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cluponka [151]
3 years ago
13

If two lightbulbs are parallel to each other, they have the same brightness. If the same two lightbulbs are put in series with e

ach other, they also have the same brightness but are dimmer than when they are in parallel. Explain why.
Physics
1 answer:
djverab [1.8K]3 years ago
6 0

Answer:

Explanation:

The bulbs becomes dimmer because effective resistance of the circuit increases since the resistance of the circuit in series connection involves adding the individual resistance compared to in parallel connection where the effective resistance = 1 / R1 + 1/R2 = R1 R2 / ( R2 + R1). The brightness remains the same because the lightbulbs must be of the same resistance. Also using ohm's law

V = IR

I = V / R

I is inversely proportional to the current in the circuit so as R, effective resistance of the circuit increases, I, current decreases  and vice versa

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Question 2 (1 point)
Tju [1.3M]

Answer:

I know someone anwsered but it would be 400M

Explanation:

i initial velocity (u)=10m/s

acceleration (a)=0

time taken (t) =40s

then distance (s)=u t +1/2 a t^2

s= u t +0 (as a is 0)

s= 10 x 40

s= 400M

7 0
3 years ago
A 0.300 kg block is pressed against a spring with a spring constant of 8050 N/m until the spring is compressed by 6.00 cm. When
natita [175]

Answer:

a) \mu_{k} = 0.704, b) R = 0.312\,m

Explanation:

a) The minimum coeffcient of friction is computed by the following expression derived from the Principle of Energy Conservation:

\frac{1}{2}\cdot k \cdot x^{2} = \mu_{k}\cdot m\cdot g \cdot \Delta s

\mu_{k} = \frac{k\cdot x^{2}}{2\cdot m\cdot g \cdot \Delta s}

\mu_{k} = \frac{\left(8050\,\frac{N}{m} \right)\cdot (0.06\,m)^{2}}{2\cdot (0.3\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot (7\,m)}

\mu_{k} = 0.704

b) The speed of the block is determined by using the Principle of Energy Conservation:

\frac{1}{2}\cdot k \cdot x^{2} = \frac{1}{2}\cdot m \cdot v^{2}

v = x\cdot \sqrt{\frac{k}{m} }

v = (0.06\,m)\cdot \sqrt{\frac{8050\,\frac{N}{m} }{0.3\,kg} }

v \approx 9.829\,\frac{m}{s}

The radius of the circular loop is:

\Sigma F_{r} = -90\,N -(0.3\,kg)\cdot (9.807\,\frac{m}{s^{2}} ) = -(0.3\,kg)\cdot \frac{v^{2}}{R}

\frac{\left(9.829\,\frac{m}{s}\right)^{2}}{R} = 309.807\,\frac{m}{s^{2}}

R = 0.312\,m

5 0
3 years ago
A cylindrical specimen of brass that has a diameter listed above, a tensile modulus of 110 GPa, and a Poisson's ratio of 0.35 is
Irina-Kira [14]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The strain experienced by the specimen is 0.00116 which is option A

Explanation:

The explanation is shown on the second uploaded image

8 0
3 years ago
In your own words explain how "Carrying Capacity" plays a vital role in today's world and our future as well
olganol [36]

Increased food production due to improved agricultural practices, control of many diseases by modern medicine and the use of energy to make historically uninhabitable areas of Earth inhabitable are examples of things which can extend carrying capacity.

8 0
3 years ago
Read 2 more answers
Two point charges are placed on the x axis.The firstcharge, q1= 8.00 nC, is placed a distance 16.0 mfromthe origin along the pos
deff fn [24]

Answer:

E = (0, 0.299) N

Explanation:

Given,

  • Charge q_1\ =\ 8.0\ nC
  • Charge q_2\ =\ 6.0\ nC
  • Distance of the first charge from the origin = (16m, 0)
  • Distance of the second charge from the origin = (-9, 0)
  • Point where the electric field required = (0, 12m)

Let \theta_1\ and\ theta_2 be the angle of the electric fields by first and second charge at the point A.

\therefore sin\theta_1\ =\ \dfrac{12}{20}\\\Rightarrow \theta_1\ =\ sin^{-1}\left (\dfrac{12}{20}\ \right )\\\Rightarrow \theta_1\ =\ 36.87^o\\\\\therefore sin\theta_1\ =\ \dfrac{12}{9}\\\Rightarrow \theta_1\ =\ sin^{-1}\left (\dfrac{12}{9}\ \right )\\\Rightarrow \theta_1\ =\ 53.13^o\\

Electric field by charge q_1 at point A,

F_1\ =\ \dfrac{kq_1}{r_1^2}\\\Rightarrow F_1\ =\ \dfrac{9\times 10^9\times 8\times 10^{-9}}{20^2}\\\Rightarrow F_1\ =\ 0.18\ N/C

Electric field by the charge q_2 at point A,

F_1\ =\ \dfrac{kq_1}{r_1^2}\\\Rightarrow F_1\ =\ \dfrac{9\times 10^9\times 6.0\times 10^{-9}}{16^2}\\\Rightarrow F_1\ =\ 0.24\ N/C

Now,

Net electric field in horizontal direction at point AF_x\ =\ F_{1x}\ +\ F_{2x}\\\Rightarrow F_x\ =\ F_1cos\theta_1\ +\ F_2cos\theta_2\\\Rightarrow F_x\ =\ 0.18\times( -cos36.87^o)\ +\ 0.24\times cos53.13^o\\\Rightarrow F_x\ =\ -0.144\ +\ 0.144\ N/C\\\Rightarrow F_x\ =\ 0\ N/C

Net electric field in vertical direction at point A.

F_y\ =\ F_{1y}\ +\ F_{2y}\\\Rightarrow F_y\ =\ F_1sin\theta_1\ +\ F_2sin\theta_2\\\Rightarrow F_y\ =\ 0.18\times sin36.87^o\ +\ 0.24\times sin53.13^o\\\Rightarrow F_y\ =\ 0.180\ +\ 0.192\\\Rightarrow F_y\ =\ 0.299\ N/C

Hence, the net electric field  at point A,

F\ =\ ( 0, 0.299 )\ N/C.

5 0
3 years ago
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