How does the efficiency of the inclined plane vary with the inclination angle?

A circuit contains two light bulbs connected in parallel. What would happen to the brightness of each light bulb if two more lig

laila [671]

Answer;

C. The brightness of each bulb would remain the same even though the total resistance of the circuit would decrease.

Explanation;

-If light bulbs are connected in parallel to a voltage source, the brightness of the individual bulbs remains more-or-less constant as more and more bulbs are added to the circuit.

-The current increases as more bulbs are added to the circuit and the overall resistance decreases. In addition, if one bulb is removed from the circuit the other bulbs do not go out. Each bulb is independently linked to the voltage source

6 0

3 years ago

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Cart a having a mass of 150 kg initially moving to the right at a speed of 8 m/s collides with cart b with a mass of 150 kg, ini

kakasveta [241]

The speed of cart b is 6m/s while the total momentum of the systmen is 4200 kg m/s

<h3>Conservation of Linear Momentum</h3>

Given Data

Mass of cart one M1 = 150kg

Initial Velocity U1 = 8m/s

Final VelocityV1 = 5 m/s

Mass of cart two M2 = 150kg

Velocity U2 = 6m/s

Applying the principle of conservation of linear momentum we have

M1U1+M2U2 = M1V1+ M2V2

a. what is the speed of cart b after collision

substituting our given data we have

150*8+ 150*6 = 150*5+150*V2

1200 + 900 = 1200+ 150V2

2100 - 1200 = 150V2

900 = 150V2

Divide both sides by 150

V2 = 900/150

V2 = 6m/s

b. what is the total momentum of the system before and after collision

Total Momentum in the system is

Total momentum = Momentum before Impact+ Momentum after Impact

Total momentum = M1U1+M2U2 + M1V1+ M2V2

Total momentum = 1200 + 900 + 1200+ 900

Total momentum = 4200 kg m/s

Learn more about Conservation of Linear Momentum here:

brainly.com/question/7538238

6 0

3 years ago

8. Three grams of Bismuth-218 decay to 0.375 grams in one hour. What is the half-

Evgen [1.6K]

Answer: 0.333 h

Explanation:

This problem can be solved using the <u>Radioactive Half Life Formula</u>:

$A=A_{o}.2^{\frac{-t}{H}}$ (1)

Where:

$A=0.375 g$ is the final amount of the material

$A_{o}=3 g$ is the initial amount of the material

$t=1 h$ is the time elapsed

$H$ is the half life of the material (the quantity we are asked to find)

Knowing this, let's substitute the values and find $h$ from (1):

$0.375 g=(3 g)2^{\frac{-1h}{H}}$ (2)

$\frac{0.375 g}{3 g}=2^{\frac{-1h}{H}}$ (3)

Applying natural logarithm in both sides:

$ln(\frac{0.375 g}{3 g})=ln(2^{\frac{-1 h}{H}})$ (4)

$-2.079=-\frac{1 h}{H}ln(2)$ (5)

Clearing $H$ :

$H=\frac{-1h}{-2.079}(0.693)$ (6)

Finally:

$h=0.333 h$ This is the half-life of the Bismuth-218 isotope

4 0

4 years ago

A huge tank of glycerine with a density of 1.260 g/cm3 is vertically stationed on a platform which is 15 m above the ground. The

EleoNora [17]

Answer:

The tank is losing $4.976*10^{-4} m^3/s$

$v_g = 19.81 \ m/s$

Explanation:

According to the Bernoulli’s equation:

$P_1 + 1 \frac{1}{2} \rho v_1^2 + \rho gh_1 = P_2 + \frac{1}{2} \rho v_2^2 + \rho gh_2$

We are being informed that both the tank and the hole is being exposed to air :

∴ P₁ = P₂

Also as the tank is voluminous ; we take the initial volume $v_1$ ≅ 0 ;

then $v_2$ can be determined as: $\sqrt{[2g (h_1- h_2)]$

h₁ = 5 + 15 = 20 m;

h₂ = 15 m

$v_2 = \sqrt{[2*9.81*(20 - 15)]$

$v_2 = \sqrt{[2*9.81*(5)]$

$v_2= 9.9 \ m/s$ as it leaves the hole at the base.

radius r = d/2 = 4/2 = 2.0 mm

(a) From the law of continuity; its equation can be expressed as:

J = $A_1v_2$

J = πr² $v_2$

J = $\pi *(2*10^{-3})^{2}*9.9$

J = $1.244*10^{-4} m^3/s$

b)

How fast is the water from the hole moving just as it reaches the ground?

In order to determine that; we use the relation of the velocity from the equation of motion which says:

v² = u² + 2gh ₂

v² = 9.9² + 2×9.81×15

v² = 392.31

The velocity of how fast the water from the hole is moving just as it reaches the ground is : $v_g = \sqrt{392.31}$

$v_g = 19.81 \ m/s$

4 0

3 years ago

Two point charges 3q and −8q (with q > 0) are at x = 0 and x = L, respectively, and free to move. A third charge is placed so

riadik2000 [5.3K]

Answer:

Explanation:

The unknown charge can not remain in between the charge given because force on the middle charge will act in the same direction due to both the remaining charges.

So the unknown charge is somewhere on negative side of x axis . Its charge will be negative . Let it be - Q and let it be at distance - x on x axis.

force on it due to rest of the charges will be equal and opposite so

k3q Q / x² =k 8q Q / (L+x)²

8x² = 3 (L+x)²

2√2 x = √3 (L+x)

2√2 x - √3 x = √3 L

x(2√2 - √3 ) = √3 L

x = √3 L / (2√2 - √3 )

Let us consider the balancing force on 3q

force on it due to -Q and -8q will be equal

kQ . 3q / x² = k3q 8q / L²

Q = 8q (x² / L²)

so charge required = - 8q (x² / L²)

and its distance from x on negative x side = √3 L / (2√2 - √3 )

3 0

3 years ago