How much current flows in the circuit if two resistors of 6 connected in parallel are
supplied with potential difference of 12 volts?​

Answer:

The amplitude of the subsequent oscillations is 13.3 cm

Explanation:

Given;

mass of the block, m = 1.25 kg

spring constant, k = 17 N/m

speed of the block, v = 49 cm/s = 0.49 m/s

To determine the amplitude of the oscillation.

Apply the principle of conservation of energy;

maximum kinetic energy of the stone when hit = maximum potential energy of spring when displaced

[tex]K.E_{max} = U_{max}\\\\\frac{1}{2} mv^2 = \frac{1}{2} kA^2\\\\where;\\\\A \ is \ the \ maximum \ displacement = amplitude \\\\mv^2 = kA^2\\\\A^2 = \frac{mv^2}{k} \\\\A = \sqrt{\frac{mv^2}{k}} \\\\A = \sqrt{\frac{1.25\ \times \ 0.49^2}{17}} \\\\A = 0.133 \ m\\\\A = 13.3 \ cm[/tex]

Therefore, the amplitude of the subsequent oscillations is 13.3 cm

Answer:

[tex]V.E=498.02m/s^2[/tex]

Explanation:

From the question we are told that:

Radius [tex]r=301Km[/tex]

Gravitational acceleration [tex]g=0.412 m/s^2[/tex]

Generally the equation for Escape velocity is mathematically given by

 [tex]V.E^2=2gr[/tex]

 [tex]V.E^2=2*0.412m/s^2*301000[/tex]

 [tex]V.E^2=248024[/tex]

 [tex]V.E=\sqrt{248024}[/tex]

 [tex]V.E=498.02m/s^2[/tex]

Answer:

Explanation:

First off, this lens is concave. Second, the image is obviously smaller, and third, the only thing that is NOT obvious, is the fact that real images are always upside down and virtual images are always right-side-up. So the choice you're looking for is D.

Answer:

D - A concave lens forms a smaller, virtual image

Explanation:

Answer:

Option (c) is correct.

Explanation:

The apparent change in the frequency of light due to the relative motion between the source and the observer is called Doppler's effect.

When the source is moving towards the observer which is at rest, the apparent frequency increases and if the observer is moving away the frequency of sound decreases.

It occurs for both light and sound.  

So, to explain the blue shift of light in the universe is due to the Doppler's effect of light.

Answer:

C. 28.2 deg

Explanation:

The horizontal range of a projectile is given as:

[tex]R = \frac{v^2Sin2\theta}{g}[/tex]

where,

R = Range

v = speed

θ = angle of launch

g = acceleration due to gravity = 9.81 m/s²

First, we will find the launch speed (v) by using the initial conditions:

R = 120 m

θ = 45°

Therefore,

[tex]120\ m = \frac{v^2Sin 90^o}{9.81\ m/s^2}\\\\v = \sqrt{(120\ m)(9.81\ m/s^2)}\\\\v = 34.31\ m/s[/tex]

Now, consider the second scenario to hit the target:

R = 100 m

Therefore,

[tex]100\ m = \frac{(34.31\ m/s)^2Sin2\theta}{9.81\ m/s^2}\\\\Sin2\theta = \frac{(100\ m)(9.81\ m/s^2)}{(34.31\ m/s)^2}\\\\2\theta = Sin^{-1}(0.833)\\\\\theta = \frac{56.44^o}{2}\\\theta = 28.22^o[/tex]

Hence, the correct option is:

C. 28.2 deg

Answer:

40 gf.

Explanation:

Please see attached photo for diagram.

In the attached photo, W is the weight of metre rule.

The weight of the metre rule can be obtained as follow:

Clockwise moment = (W×10) + (10×35)

Clockwise moment = 10W + 350

Anticlock wise moment = 25 × 30

Anticlock wise moment = 750

Clockwise moment = Anticlock wise moment

10W + 350 = 750

Collect like terms

10W = 750 – 370

10W = 400

Divide both side by 10

W = 400/ 10

W = 40 gf

Thus, the weight of the metre rule is 40 gf

Answer:

40 gf

Explanation:

The balance point of the uniform meter rule with the suspended weights = 40 cm = The pivot point

The location where the 25 gf weight is suspended = 10 cm

The location where the 10 gf weight is suspended = 75 cm

Let W represent the weight of the meter rule.

