A 10kg mass is moving to the right with a velocity of 15 m/s. A 13kg mass

Which contains the most moles: 10 g of hydrogen gas, 100 g of carbon, or 50 g of lead?

Stolb23 [73]

Answer:

Carbon

Explanation:

We know that

Number of moles = mass/molecular mass

So

Hydrogen: given as H2 with 10g an atomic number of 1 so MW( 1x 2= 2)

Number of moles = Weight/MW

= 10/2

= 5

Carbon: atomic number 12 C and mass 100

Number of moles = Weight/MW

= 100/12

= 8.33

Lead: atomic num 207 and mW 207 pb

Number of moles = Weight/MW

= 50/207

= 0.241

Therefore Carbon has the most moles.

5 0

3 years ago

A manufacturer provides a warranty against failure of a carbon steel product within the first 30 days after sale. Out of 1000 so

hodyreva [135]

Answer:A) Risk(R)= $1000

B) There is justification for spending an additional cost of $100 to prevent a corrosion whose consequence in monetary terms is $1000

Explanation:R= Risk,

P=Probability of failure

C= Consequence of failure

Mathematically, R=P ×C

10 out of 1000 carbon-steal products failed

Probability of failure= 10/1000 =0.01

The consequence of failure by corrosion given in monetary term =$100,000

Risk of failure = 0.01 × $100,000

R=$1000

4 0

3 years ago

A 420-turn circular coil with an area of 0.0650 m2 is mounted on a rotating frame, which turns at a rate of 22.3 rad/s in the pr

Ivenika [448]

Answer:

33.48 V

Explanation:

Parameters given:

Number of turns, N = 420

Magnetic field strength, B = 0.055 T

Area, A = 0.065 m²

Angular velocity, ω = 22.3 rad/s

EMF induced in a coil is given as:

EMF = -dΦ/dt

where Φ = magnetic flux

Magnetic flux, Φ, is given as:

Φ = B * N * A * cosωt

EMF = -d( B * N * A * cosωt) / dt

EMF = B * N * A * ω * sinωt

where ωt = 90°

Therefore:

EMF = 0.055 * 420 * 0.065 * 22.3 * sin90°

EMF = 33.48 V

7 0

3 years ago

An object of mass 6 kg. is resting on a horizontal surface. A horizontal force

son4ous [18]

Answer:

a) The work done by the applied force is 1500 joules.

b) The kinetic energy of the block after 10 seconds is 1200 joules.

c) The magnitude of the force of friction is 3 newtons and its direction is against motion.

d) 300 joules of energy are lost during motion.

Explanation:

a) Since the object has a constant mass, on which a constant horizontal force is exerted. The work done by the force ( $W$ ), measured in joules, is defined by the following expression:

$W = F\cdot \Delta x$ (1)

Where:

$F$ - Force, measured in newtons.

$\Delta x$ - Distance, measured in meters.

If we know that $F = 15\,N$ and $\Delta x = 100\,m$ , then the work done by the force exerted on the object is:

$W = (15\,N)\cdot (100\,m)$

$W = 1500\,J$

The work done by the applied force is 1500 joules.

b) At first we need to calculate the net acceleration of the object ( $a$ ), measured in meters per square second. By assuming a constant acceleration, we use the following kinematic formula:

$\Delta x = v_{o}\cdot t +\frac{1}{2}\cdot a\cdot t^{2}$ (2)

Where $v_{o}$ is the initial velocity of the object, measured in meters per second.

We clear the acceleration within the equation above:

$\frac{1}{2}\cdot a \cdot t^{2} = \Delta x-v_{o}\cdot t$

$a = \frac{2\cdot (\Delta x - v_{o}\cdot t)}{t^{2}}$

If we know that $\Delta x = 100\,m$ , $v_{o} = 0\,\frac{m}{s}$ and $t = 10\,s$ , then the net acceleration experimented by the object is:

$a = \frac{2\cdot \left[100\,m-\left(0\,\frac{m}{s} \right)\cdot (10\,s)\right]}{(10\,s)^{2}}$

$a = 2\,\frac{m}{s^{2}}$

By the 2nd Newton's Law, we construct the following equation of equilibrium under the consideration of a friction force acting against the motion of the object:

$\Sigma F = F - f = m\cdot a$ (3)

Where:

$F$ - External force exerted on the object, measured in newtons.

$f$ - Kinetic friction force, measured in newtons.

If we know that $F = 15\,N$ , $m = 6\,kg$ and $a = 2\,\frac{m}{s^{2}}$ , the kinetic friction force is:

$f = F-m\cdot a$

$f = 15\,N-(6\,kg)\cdot \left(2\,\frac{m}{s^{2}} \right)$

$f = 3\,N$

The work done by friction ( $W'$ ), measured in joules, is:

$W' = f\cdot \Delta x$ (4)

$W' = (3\,N) \cdot (100\,m)$

$W' = 300\,J$

And the net work experimented by the object is:

$\Delta W = 1500\,J - 300\,J$

$\Delta W = 1200\,J$

By the Work-Energy Theorem we understand that change in translational kinetic energy ( $\Delta K$ ), measured in joules, is equal to the change in net work. That is: