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valina [46]
3 years ago
11

Examine the lightbulbs in the circuit below. Write a sentence explaining what would happen if lightbulb A burned out. Repeat thi

s for lightbulbs B, C, and D.

Physics
1 answer:
tatiyna3 years ago
8 0

If the lightbulb A in the circuit shown in the image burned out, the path for the  current to flow is disrupted because one of its terminals is connected direct to the source. So, there will be no current through the lightbulbs B, C, and D, and they will turn off. Similarly it will happen, if the lightbulb D burned out.

If the lightbulb B burned out the current will continue circulating through the lightbulbs A, C, and D, because lightbulb B is connected in parallel. Similarly it will happen, if the lightbulb C burned out.

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If a 5.5 kg object experiences 15 N of force for .15 seconds what is the speed change
Nady [450]

The speed change : Δv = 0.41 m/s

<h3>Further explanation</h3>

Given

mass = 5.5 kg

Force = 15 N

time = 0.15 s

Required

the speed change

Solution

Newton 2nd's law

Impulse and momentum

F = m.a

F = m . Δv/t

F.t = m.Δv

Input the value :

15 N x 0.15 s = 5.5 kg x Δv

Δv = 0.41 m/s

8 0
2 years ago
A driver entering the outskirts of a city takes her foot off the accelerator so that the car slows down from 90 km/h to 50 km/h
Varvara68 [4.7K]

Answer:

Explanation:

a = (vf - vi) / t

a = (50 - 90) / 10.0

a = -4 km/h/s(1000 m/km / 3600 s/h)

a = - 1.11 m/s²

5 0
2 years ago
The potential energy stored in the compressed spring of a dart gun, with a spring constant of 32.50 N/m, is 0.640 J. Find by how
liraira [26]

Answer:

A

   x = 0.198456 \ m

B

    h  =  1.3061 \  m

C

 v =  5.06 \  m/s

D

  d = 4.0273 \  m

Explanation:

Considering the first question

From the question we are told that

   The spring constant is  k  =  32.50 N/m

    The potential energy is  PE  =  0.640 \ J

Generally the potential  energy stored in spring  is mathematically represented as   PE  =  \frac{1}{2}  *  k  *  x^2

=>    0.640=  \frac{1}{2}  * 32.50  *  x^2  

=>    x = \sqrt{0.03938}  

=>    x = 0.198456 \ m  

Considering the second question

 From the question we are told that

   The mass of the dart is  m =  0.050 kg

Generally from the law of energy conservation

         PE =  mgh

=>       0.640   =  0.050 *  9.8  *  h

=>      h  =  1.3061 \  m

Considering the third  question

   The height at which the dart was fired horizontally is  H  =   3.90\  m

Generally  from the law of energy conservation

         PE = KE

Here  KE is kinetic energy of the dart which is mathematical represented as

     KE  =  \frac{1}{2}  *  mv^2

=>      0.640 =  \frac{1}{2}  * 0.050 *  v^2

=>       v^2 = 25.6

=>       v =  5.06 \  m/s

Considering the fourth question

Generally the total time of flight of the dart is mathematically represented as

       t  =  \frac{ 2 *  H }{g}

=>     t  =  \frac{ 2 * 3.90 }{9.8 }

=>     t  =  0.7959 \ s

Generally the  horizontal distance from the equilibrium position to the ground is  mathematically represented as

       d =  v  *   t

=>     d = 5.06  *   0.7959

=>     d = 4.0273 \  m

5 0
2 years ago
A child whirls a 3.00 kg ball on a string .50 m from the axis of rotation in a horizontal circle. The ball makes 1 revolution in
melamori03 [73]

Answers:

a) 0.5 m/s^{2}

b) 1.5 N

Explanation:

a) The centripetal acceleration a_{c} of an object moving in a uniform circular motion is given by the following equation:  

a_{c}=\omega^{2} r  

Where:

\omega=1 \frac{rev}{s} is the angular velocity of the ball

r=0.5 m is the radius of the circular motion, which is equal to the length of the string

Then:

a_{c}=(1 \frac{rev}{s})^{2} 0.5 m  

a_{c}=0.5 m/s^{2} This is the centripetal acceleration of the ball

b) On the other hand, in this circular motion there is a force (centripetal force F) that is directed towards the center and is equal to the tension (T) in the string:

F=T=m. a_{c}

Where m=3 kg is the mass of the ball

Hence:

T=(3 kg)(0.5 m/s^{2})

T=1.5 N This is the tension in the string

7 0
3 years ago
You are asked to find a set of dimensionless parameters to organize data from a laboratory experiment, in which a tank is draine
mihalych1998 [28]

Answer:

Step 1

Given: A tank is drained through an orifice from initial liquid level (h₀). The time(r), to  drain the tank, depends on tank diameter (D), orifice diameter (d), acceleration due to gravity (g), liquid density (ρ) and liquid viscosity  (μ).

r = f (h₀, D, d, g, ρ, μ)

Step 2

There are totally seven parameters r, h₀, D, d, g, ρ and μ

∴ n = 7

The primary variables are M, L, T

∴ m = 3

So, the number of repeating variables are, r = 3

Select the three repeating variables are ρ, d, g

Then according to Buckingham, there will be n-m (⇒ 7-3 = 4) dimensionless groups given by π₁, π₂, π₃ and π₄

So, The number of π terms = 4

Explanation:

See attached images for step 3 to 7

6 0
3 years ago
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