We have that the location of the application of the weight of the meter rule is at the center, 50 cm mark, point

Given that the meter rule is balanced, and taking moment about the pivot point, we have;

The moment om the left hand side, LHS, of the pivot point = The moment on the right hand side, RHS, of the pivot point

The moment on the LHS = 25 gf × (40 cm - 10 cm) = 750 gf·cm

The moment on the RHS = W × (50 cm - 40 cm) + 10 gf × (75 cm - 40 cm)

The moment on the RHS = W·(10 cm)  + 350 gf·cm

∴ 750 gf·cm = W·(10 cm) + 350 gf·cm

W·(10 cm) = 750 gf·cm - 350 gf·cm = 400 gf·cm

W = (400 gf·cm)/(10 cm) = 40 gf

The weight of the meter scale (rule), W = 40 gf.

Answer:

A. The size of the ice patches

Explanation:

In an experiment, the control variable also known as the CONSTANT is the variable that must be kept uniform or the same for all groups throughout the experiment in order not to influence the outcome of the experiment.

According to the experiment described in this question, the effect of different salts on melting points is investigated by a student. Sodium chloride (NaCl), Calcium Chloride (CaCl2), Potassium Carbonate (KCO3) and inert sand are the four types of salt used. The volume of the ice used and melted was finally measured. This means that the SIZE OF THE ICE PATCHES USED is the control variable of the experiment as the same size was used for all groups throughout.

Answer:

the size of the ice patches

Explanation:

Answer:

i) a₁ = -g (sin θ + μ cos θ), x = v₀² / 2a₁

ii) W = mg L sin  θ ,  iii)     Wₙ = 0

iv)  W = - μ m g  L cos  θ x

Explanation:

With a drawing this exercise would be clearer, I understand that you have a block on a ramp and it is subjected to some force that makes it rise, for example the tension created by a descending block.

The movement is that when the system is released, the tension forces are greater than the friction and the component of the weight and therefore the block rises up the ramp

At some point the tension must become zero, when the hanging block reaches the ground, as the block has a velocity it rises with a negative acceleration to a point and stops where the friction force and the weight component would be in equilibrium along the way. along the plane

i) Let's use Newton's second law

the reference system is with the x axis parallel to the ramp

Axis y

      N - W cos θ = 0

X axis

      T - W sin θ - fr = ma

the friction force is

      fr = μ N

      fr = μ mg cos θ

we substitute

      T - m g sin sin θ - μ mg cos θ = m a

      a = T / m - g (sin θ + μ cos θ)

With this acceleration we can find the height that the block reaches, this implies that at some point the tension becomes zero, possibly when a hanging block reaches the floor.

      T = 0

       a₁ = -g (sin θ + μ cos θ)

       v² = v₀² - 2a1 x

       v = 0       at the highest point

       x = v₀² / 2a₁

ii) the work of the gravitational force is

       W = F .d

       W = mg sin  θ   L

iii) the work of the normal force

the force has 90º with respect to the displacement so cos 90 = 0

         Wₙ = 0

iv) friction force work

friction force always opposes displacement

         W = - fr d

         W = - μ m g cos  θ L

Answer:

2 more neutrons

Explanation:

To obtain the answer to the question, let us calculate the number of neutrons in carbon–14 and standard carbon (i.e carbon–12). This can be obtained as follow:

For carbon–14:

Mass number = 14

Proton number = 6

Neutron number =?

Mass number = Proton + Neutron

14 = 6 + Neutron

Collect like terms

14 – 6 = Neutron

8 = Neutron

Neutron number = 8

For carbon–12:

Mass number = 12

Proton number = 6

Neutron number =?

Mass number = Proton + Neutron

12 = 6 + Neutron

Collect like terms

12 – 6 = Neutron

6 = Neutron

Neutron number = 6

SUMMARY:

Neutron number of carbon–14 = 8

Neutron number of carbon–12 = 6

Finally, we shall determine the difference in the neutron number. This can be obtained as follow:

Neutron number of carbon–14 = 8

Neutron number of carbon–12 = 6

Difference =?

Difference = (Neutron number of carbon–14) – (Neutron number of carbon–12)

Difference = 8 – 6

Difference = 2

Therefore, carbon–14 has 2 more neutrons than standard carbon (i.e carbon–12)

Answer:

parallax

Due to foreshortening, nearby objects show a larger parallax than farther objects when observed from different positions, so parallax can be used to determine distances. To measure large distances, such as the distance of a planet or a star from Earth, astronomers use the principle of parallax.

