What type of lamp is used in a spectrophotometer to produce visible light?

How much energy is need to raise 50 kg of water from 45 c to 80c?

Dmitriy789 [7]

Based on your problem, what you are looking for is the quantity of heat. To solve for it, you will need this formula:

Q = mc(T2-T1)

Where: Q = Quantity of heat
m = mass of the substance
c = Specific heat
T2 = Final temperature
T1 = Initial temperature

Now the specific heat of water is 4.184 J/(g°C), meaning that is how much energy is required to raise the temperature of 1g of liquid water by 1 degree Celsius.

Since your mass is in kilograms, let us convert that into grams, which will be equal to 50,000 grams. Now we can put our given into the equation:

Q = mc(T2-T1)
= 50,000g x 4.184 J/(g°C) x (80°C - 45°C)
= 50,000 g x 4.184 J/(g°C) x 35°C
= 7,322,000 J or 7,322 kJ or 7.322 MJ

8 0

3 years ago

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. A light bulb glows because of it’s resistance, and the brightness of the bulbincreases with the electrical power delivered to

sesenic [268]

Complete Question

The complete question is shown in the first uploaded image

Answer:

a

When the both bulb are in the circuit bulb B glows equally brighter to bulb A

This because the power delivered to the both bulb are equal

b

The bulb A on the right will glow brighter than the bulb A on the left due to the fact that the power supplied to bulb A on the right is higher than that gotten by bulb A on the left.

Explanation:

From the question we are been told that the two bulbs are identical

So their resistance denoted by R is the same

Considering the left circuit where the two bulbs are connected in series which mean that the same current is passing through them

$R_A =R_B =R$

$i_A = i_B =i$

$R_{eq} = R_1 +R_2 = 2R$

$i = \frac{V}{2R}$

The power that is been deposited on the circuit is evaluated as

$P_A = i^2R$

$P_A = \frac{V^2}{4R}$

$P_B = i^2R$

$P_B = \frac{V^2}{4R}$

For the fact that the power deposited on the bulbs are the same they will glow equally

When B is now removed and only A is left

$R_{eq} = R_A = R$

$i = \frac{V}{R}$

$P'_A = i^2R$

$P'_A = \frac{V^2}{R}$

For the fact that its only bulb A that is on that right circuit the power delivered to it would be greater compared to the left circuit bulb A

7 0

3 years ago

A wheel 2.00 m in diameter lies in a vertical plane androtates about its central axis with a constant angularacceleration of 4.0

quester [9]

Answer: a) 20.8 rad/s, b) 20.8 m/s, c) 432.67 m/s²,

d)54.08 rad.

Explanation: diameter of wheel is 2m hence radius is 2/2 = 1m.

Constant angular acceleration = α = 4.00 rad/s²

Since the motion of the wheel is of a constant angular acceleration, hence newton's laws of motion is applicable.

Note that the body is starting from rest hence, initial angular velocity (ω0) is zero.

Time taken = 5.2s

a)

Recall that

ω = ω0 + αt

ω = 0 + 4(5.2)

ω = 20.8 rad/s.

b)

Tangential speed (v) is the linear speed and is given as

v = ωr

Where r is the radius of the wheel.

Note that ω = 20.8 rad/s

v = 20.8 × 1

v = 20.8 m/s.

c)

Total acceleration = √(a*)² + (a')²

Where a* is the radial component of acceleration = v²/r and a' is the vertical component of acceleration = αr.

Radial component of acceleration = v²/r = 20.8²/1 = 432.64 rad/s²

Vertical component of acceleration = αr = 4 × 1 = 4m/s²

Total acceleration = √(432.64)² + (4)²

Total acceleration = √187,177.3696 + 16

Total acceleration = √187,193.3696

Total acceleration = 432.67 m/s²

d)

θ = ω0t + αt²/2

But ω0 = 0 and θ = angular displacement ( angular position)

θ = 4(5.2)²/2

θ = 54.08 rad.

8 0

3 years ago

EASY BRAINLIEST!!URGENT PLEASE HELP.

Gekata [30.6K]

Answer:

False

Explanation:

Contrary to popular belief, machines do not increase the amount of work that is done. They just change how the work is done. Machines make work easier by increasing the amount of force that is applied, increasing the distance over which the force is applied, or changing the direction in which the force is applied.

6 0

3 years ago

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A 0.500 kg mass is oscillating on a spring with k = 330 N/m. The total energy of its oscillation is 3.24 J. What is the amplitud

Oliga [24]

Answer:

0.140

Explanation:

may i be marked brainliest

4 0

3 years ago