All categories

3 years ago

True or false: Equilibrium occurs when the quantity supplied exceeds the quantity demanded.

2 answers:

7 0

This is an economic question, just to add. This statement is wrong. Equilibrium is the point where quantity supplied meets the demand.
Supply and Demand are equal

Anastasy [175]3 years ago

4 0

False . IT IS FALSE. WE ARE TO LAZY TO READ ALL THAT.^^^^^

You might be interested in

A 5kg box slides across the floor with an initial velocity of 5m/s. If the coefficient of kinetic friction between the box and t

4vir4ik [10]

The net force on the block perpendicular to the floor is

∑ F[perp] = F[normal] - mg = 0

so that

F[normal] = (5 kg) g = 49 N

Then

F[friction] = 0.1 F[normal] = 4.9 N

so that the net force parallel to the floor is

∑ F[para] = -4.9 N = (5 kg) a

Solve for the acceleration a :

a = (-4.9 N) / (5 kg) = -0.98 m/s²

Starting with an initial velocity of 5 m/s, the box comes to a stop after time t such that

0 = 5 m/s - (0.98 m/s²) t

⇒ t ≈ 5.1 s

5 0

2 years ago

2.85 A police car is traveling at a velocity of 18.0 m/s due north, when a car zooms by at a constant velocity of 42.0 m/s due n

katrin2010 [14]

Answer:

11.1 s

Explanation:

Speed of the police car as given = v = 18 m/s

Speed of the car = V = 42 m/s

Reaction time = t = 0.8 s

Distance traveled by the police car during the reaction time = d₁= 0.8 x 18 = 14.4 m

Distance traveled by speeding car = d₂ =0.8 x 42 = 33.6 m

Acceleration of the police car = a = 5 m/s/s

The police car can catch the speeding car only if it travels a distance equal to the speeding car in a time t.

Distance traveled by the police car = D = d₁ + v t +0.5 at², according to the kinematic equation.

⇒ D = 14.4 + 18 t + 0.5 (5) t²

⇒ D = 14.4 + 18 t+2.5 t² → (1)

For the speeding car, distance traveled is D = 33.6 + 42 t, since it is constant velocity. Substitute for D from the above equation (1).

⇒ 14.4 + 18 t+2.5 t²= 33.6 + 42 t

⇒ 2.5 t² -24 t - 19.2 = 0

⇒ t = 10.3 s

Total time = t +0.8 s

⇒ Time taken for the police car to reach the speeding car = 10.3+0.8= 11.1 s

5 0

3 years ago

A parallel-plate vacuum capacitor has 8.38 J of energy stored in it. The separation between the plates is 2.30 mm. If the separa

Elanso [62]

Answer:

Explanation:

plate separation = 2.3 x 10⁻³ m

capacity C₁ = ε A / d

= ε A / 2.3 x 10⁻³

C₂ = ε A / 1.15 x 10⁻³

$\frac{C_2}{C_1}$ = $\frac{2.3}{1.15}$

a ) when charge remains constant

energy = $\frac{q^2}{2C}$

q is charge and C is capacity

energy stored initially E₁= $\frac{q^2}{2C_1}$

energy stored finally E₂ = $\frac{q^2}{2C_2}$

$\frac{E_1}{E_2} = \frac{C_2}{C_1}$ = $\frac{2.3}{1.15}$

$E_2$ = $\frac{1.15}{2.3 } \times E_1$

= $\frac{1.15}{2.3 } \times 8.38$

= 4.19 J

b )

In this case potential diff remains constant

energy of capacitor = 1/2 C V²

energy is proportional to capacity as V is constant .

$\frac{E_2}{E_1} = \frac{C_2}{C_1}$

$\frac{E_2}{8.38} = \frac{2.3}{1.15}$

$E_2$ = 16.76 .

8 0

3 years ago

the chef was careless and put the aluminum cookie sheet directly on the hot stove, which melted 0.05 kg of the aluminum. how muc

sergejj [24]

Heat required to melt 0.05 kg of aluminum is 28.7 kJ.

<h3>What is the energy required to melt 0.05 kg of aluminum?</h3>

The heat energy required to melt 0.05 kg of aluminum is obtained from the heat capacity of aluminum and the melting point of aluminum.

The formula to be used is given below:

Heat required = mass * heat capacity * temperature change

Assuming the aluminum sheet was at room temperature initially.;

Room temperature = 25 °C

Melting point of aluminum = 660.3 °C

Temperature difference = (660.3 - 25) = 635.3 903

Heat capacity of aluminum = 903 J/kg/903

Heat required = 0.05 * 903 * 635.3

Heat required = 28.7 kJ

In conclusion, the heat required is obtained from the heat change aluminum and the mass of the aluminum melted.

Learn more about heat capacity at: brainly.com/question/21406849

#SPJ1

3 0

2 years ago

Confirm if this is correct or not. If it isn't correct, please correct it.

kow [346]

Answer:

d = 421.83 m

Explanation:

It is given that,

Height, h = 396.9 m

Horizontal speed, v = 46.87 m/s

We need to find the distance traveled by the ball horizontally. Let t is the time taken by the ball. Using second equation of motion for vertical direction. So,

$396.9=0\times t+\dfrac{1}{2}\times 9.8 t^2\\\\t=9\ s$

Now d is the distance covered by the cannonball. So,

$d=vt\\\\d=46.87\times 9\\\\d=421.83\ m$

Hence, this is the required solution.

3 0

3 years ago