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3 years ago

In terms of instrument sensitivity, explain why he choice of wavelength on the colorimeter is important

1 answer:

8 0

Colorimeters function by passing colored light through a specific optical filter. The choice of wavelength on the colorimeter is important in terms of instrument sensitivity since the device gives the most sensitivity which can be used as a test factor in determining the lower detection limit.

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Which interaction of nature is weakest?

LekaFEV [45]

Electromagnetic would be the weakest interaction of nature.

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3 years ago

In order to determine the velocity, you must know

Naddik [55]

You need to know the speed and direction of object

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3 years ago

Read 2 more answers

Two blocks a and b ($m_a>m_b$) are pushed for a certain distance along a frictionless surface. how does the magnitude of the

Yuki888 [10]

Answer:

the magnitude of the work done by the two blocks is the same.

Explanation:

The work done by block a on block b is given by:

$W_a = F_a d$

where Fa is the force exerted by block a on block b, and d is the distance they cover.

The work done by block b on block a is given by:

$W_b = F_b d$

where Fb is the force exerted by block b on block a, and d is still the distance they cover.

For Newton's third law, the force exerted by block a on block b is equal to the force exerted by block b on block a, therefore

$F_a = F_b$

and so

$W_a=W_b$

3 0

3 years ago

Your job is to lift 30 kgkg crates a vertical distance of 0.90 mm from the ground onto the bed of a truck. For related problem-s

Gekata [30.6K]

Answer:

The number of crates is 84580.

Explanation:

mass, m = 30 kg

height, h = 0.9 mm

Power, P = 0.5 hp = 0.5 x 746 W = 373 W

time, t = 1 minute = 60 s

Let the number of crates is n.

Power is given by the rate of doing work.

$P = \frac{n m gh}{t}\\\373 =\frac{n\times 30\times9.8\times 0.9\times 10^{-3}}{60}\\\\n =84580$

7 0

3 years ago

Resistors 1 and 2− R1 = 50 Ω , R2 = 90 Ω − are connected in series to a 6.0-V battery. Part APart complete What is the potential

kondor19780726 [428]

Answer:

Part A: The voltage across resistor R1 is approximately $\rm 2.1 \; V$ .

Part B: When the value of resistor R1 decreases, the current in this circuit will increase.

Part C: When the value of resistor R1 decreases, the voltage across resistor R1 will decrease.

Explanation:

Resistor R1 and and R2 are connected in series. That's equivalent to a single resistor of $R_1 + R_2 = 50 + 90 = 140\; \Omega$ . The voltage across the two resistor, combined, is equal to $\rm 6\; V$ . Hence by Ohm's Law, the current through the circuit will be equal to $\rm \dfrac{6\; V}{140\; \Omega} = \dfrac{3}{70}\; A$ .

These two resistors are connected in series. The voltage across each of them might differ. However, the current through each of them should both be equal to the current through the circuit. In this case, the current through both R1 and R2 should be equal to $\rm \dfrac{3}{70}\; A$ . Apply Ohm's Law (again) to find the voltage across R1:

$V = I \cdot R = \dfrac{3}{70} \times 50 \approx \rm 2.1\; V$ .

Since the equivalent resistance is equal to $R_1 + R_2$ , when the value of $R_1$ decreases, the equivalent resistance will also decrease. By Ohm's Law, $I = \dfrac{V}{R}$ . When the value of the denominator ( decreases, the value of the quotient, $I$ the current through the circuit, will increase.

Keep in mind that if two resistors are connected in series,

$I(R_1) = I(\text{Circuit}) = I(R_2)$ .

The resistance of R1 decreases, while the current through it increases. Applying Ohm's Law on R1 won't give much useful information. However, since the resistance of R2 stays the same, the voltage across it will increase when its current increases (again by Ohm's Law.)

Again, since the two resistors are connected in series,

$V(R_1) + V(R_2) = V(\text{Circuit}) = \rm 6 \; V$ ,

when the voltage across R2 increases, the voltage across R1 will decrease.

4 0

3 years ago