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fomenos
3 years ago
7

Determine the specific heat of iron if 6.1 j of energy are needed to warm 1.50 g of iron from 20.0◦c to 29.0◦c?

Physics
1 answer:
ratelena [41]3 years ago
8 0
As the iron core isprogressively laid down, the mechanism of iron oxidation changes from a protein dominated process with H202 being the primary product of O2 reduction to a mineral surface dominated process where H20 is the primary product. These results emphasize the importance of the apoferritin shell in facilitating iron oxidation in the early stage of iron deposition prior to significant development of the polynuclear iron core. 
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Two cars are traveling along a straight line in the same direction, the lead car at 25 m/s and the other car at 35 m/s. At the m
Phoenix [80]

Answer:

a. t_1=12.5\ s

b. a_2=-13.61\ m.s^{-2}  must be the minimum magnitude of deceleration to avoid hitting the leading car before stopping

c. t_2=2.5714\ s is the time taken to stop after braking

Explanation:

Given:

  • speed of leading car, u_1=25\ m.s^{-1}
  • speed of lagging car, u_{2}=35\ m.s^{-1}
  • distance between the cars, \Delta s=45\ m
  • deceleration of the leading car after braking, a_1=-2\ m.s^{-2}

a.

Time taken by the car to stop:

v_1=u_1+a_1.t_1

where:

v_1=0 , final velocity after braking

t_1= time taken

0=25-2\times t_1

t_1=12.5\ s

b.

using the eq. of motion for the given condition:

v_2^2=u_2^2+2.a_2.\Delta s

where:

v_2= final velocity of the chasing car after braking = 0

a_2= acceleration of the chasing car after braking

0^2=35^2+2\times a_2\times 45

a_2=-13.61\ m.s^{-2} must be the minimum magnitude of deceleration to avoid hitting the leading car before stopping

c.

time taken by the chasing car to stop:

v_2=u_2+a_2.t_2

0=35-13.61\times t_2

t_2=2.5714\ s  is the time taken to stop after braking

7 0
3 years ago
A 150-W lamp is placed into a 120-V ac outlet. What is the peak current?
devlian [24]

Answer:

Explanation:

Formula

W = I * E

Givens

W = 150

E = 120

I = ?

Solution

150 = I * 120   Divide by 120

150/120 = I

5/4  = I

I = 1.25

Note: This is an edited note. You have to assume that 120 is the RMS voltage in order to go any further. That means that the peak voltage is √2 times the size of 120. The current has the same note applied to it. If the voltage is its rms value, then the current must (assuming the properties of the bulb do not change)

On the other hand, if the voltage is the peak value at 120 then 1.25 will be correct.

However I would go with the other answerer's post and multiply both values by  √2

3 0
3 years ago
Read 2 more answers
A loaded ore car has a mass of 950 kg. and rolls on rails ofnegligible friction. It starts from rest ans is pulled up a mineshaf
stiks02 [169]

(a) 10241 W

In this situation, the car is moving at constant speed: this means that its acceleration along the direction parallel to the slope is zero, and so the net force along this direction is also zero.

The equation of the forces along the parallel direction is:

F - mg sin \theta = 0

where

F is the force applied to pull the car

m = 950 kg is the mass of the car

g=9.8 m/s^2 is the acceleration of gravity

\theta=30.0^{\circ} is the angle of the incline

Solving for F,

F=mg sin \theta = (950)(9.8)(sin 30.0^{\circ})=4655 N

Now we know that the car is moving at constant velocity of

v = 2.20 m/s

So we can find the power done by the motor during the constant speed phase as

P=Fv = (4655)(2.20)=10241 W

(b) 10624 W

The maximum power is provided during the phase of acceleration, because during this phase the force applied is maximum. The acceleration of the car can be found with the equation

v=u+at

where

v = 2.20 m/s is the final velocity

a is the acceleration

u = 0 is the initial velocity

t = 12.0 s is the time

Solving for a,

a=\frac{v-u}{t}=\frac{2.20-0}{12.0}=0.183 m/s^2

So now the equation of the forces along the direction parallel to the incline is

F - mg sin \theta = ma

And solving for F, we find the maximum force applied by the motor:

F=ma+mgsin \theta =(950)(0.183)+(950)(9.8)(sin 30^{\circ})=4829 N

The maximum power will be applied when the velocity is maximum, v = 2.20 m/s, and so it is:

P=Fv=(4829)(2.20)=10624 W

(c) 5.82\cdot 10^6 J

Due to the law of conservation of energy, the total energy transferred out of the motor by work must be equal to the gravitational potential energy gained by the car.

The change in potential energy of the car is:

\Delta U = mg \Delta h

where

m = 950 kg is the mass

g=9.8 m/s^2 is the acceleration of gravity

\Delta h is the change in height, which is

\Delta h = L sin 30^{\circ}

where L = 1250 m is the total distance covered.

Substituting, we find the energy transferred:

\Delta U = mg L sin \theta = (950)(9.8)(1250)(sin 30^{\circ})=5.82\cdot 10^6 J

8 0
3 years ago
What is the Morgan-Keenan Spectral Classification system?
Angelina_Jolie [31]

Answer: A method of classifying stars by their temperatures and compositions.

Explanation:

4 0
3 years ago
What is The common value for the speed of light in a vacuum
Talja [164]
The most common value for the speed of light is 3*10^8 meters/second.

A more accurate number is <span>299 792 458 m/second, but that number is hardly ever used.</span>
4 0
3 years ago
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