Answer:

$900

Explanation:

Step 1: Our output value is 9000.

Step 2: We represent the unknown value with x.

Step 3: From step 1 above,$9000=100\%$

Step 4: Similarly, x=10%

Step 5: This results in a pair of simple equations:

$9000=100

Step 6: By dividing equation 1 by equation 2 and noting that both the RHS (right hand side) of both

equations have the same unit (%); we have

\frac{9000}{x}=\frac{100\%}{10\%}

Step 7: Again, the reciprocal of both sides gives

\frac{x}{9000}=\frac{10}{100}$

\Rightarrow x=900$

Therefore, $10\%$ of $9000$ is $900$

unknown value with x

step 1 above,$9900=100%

similarly ,x=10%

Answer:

0.25s

Explanation:

5/20 = 0.25s

This might be correct

Answer:

(a) 3.93 m/s

(b) 861.66 m

Explanation:

A = 4.5 m/s  [27° S of W] for 3.0 minutes

B = 3.5 m/s [35° S of E] for 4.1 minutes

Distance A = 4.5 x 3 x 60 = 810  m

Distance B = 3.5 x 4.1 x 60 = 861 m

(a) The average speed is defined as the ratio of the total distance to the total time.

Total distance, d = 810 + 861 = 1671 m

total time, t = 3 + 4.1 = 7.1 minutes = 7.1 x 60 = 426 seconds

The average speed is

[tex]v=\frac{1671}{426}=3.93 m/s[/tex]

(b)

[tex]\overrightarrow{A} = 810(- cos 27 \widehat{i} - sin 27 \widehat{j})=- 721.7 \widehat{i} - 367.7 \widehat{j}\\\\\overrightarrow{B} = 861( cos 35 \widehat{i} - sin 35 \widehat{j})= 705.3 \widehat{i} - 493.8 \widehat{j}\\\\\overrightarrow{C} = (- 721.7 + 705.3) \widehat{i} - (367.7 + 493.8) \widehat{j} \\\\\overrightarrow{C}= - 16.4 \widehat{i} - 861.5 \widehat{j}[/tex]

The magnitude is

[tex]C =\sqrt{16.4^2+861.5^2} = 861.66 m[/tex]

Answer:

equilibrium

Explanation:

acceleration will remain constant

Answer:

T₂ = 305.17 K

Explanation:

Given that,

Heat, Q = 6000 J

Mass, m = 200 gram

Initial temperature, T₁ = 25° C

We need to find its final temperature. Let it is T₂.

We know that,

[tex]Q=mc\Delta T[/tex]

Where

c is the specific heat of water, c = 4.18 J/g°C

So,

[tex]6000=200\times 4.18\times (T_2-298)\\\\\dfrac{6000}{200\times 4.18}=(T_2-298)\\\\7.17=(T_2-298)\\\\7.17+298=T_2\\\\T_2=305.17\ K[/tex]

So, the final temperature is equal to 305.17  K.

Answer:

Explanation:

The equation for the electric field is

[tex]E=\frac{kQ}{r^2}[/tex] so filling in:

[tex]E=\frac{9.00*10^9(.0025)}{(.0055)^2}[/tex] which in the end gives you

E = 7.4 × 10¹¹, choice A

Answer:

It is proved.

Explanation:

PV = n R T

where,  P is the pressure, V is the volume, n is the number of moles, R is the gas constant, T is  the absolute temperature.

Unit of pressure is Pascal= Newton/square meter = [tex][ML^{-1}T^{-2}][/tex]

Unit of volume is cubic meter = [tex][L^{3}][/tex]

So, the unit of P V

[tex][ML^{-1}T^{-2}] \times [L^{3}]\\\\[ML^2T^{-2}}[/tex]

[tex][ML^{2}T^{-2}][/tex]

It is same as the unit of energy.

hence proved.  

