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2 years ago

10

Complete combustion of 1.0 metric ton of coal (assuming pure carbon) to gaseous carbon dioxide releases 3.3 x 1010 j of heat. co

nvert this energy to kilocalories

1 answer:

netineya [11]2 years ago

6 0

Keeping in mind that the conversion between calories and Joules is
$1 cal = 4.186 J$
we can write the conversion factor using the kilocalories:
$1 kcal = 4186 J$

The energy released in our problem is
$E=3.3 \cdot 10^{10} J$
so we can set a simple proportion to find its equivalent in kcal:
$1 kcal: 4186 J = x: 3.3 \cdot 10^{10} J$
from which we find:
$x= \frac{3.3 \cdot 10^{10} J \cdot 1 kcal}{4186 J} =7.88 \cdot 10^6 kcal$

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Two small diameter, 10gm dielectric balls can slide freely on a vertical channel each carry a negative charge of 1microcoulomb.

dimulka [17.4K]

Answer:

The distance of separation is $d = 0.092 \ m$

Explanation:

The mass of the each ball is $m= 10 g = 0.01 \ kg$

The negative charge on each ball is $q_1 =q_2=q = 1 \mu C = 1 *10^{-6} \ C$

Now we are told that the lower ball is restrained from moving this implies that the net force acting on it is zero

Hence the gravitational force acting on the lower ball is equivalent to the electrostatic force i.e

$F = \frac{kq_1 * q_2}{d}$

=> $m* g = \frac{kq_1 * q_2}{d}$

here k the the coulomb's constant with a value $k = 9*10^{9} \ kg\cdot m^3\cdot s^{-4}\cdot A^2.$

So

$0.01 * 9.8 = \frac{ 9*10^9 *[1*10^{-6} * 1*10^{-6}]}{d}$

$d = 0.092 \ m$

5 0

2 years ago

Physics in motion unit 6a the nature of waves

mylen [45]

What’s the question?

7 0

2 years ago

A proton of mass mp is released from rest just above the lower plate and reaches the top plate with speed vp. An electron of mas

vodka [1.7K]

Answer:

$v_e=\sqrt{\frac{m_pv_p^2}{m_e}}$

Explanation:

You can consider that the force that acts over the proton is the same to the force over the electron. This is because the electric force is given by:

$F=qE$

$F_p=F_e$

where E is the constant electric field between the parallel plates, and is the same for both electron and proton. Also, the charge is the same.

by using the Newton second law for the proton, and by using kinematic equation for the calculation of the acceleration you can obtain:

$m_pa_p=qE\\\\a_p=\frac{v_p^2}{2d}\\\\\frac{m_pv_p^2}{2d}=qE$

(it has been used that vp^2 = v_o^2+2ad) where d is the separation of the plates, ap the acceleration of the proton, vp its velocity and mp its mass.

By doing the same for the electron you obtain:

$\frac{m_ev_e^2}{2d}=qE$

we can equals these expressions for both proton and electron, because the forces qE are the same:

$\frac{m_pv_p^2}{2d}=\frac{m_ev_e^2}{2d}\\\\v_e=\sqrt{\frac{m_pv_p^2}{m_e}}$

4 0

2 years ago

The famous cliff divers of Acapulco leap from a perch 35 m above the ocean. How fast are they moving when they reach the surface

Rus_ich [418]

1) 26.2 m/s

The mechanical energy of the divers at any point of their vertical motion is sum of the kinetic energy and the gravitational potential energy:

$E=K+U = \frac{1}{2}mv^2 + mgh$

where

m is the mass of the diver

v is the speed

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

h is the height above the water

When the diver is on the cliff, v = 0 (he is at rest), so K=0 and the initial mechanical energy is just potential energy:

$E_i = mgh$

where h=35 m is the height of the cliff.

When the diver hits the water above, h = 0, so U=0 and the final mechanical energy is just kinetic energy:

$E_f = \frac{1}{2}mv^2$

since the total mechanical energy is conserved, we have

$E_i = E_f\\mgh = \frac{1}{2}mv^2$

And solving the equation for v, we find the speed when they reach the surface of the water:

$v=\sqrt{2gh}=\sqrt{2(9.8 m/s^2)(35 m)}=26.2 m/s$

2) It is converted into thermal energy of the water

When the diver enters the water, he suddenly feels another force acting against the motion of the diver: the resistance of the water. The resistance of the water acts upward, slowing down the diver until he stops.

In this process, the speed of the diver (v) decreases, and therefore the kinetic energy of the diver decreases as well, until it becomes zero.

However, this does not mean that the conservation of energy has been violated. In fact, the kinetic energy of the diver has been converted into thermal energy of the molecules of water surrounding the diver.

8 0

2 years ago

You are driving downhill on a rural road with a 3% grade at a speed of 45 mph. While playing on the side of the road, a child ac

Dennis_Churaev [7]

Answer: a) 95.07m b) 81.88 m

Explanation:

a)

For finding the distance when vehicle is going downhill we have the formula as:

Stop sight distance= Velocity*Reaction time + Velocity² / 2*g*(f constant- Grade value)

Now by AASHTO, we have for v= 45 mph= 72.4 kph, f= 0.31

Reaction time= 0.28

So putting values we get

Stop sight distance= 0.28*72.4 *1 + $\frac{(0.28*72.4)^{2} }{2*9.81*(0.31-0.03)}$

Stop sight distance= 95.07 m

b)

For finding the distance when vehicle is going uphill we have the formula as:

Stop sight distance= Velocity*Reaction time + Velocity² / 2*g*(f constant- Grade value)

Now by AASHTO, we have for v= 45 mph= 72.4 kph, f= 0.31

Reaction time= 0.28

So putting values we get

Stop sight distance= 0.28*72.4 *1 + $\frac{(0.28*72.4)^{2} }{2*9.81*(0.31+0.03)}$

Stop sight distance= 81.88 m

5 0

2 years ago