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Musya8 [376]
3 years ago
8

Please help if you can.

Physics
1 answer:
VMariaS [17]3 years ago
3 0

Initially there were 10 bulbs of 60 Watt power

So total power of all bulbs = 60 * 10 = 600 W

now each bulb used for 4 hours daily

so total energy consumed daily

E = P * t

E = 600 * 4 = 2400 Wh

E = 2.4 kWh

now we have total power consumed in 1 year

E = 365 * 2.4 = 876 kWh

cost of electricity = 10 cents/ kWh

so total cost of energy for one year

P_1 = 876 * 10 = 8760 cents = $87.60

Now if all 60 Watt bulbs are replaced by 30 Watt bulbs

So total power of all bulbs = 30 * 10 = 300 W

now each bulb used for 4 hours daily

so total energy consumed daily

E = P * t

E = 300 * 4 = 1200 Wh

E = 1.2 kWh

now we have total power consumed in 1 year

E = 365 * 1.2 = 438 kWh

cost of electricity = 10 cents/ kWh

so total cost of energy for one year

P_1 = 438 * 10 = 4380 cents = $43.80

total money saved in 1 year

Savings = 87.6 - 43.8 = $43.80

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Liula [17]

Answer:

a) v₁fin = 3.7059 m/s   (→)

b) v₂fin = 1.0588 m/s     (→)

Explanation:

a) Given

m₁ = 0.5 Kg

L = 70 cm = 0.7 m

v₁in = 0 m/s   ⇒  Kin = 0 J

v₁fin = ?

h<em>in </em>= L = 0.7 m

h<em>fin </em>= 0 m   ⇒    U<em>fin</em> = 0 J

The speed of the ball before the collision can be obtained as follows

Einitial = Efinal

⇒ Kin + Uin = Kfin + Ufin

⇒ 0 + m*g*h<em>in</em> = 0.5*m*v₁fin² + 0

⇒ v₁fin = √(2*g*h<em>in</em>) = √(2*(9.81 m/s²)*(0.70 m))

⇒ v₁fin = 3.7059 m/s   (→)

b)  Given

m₁ = 0.5 Kg

m₂ = 3.0 Kg

v₁ = 3.7059 m/s    (→)

v₂ = 0 m/s

v₂fin = ?

The speed of the block just after the collision can be obtained using the equation

v₂fin = 2*m₁*v₁ / (m₁ + m₂)

⇒  v₂fin = (2*0.5 Kg*3.7059 m/s) / (0.5 Kg + 3.0 Kg)

⇒  v₂fin = 1.0588 m/s     (→)

7 0
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What is the charge on a hypothetical ion with 35 protons and 37 electrons?
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5 0
3 years ago
The car in the figure travels at a constant speed along the road shown. Draw vector showing its acceleration at the point C if t
yuradex [85]

Answer:

The vector form is as shown in the attachment

Explanation:

The figure as shown in the diagram, indicates that the car is moving along the road at a constant speed. Centripetal acceleration comes into play for an object moving in a circular motion at uniform speed. The centripetal acceleration is the acceleration experienced by an object while in uniform circular motion.

Mathematically from centripetal acceleration; a = v2/r

The equation shows that there is an inverse relationship between the acceleration and the radius of curvature as such the radius of curvature at the point A will be more than the radius of curvature at the point C, this shows that the centripetal acceleration at point C will be more than the centripetal acceleration at point A.

The attachment shows the figure and the representation in vectorial form.

6 0
3 years ago
A small 1.0 kg steel ball rolls west at 3.0 m/s collides with a large 3.0 kg ball at rest. After the collision, the small ball m
77julia77 [94]

Answer:

The direction of the momentum of the large ball after the collision with respect to east is 146.58°.

Explanation:

Given that,

Mass of large ball = 3.0 kg

Mass of steel ball = 1.0 kg

Velocity = 3.0 kg

After collision,

Velocity = 2.0 m/s

Using conservation of momentum

m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}

3.0\times0+1.0\times(3.0)(-i)=1.0\times2(-j)+3.0\times v_{2}

-3i+2j=3.0\times v_{2}

v_{2}=-i+0.66j

The direction of the momentum

tan\theta=\dfrac{0.66}{-1}

\theta=tan^{-1}\dfrac{0.66}{-1}

\theta=-33.42^{\circ}

The direction of the momentum with respect to east

\theta=180-33.42=146.58^{\circ}

Hence, The direction of the momentum of the large ball after the collision with respect to east is 146.58°.

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