X + 10 times X = 50 what is the answer

Ann [662]

Well basically in physics and chemistry, the law of conservation of energy states that the total energy of an isolated system remains constant, it is said to be conserved over time. This law means that energy can neither be created nor destroyed, it can only be transformed or transferred from one form to another.

Hope this helps (:

5 0

4 years ago

Which of the following is a difference between a mercury barometer and an aneroid barometer?

melomori [17]

They are made of different substances

7 0

3 years ago

Three crates with various contents are pulled by a force Fpull=3615 N across a horizontal, frictionless roller‑conveyor system.

SIZIF [17.4K]

The question is incomplete. Here is the complete question.

Three crtaes with various contents are pulled by a force Fpull=3615N across a horizontal, frictionless roller-conveyor system.The group pf boxes accelerates at 1.516m/s2 to the right. Between each adjacent pair of boxes is a force meter that measures the magnitude of the tension in the connecting rope. Between the box of mass m1 and the box of mass m2, the force meter reads F12=1387N. Between the box of mass m2 and box of mass m3, the force meter reads F23=2304N. Assume that the ropes and force meters are massless.

(a) What is the total mass of the three boxes?

(b) What is the mass of each box?

Answer: (a) Total mass = 2384.5kg;

(b) m1 = 915kg;

m2 = 605kg;

m3 = 864.5kg;

Explanation: The image of the boxes is described in the picture below.

(a) The system is moving at a constant acceleration and with a force Fpull. Using Newton's 2nd Law:

$F_{pull}=m_{T}.a$

$m_{T}=\frac{F_{pull}}{a}$

$m_{T}=\frac{3615}{1.516}$

$m_{T}=2384.5$

Total mass of the system of boxes is 2384.5kg.

(b) For each mass, analyse each box and make them each a free-body diagram.

For $m_{1}$ :

The only force acting On the $m_{1}$ box is force of tension between 1 and 2 and as all the system is moving at a same acceleration.

$m_{1} = \frac{F_{12}}{a}$

$m_{1} = \frac{1387}{1.516}$

$m_{1}$ = 915kg

For $m_{2}$ :

There are two forces acting on $m_{2}$ : tension caused by box 1 and tension caused by box 3. Positive referential is to the right (because it's the movement's direction), so force caused by 1 is opposing force caused by 3:

$m_{2} = \frac{F_{23}-F_{12}}{a}$

$m_{2} = \frac{2304-1387}{1.516}$

$m_{2}$ = 605kg

For $m_{3}$ :

$m_{3} = m_{T} - (m_{1}+m_{2})$

$m_{3} = 2384.5-1520.0$

$m_{3}$ = 864.5kg

8 0

3 years ago

A 15.0-Ω resistor and a coil are connected in series with a 6.30-V battery with negligible internal resistance and a closed swit

Aloiza [94]

Answer:

0.04328 H

0.00288 seconds

0.01326 seconds

Explanation:

V = Voltage = 6.3 V

R = Resistance = 15 Ω

L = Inductance

I = Decayed current = 0.21 A

t = Time to decay = 2 ms

Maximum Current is given by

$I_0=\frac{V}{R}\\\Rightarrow I_0=\frac{6.3}{15}\\\Rightarrow I_0=0.42\ A$

Current in the circuit is given by

$I=I_0exp\frac{-Rt}{L}\\\Rightarrow ln\frac{I_0}{I}=\frac{Rt}{L}\\\Rightarrow L=\frac{Rt}{ln\frac{I_0}{I}}\\\Rightarrow L=\frac{15\times 2\times 10^{-3}}{ln\frac{0.42}{0.21}}\\\Rightarrow L=0.04328\ H$