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saw5 [17]
3 years ago
15

A recent home energy bill indicates that a household used 475475 kWh (kilowatt‑hour) of electrical energy and 135135 therms for

gas heating and cooking in a period of 1 month. Given that 1.00 therm is equal to 29.329.3 kWh, how many milligrams of mass would need to be converted directly to energy each month to meet the energy needs for the home?
Physics
1 answer:
faltersainse [42]3 years ago
5 0

Answer:

Explanation:

Given that,.

A house hold power consumption is

475 KWh

Gas used is

135 thermal gas for month

Given that, 1 thermal = 29.3 KWh

Then,

135 thermal = 135 × 29.3 = 3955.5 KWh

So, total power used is

P = 475 + 3955.5

P =4430.5 KWh

Since 1 hr = 3600 seconds

So, the energy consumed for 1hr is

1KW = 1000W

P = energy / time

Energy = Power × time

E = 4430.5 KWhr × 1000W / KW × 3600s / hr

E = 1.595 × 10^10 J

So, using Albert Einstein relativity equation

E = mc²

m = E / c²

c is speed of light = 3 × 10^8 m/s

m = 1.595 × 10^10 / (3 × 10^8)²

m = 1.77 × 10^-7 kg

Then,

1 kg = 10^6 mg

m = 1.77 × 10^-7 kg × 10^6 mg / kg

m = 0.177mg

m ≈ 0.18 mg

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A 7.00-kg object accelerates from rest to a final velocity in 55 seconds. If the magnitude of the
Len [333]

Answer:

The final velocity of the object is 330 m/s.

Explanation:

To solve this problem, we first must find the acceleration of the object.  We can do this using Newton's Second Law, given by the following equation:

F = ma

If we plug in the values that we are given in the problem, we get:

42 = 7 (a)

To solve for a, we simply divide both sides of the equation by 7.

42/7 = 7a/7

a = 6 m/s^2

Next, we should write out all of the information we have and what we are looking for.

a = 6 m/s^2

v1 = 0 m/s

t  = 55 s

v2 = ?

We can use a kinematic equation to solve this problem.  We should use:

v2 = v1 + at

If we plug in the values listed above, we should get:

v2 = 0 + (6)(55)

Next, we should solve the problem by performing the multiplication on the right side of the equation.

v2 = 330 m/s

Therefore, the final velocity reached by the object is 330 m/s.

Hope this helps!

7 0
3 years ago
In a "Rotor-ride" at a carnival, people rotate in a vertical cylindrically walled "room." If the room radius is 5.5 m, and the r
Crank

Answer:

0.181

Explanation:

We can convert the 0.5 rps into standard angular velocity unit rad/s knowing that each revolution is 2π:

ω = 0.5 rps = 0.5*2π = 3.14 rad/s

From here we can calculate the centripetal acceleration

a_c = \omega^2r = 3.14^2*5.5 = 54.3 m/s^2

Using Newton 2nd law we can calculate the centripetal force that pressing on the rider, as well as the reactive normal force:

F = N = a_cm = 54.3 m

Also the friction force and friction acceleration

F_f = N\mu = 54.3 m \mu N

a_f = F_f / m = 54.3 \mu

For the rider to not slide down, friction acceleration must win over gravitational acceleration g = 9.81 m/s2:

g = a_f = 54.3 \mu

9.81 = 54.3 \mu

\mu = 9.81 / 54.3 = 0.181

6 0
3 years ago
ShowWwww workKkkkk pleaseeeeeeeeeee
Bogdan [553]
Ok so basically you have to look at the work and then look at the work and then it will show
6 0
3 years ago
An Airplane taxis onto the runway going at 10 m/s. If it can accelerate steady at 3
kirza4 [7]

Answer: 1333 m

the length of runway it will need is S = \frac{90^{2}-10^{2}  }{2.3}=1333 (m)

Explanation:

4 0
3 years ago
Sobre un gas contenido en un cilindro provisto de un pistón se realiza un trabajo de 7000 Joules, mediante un proceso isotérmico
natita [175]

Answer:

En un proceso isotérmico, es decir, la temperatura no cambia, el trabajo puede escribirse como:

W = n*R*T*Ln(P1/P2)

Donde P1 es la presión inicial y P2 la presión final.

Donde las cantidades:

n =  número de moles

R = constante de los gases ideales

T = temperatura no cambian.

Y sabemos que la ecuación de la energía interna es:

U = C*n*R*T

Donde C es otra constante que depende del gas.

De aca, podemos concluir que ninguna de estas variables cambia en nuestro proceso, por lo que la variación de la energía interna es cero.

U2 - U1 = 0

b) Para el calor cedido o absorbido, la formula básica es:

ΔQ = C*(T2 - T1)

Donde ΔQ es el calor absorbido o cedido por el gas, C es una constante que depende del gas, T2 es la temperatura final del gas y T1 es la temperatura inicial del gas.

Como la temperatura no cambia en el proceso, entonces:

T2 = T1

ΔQ = C*(T2 - T1) = C*0 = 0

No hay calor absorbido ni cedido.

c) Podemos concluir que en un proceso isotérmico la energía interna no cambia, y no hay un intercambio de calor.

8 0
3 years ago
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