Answer:

English only

Explanation:

When solving problems related to Electric Fields, care must be taken about symmetries. In our particular case when we take a look to at the drawings of the attached file, we realize:

1.-By symmetry each dx associated at a, has an opposite dx with point b as reference. The respective dE ( the charge is uniform ) is the same, as the charge of the wire is positive the force and the Field on a test charge (+) located at h will be upward, therefore the components dEx will cancel each other and the Electric Field becomes E = Ey = ∫ 2×dE× cosθ

The solutions:

A) Ey = 4623 N/C

B) Ey = 19.34 N/C

E = Ey = ∫ 2×dE× cosθ

Here     cosθ   = h/ d   ⇒  cosθ = h/√h² + x²      dE = K× dQ / d²

d² = h² + x²

k = 8.9 ×10⁹ Nm²C⁻²  ;   dQ = λ×dx     λ = 150×10⁻⁹ C    h = 0.08 m

Then by substitution

Ey =  2 ∫[K× λ×dx/ (h² + x²) ] × h / √h² + x²

reordering that equation:

Ey = 2×K×λ×h ∫ dx / [√ ( h² + x² ) ]³          (2)

To solve the integral we make use of a change of variables

x = h × tanα     then   x² = h² ×tan²α   and  dx = h× sec²α dα

plugging that values in equation (2)

Ey  =  2×K×λ×h ∫  h× sec²α× dα / [√ ( h² + h²tan²α)]³

Ey  = 2×K×λ×h² ∫ sec²α× dα / [ h × √ (1 + tan²α)]³            1 + tan²α = sec²α

Ey = 2×K×λ×h²× ∫ (sec²α / h³× sec³α )×dα

Ey = 2×K×λ/h × ∫ ( 1 / secα dα

Ey = 2×K×λ/h × sinα             now we αneed to come back to our original variables:

as   x = h × tanα         tanα = x/h   then x is the opposite leg in a right triangle  and h the adjacent one then the hypothenuse is √ (h² + x²)         then    sin α = x/ √ (h² + x²)      

Ey = 2×K×λ/h × x/ √ (h² + x²) |₀⁰°⁰⁵

Ey  = 2×8.9×10⁹× 150×10⁻⁹× 5×10⁻²/8× 10⁻²× √ 10⁻² ( 8 + 5 )   N/C

Ey = 4623 N/C

To answer the second question again we will make use of symmetries if you look at drawing ( Figure 2 ) you see that again the components in direction of x-axis cancel each other and the components in y-axis direction will add. Then

Ey = ∫ dE× cosθ

following the same procedure  we will find:

Ey = ∫ [K×λ × dl/d²] × h/ d

The importan point here is that the radius of the circle is

2×π×r = 0.01      ( the length of the wire)  ⇒  r = 0.16×10⁻² m

And we need to take into account that the integration is over the circle and the length of the circle is 0.01 m or ××2×π×r. All other factors are constant. Then by substitution

Ey = [K×λ ×h×  / ( √ r² + h²)³ ] × 10⁻²    N/C

Ey = 8.9 × 10⁹ × 150× 10⁻⁹ × 6× 10⁻² × 10⁻² / √ 10⁻² ( 0.16 + 6)

Ey = 0.8 × 10² / 6

Ey = 19.34 N/C

Answer:

D)7 1/2 or 15/2

Explanation:

Let's calculate Combined resistance of the parallel first

1/Rt= 1/2+1/6=4/6

Rt=6/4 which is also equal with 3/2

Now let's add it with the series one

Rt= 6+3/2

=15/2 And when we put that un a mixed fraction 7 1/2

Answer:

The answer is "physics ".

Explanation:

Physics is the branch of science that addresses the properties of crystalline and its interaction with the fundamental elements of the universe. It covers subjects ranging in quantum mechanics with extremely little ones with quantum mechanics to the whole cosmos. You must be constant whether you like it or not, thus everyone must learn physics, irrespective of whether they're in Uganda, and plenty of other countries should have physics to dare study.

Answer:

The correct option is a) 10 mph, 0 mph.

Explanation:

1. The average speed (S) is a magnitude given by:

[tex] S = \frac{D}{T} [/tex]  

Where:

D: is the total distance = 1 mi

T: is the total time = 6 min

[tex] S = \frac{D}{T} = \frac{1 mi}{6 min}*\frac{60 min}{1 h} = 10 mph [/tex]

Hence, the average speed is 10 mph.

