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BartSMP [9]
3 years ago
7

A 120 resistor a 60 ohm resistor and a 40 ohm resistor are connected in parallel to a 120 volt power source. what is the current

running through the entire circuit?
Physics
1 answer:
larisa [96]3 years ago
6 0

Answer:

6 A

Explanation:

First of all, we need to calculate the equivalent resistance of the circuit. The three resistors are connected in parallel, so their equivalent resistance is given by:

\frac{1}{R_T}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}=\frac{1}{120 \Omega}+\frac{1}{60 \Omega}+\frac{1}{40 \Omega}=\frac{3+2+1}{120 \Omega}=\frac{6}{120 \Omega}\\R_T = \frac{120}{6} \Omega

And now we can use Ohm's law to find the current in the circuit:

V=R_T II=\frac{V}{R_T}=\frac{120 V}{\frac{120}{6}\Omega}=6 A

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A student drops two metallic objects into a 120-g steel container holding 150 g of water at 25°C. One object is a 206-g cube of
spayn [35]

Answer:

Mass of the aluminium chunk = 278.51 g

Explanation:

For an isolated system as given the energy lost and gains in the system will be zero therefore sum of all transfer of energy will be zero,as the temperature will also remain same

A specific heat formula is given as                  

Energy Change = Mass of liquid x Specific Heat Capacity x Change in temperature

                                       Q =  m×c×ΔT

                        Heat gain by aluminium + heat lost by copper  = 0    (1)

For Aluminium:

      Q = m\times0.897\frac{J}{g.k}\times(25-5)

      Q = m x 17.94 joule

For Copper:

Q= 206g\times0.385\frac{J}{g.k} \times(88-25)

       Q= 4996.53 Joule

from eq 1

     m x 17.94 = 4996.53

     mass of aluminium = \frac{4996.53}{17.94} g

    Mass of the aluminium chunk = 278.51 g

                         

3 0
3 years ago
Suppose you run into a wall at 4.5 meters per second (about 10 mph). Let's say the wall brings you to a complete stop in 0.5 sec
STatiana [176]

Answer with Explanation:

We are given that

Initial velocity,u=4.5 m/s

Time=t =0.5 s

Final velocity=v=0m/s

We have to find the deceleration and estimate the force exerted by wall on you.

We know that

Acceleration=\frac{v-u}{t}

Using the formula

Acceleration=a=\frac{0-4.5}{0.5}

deceleration=a=-9m/s^2

We know that

Force =ma

Using the formula and suppose mass  of my body=m=40 kg

The force exerted by wall on you

Force=40\times (9)=360N

3 0
2 years ago
A small block on a frictionless, horizontal surface has a mass of 0.0250 kg. It is attached to a massless cord passing through a
frosja888 [35]

Answer:

a) Yes

b) 7 rad/s

c) 0.01034 J

Explanation:

a)

Yes the angular momentum of the block is conserved since the net torque on the block is zero.

b)

m = mass of the block = 0.0250 kg

w₀ = initial angular speed before puling the cord = 1.75 rad/s

r₀ = initial radius before puling the cord = 0.3 m

w = final angular speed after puling the cord = ?

r = final  radius after puling the cord = 0.15 m

Using conservation of angular momentum

m r₀² w₀ = m r² w

r₀² w₀ = r² w

(0.3)² (1.75) = (0.15)² w

w = 7 rad/s

c)

Change in kinetic energy is given as

ΔKE = (0.5) (m r² w² - m r₀² w₀²)

ΔKE = (0.5) ((0.025) (0.15)² (7)² - (0.025) (0.3)² (1.75)²)

ΔKE = 0.01034 J

6 0
3 years ago
Waves A and B have the same speed, but wave A has a shorter wavelength. Which wave has the higher frequency?
Nuetrik [128]
Frequency and wavelength are inversely proportional.
A shorter wavelength implies a higher frequency.
3 0
3 years ago
A white billiard ball with mass mw = 1.37 kg is moving directly to the right with a speed of v = 3.14 m/s and collides elastical
Ira Lisetskai [31]

That latest value for the Angle is in Grads, not in Kilograms.

Apply law of conservation of momentum along vertical direction.

m_1v_1sin\theta_1-m_2v_2sin\theta_2=0

v_2=\frac{v_1sin\theta_1}{sin\theta_2}=\frac{sin54}{sin36}v_1 = 1.376v_1

Apply law of conservation of momentum along the horizontal direction

m_1u_1=m_1v_1cos\theta_1+m_2v_2cos\theta_2

u_1=v_1(cos\theta_1+1.376cos\theta_2)

u_1=v_1(cos(54)+1.376cos(36))

u_1=(1.7) v_1

v_1= \frac{3.14}{1.7}=1.84m/s

The second ball velocity is v_2 = (1.376)(1.84)=2.531m/s

The magnitud of final total momentum is

m(v_1+v_2)=(1.37)(2.531+1.84)=5.98kgm/s

The magnitude of final energy is

\frac{1}{2}m(v^2_1+v^2_2)=\frac{1}{2}(1.37)(2.531^2+1.84^2)=6.07J

6 0
3 years ago
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