy /dt

dy / dt = -32t

when t = 3 s

v = -32(3)

v = -96 m/s

5 0
4 years ago
A ray of monochromatic light ( f = 5.09 × 10^14 Hz) passes from air into Lucite at an angle
harina [27]
     Let us consider the air with the index 1 and the lucite with index 2. Using the Snell's Secound Law, we have:

\frac{sen\O_{2}}{sen\O_{1}} = \frac{n_{2}}{n_{1}} \\ sen\O_{2}= \frac{n_{2}*sen\O_{1}}{n_{1}}
 
     Entering the unknowns, remembering that the air refrective index is 1 and the lucite refrective index is 1.5, comes:

sen\O_{2}= \frac{n_{2}*sen\O_{1}}{n_{1}} \\ sen\O_{2}= \frac{1.5* \frac{1}{2} }{1} \\ sen\O_{2}=0.75
  
     Using the arcsin properties, we get:

sen\O_{2}=0.75 \\ arcsin(0.75)=\O_{2} \\ \boxed {\O_{2}=48.59^o}

Obs: Approximate results, and the drawing is attached

If you notice any mistake in my english, let me know, because i am not native.

6 0
3 years ago
5. An ion consist of 12 electrons and 11 proton. What is the charge<br> on the ion?
sammy [17]

Answer:

-1

Explanation:

Electrons have a negative charge and protons have a positive charge. (+11) + (-12) = -1

5 0
3 years ago
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