2. The average velocity is a vector:

[tex] V = \frac{\Delta d}{\Delta t} = \frac{d_{f} - d_{i}}{t_{f} - t_{i}} [/tex]

Where:

[tex]d_{f}[/tex]: is the final distance                                    

[tex]d_{i}[/tex]: is the initial distance  

[tex]t_{f}[/tex]: is the final time                

[tex]t_{i}[/tex]: is the initial time

Since Juanita ran one mile around her school track, the final position is the same that the initial position, so the magnitude of the average velocity is zero.                                               

Therefore, the correct option is a) 10 mph, 0 mph.

I hope it helps you!          

Answer:

150 kJ

Explanation:

Applying,

ΔH = Energy level of Product(E') - Energy level of reactant(E)

Where ΔH = enthalpy change, E' and E = energy level of the product and the reactant respectively

ΔH = E'-E............. Equation 1

From the diagram,

Given: E' = 200 kJ, E = 50 kJ

Substitute these values into equation 1

ΔH = 200-50

ΔH = 150 kJ

Hence the enthalpy change for the reaction is 150 kJ

Answer:

The height balls rise above the collision point, is approximately 7.37 meters

Explanation:

The given parameters just before the collision are;

The mass, m₁ and velocity, v₁ of the ball moving upward are;

m₁ = 3.0 kg, v₁ = 22 m/s

The mass, m₂ and velocity, v₂ of the ball moving downward are;

m₂ = 1.3 kg, v₂ = -11 m/s (downward motion)

The type of collision = Inelastic collision

We note that the momentum is conserved for inelastic collision

Let, [tex]v_f[/tex], represent the final velocity of the balls after collision, we have;

∴ Total initial momentum = Total final momentum

m₁·v₁ + m₂·v₂ = (m₁ + m₂)·[tex]v_f[/tex]

Therefore, we get;

m₁·v₁ + m₂·v₂ = 3.0 kg × 22 m/s + 1.3 kg × (-11) m/s = 51.7 kg·m/s

(m₁ + m₂)·[tex]v_f[/tex] = (3.0 kg + 1.3 kg) ×

∴ 51.7 kg·m/s = 4.3 kg × [tex]v_f[/tex]

[tex]v_f[/tex] = (51.7 kg·m/s)/4.3 kg ≈ 12.023 m/s

The final velocity, [tex]v_f[/tex] ≈ 12.023 m/s

The maximum height, h, the combined balls will rise from the point of collision, moving upward at a velocity of [tex]v_f[/tex] ≈ 12.023 m/s, is given from the kinetic equation of motion, v² = u² - 2·g·h, as found follows

At maximum height, we have;

[tex]h_{max} = \dfrac{v_f^2}{2 \cdot g }[/tex]

Therefore;

[tex]h_{max} \approx \dfrac{12.023^2}{2 \times 9.81 } \approx 7.37[/tex]

The height the combined two balls of putty rise above the collision point, [tex]h_{max}[/tex] ≈ 7.37 m.

Answer:

1.the friction of air, gravity2.gravity

Answer:

The softball experiences an applied force as a result of Amy’s throw. As the ball moves, it experiences drag from the air it passes through. It also experiences a downward pull because of gravity.

Explanation:

Plato

Answer: hi your question is incomplete  attached below is the complete question

answer :

magnitude of acceleration :  | a | = g = 9.81 m/s^2

direction : a = - g j

Explanation:

Neglecting Air resistance

magnitude of acceleration :

| a | = g = 9.81 m/s^2

Direction of acceleration

a = - g j  ( given that the direction of acceleration is against the acceleration due to gravity i.e. in the opposite direction )

After half of a revolution ...

==> Distance = π•r

==> Displacement = 2•r

Answer:

Heat required = mass× latent heat Q = 0.15 × 871 ×

The heat required to evaporate 0.15 kg of lead at 1750°C will be 130,650 J.

The movement of energy from a hot to a cold item is characterized as heat. Heat energy flows from a hot material to a cold one.

This occurs because faster-vibrating molecules transmit their energy to slower-vibrating ones.

The given data in the problem is;

m is the mass of lead =  0.15 kg  

T is the temperature = 1750°C,

The latent heat of vaporization for lead is, [tex]\rm L_V[/tex] = 871 × 10³ J/kg.

The heat is found as;

[tex]\rm Q= m \times L_V \\\\ \rm Q= 0.15 \times 871 \times 10^3 \\\\ Q=130,650 \ J[/tex]

Hence the heat required to evaporate 0.15 kg of lead at 1750°C will be 130,650 J.

To learn more about the heat refer to the link;

brainly.com/question/1429452