Which is greater, the mass of the compounds before a chemical reaction or the mass of the compounds after a chemical reaction?

A ball rolls from 10 m to -25 m in 2.5 seconds. What was its average velocity? (Units = m/s)

marissa [1.9K]

Answer:

v = -14 m/s

Explanation:

Given that,

Initial location of the ball, X₁ = 10 m

Final position of the ball, X₂ = -25 m

Time taken to travel is, t = 2.5 s

The average velocity of the ball is given by the formula,

V = X₂ - X₁ / t m/s

Substituting the values in the above equation,

V = -25 - 10 / 2.5

= -14 m/s

The negative sign in the velocity indicates that ball rolls in the opposite direction.

Hence, the average velocity of the ball is v = -14 m/s

8 0

3 years ago

Calculate the electric field at the center of a square

pantera1 [17]

Answer:

$E_y=1175510.2\ N.C^{-1}$

The Magnitude of electric field is in the upward direction as shown directly towards the charge $q_1$ .

Explanation:

Given:

side of a square, $a=52.5\ cm$

charge on one corner of the square, $q_1=+45\times 10^{-6}\ C$

charge on the remaining 3 corners of the square, $q_2=q_3=q_4=-27\times 10^{-6}\ C$

Distance of the center from each corners $=\frac{1}{2} \times diagonals$

$diagonal=\sqrt{52.5^2+52.5^2}$

$diagonal=74.25\cm=0.7425\ m$

∴Distance of center from corners, $b=0.3712\ m$

Now, electric field due to charges is given as:

$E=\frac{1}{4\pi\epsilon_0}\times \frac{q}{b^2}$

For charge $q_1$ we have the field lines emerging out of the charge since it is positively charged:

$E_1=9\times 10^9\times \frac{45\times 10^{-6}}{0.3712^2}$

$E_1=2938775.5\ N.C^{-1}$

Force by each of the charges at the remaining corners:

$E_2=E_3=E_4=9\times 10^9\times \frac{27\times 10^{-6}}{0.3712^2}$

$E_2=E_3=E_4=1763265.3\ N.C^{-1}$

Now, net electric field in the vertical direction:

$E_y=E_1-E_4$

$E_y=1175510.2\ N.C^{-1}$

Now, net electric field in the horizontal direction:

$E_y=E_2-E_3$

$E_y=0\ N.C^{-1}$

So the Magnitude of electric field is in the upward direction as shown directly towards the charge $q_1$ .

8 0

2 years ago

An inductor is connected to a 26.5 Hz power supply that produces a 41.2 V rms voltage. What minimum inductance is needed to keep

alexira [117]

Answer:

The minimum inductance needed is 2.78 H

Explanation:

Given;

frequency of the AC, f = 26.5 Hz

the root mean square voltage in the circuit, $V_{rms}$ = 41.2 V

the maximum current in the circuit, I₀ = 126 mA

The root mean square current is given by;

$I_{rms} = \frac{I_o}{\sqrt{2} } \\\\I_{rms} = \frac{126*10^{-3}}{\sqrt{2} }\\\\I_{rms} =0.0891 \ A$

The inductive reactance is given by;

$X_l = \frac{V_{rms}}{I_{rms}} \\\\X_l= \frac{41.2}{0.0891}\\\\X_l = 462.4 \ ohms$

The minimum inductance needed is given by;

$X_l = \omega L\\\\X_l = 2\pi fL\\\\L = \frac{X_l}{2\pi f}\\\\L = \frac{462.4}{2\pi *26.5}\\\\L = 2.78 \ H$

Therefore, the minimum inductance needed is 2.78 H

7 0

2 years ago

8. When you see a stopped local bus (1 point)

elena55 [62]

Answer: be alert for pedestrians near the bus.

Explanation: Due to road accidents many Governments around the world has adopted and put in place certain rules and regulations with regards to road safety, this is so to prevent the or reduce the chances of accidents happening.

Road safety rules are rules and guidelines put in place by Government in order to prevent road accidents and maintain a free flow of traffic. An example of such rules is 'be alert for pedestrians near the bus ' when approaching a local bus that is stopped.

5 0

2 years ago

Consider a large tank holding 1000 L of pure water into which a brine solution of salt begins to owat a constant rate of 6L/min.

lbvjy [14]

Answer:

T = 693.147 minutes

Explanation:

The tank is being continuously stirred. So let the salt concentration of the tank at some time t be x in units of kg/L.

Therefore, the total salt in the tank at time t = 1000x kg

Brine water flows into the tank at a rate of 6 L/min which has a concentration of 0.1 kg/L

Hence, the amount of salt that is added to the tank per minute = $(6\times0.1)kg/min=0.6kg/min$

Also, there is a continuous outflow from the tank at a rate of 6 L/min.

Hence, amount of salt subtracted from the tank per minute = 6x kg/min

Now, the rate of change of salt concentration in the tank = $\frac{dx}{dt}$

So, the rate of change of salt in the tank can be given by the following equation,

$1000\frac{dx}{dt} =0.6-6x$

or, $\int\limits^{0.05}_0 {\frac{1000}{0.6-6x} } \, dx =\int\limits^T_0 {} \, dt$

or, T = 693.147 min (time taken for the tank to reach a salt concentration

of 0.05 kg/L)

3 0

3 years ago