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3 years ago

At what point on this position-time graph is the instantaneous velocity equal to zero?

Physics

2 answers:

denis-greek [22]3 years ago

7 0

The correct answer is D.

The instantaneous velocity of a body is given by the gradient of the tangent drawn to the point in the position- time graph.

The tangents drawn to points A and B slopes upwards and thus have a positive gradient, showing that the particle has velocity in the positive direction.

The tangent drawn to point E slopes down wards and hence its gradient is negative. The velocity of the particle is also finite at E, but in opposite direction to that at A and B.

However, point D lies at the top of the peak in the position- time graph. A tangent drawn to the point D is flat and parallel to the time axis. The gradient of the tangent is zero, implying that the velocity of the [article at D is zero.

Thus, from the position- time graph, it can be seen that the velocity of the article at D is zero.

Elena L [17]3 years ago

5 0

Answer:

its d

Explanation:

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Q 1 . How many significant figures are in the following measurement? 0.0009(1 point)

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Here we have some questions about experimental errors.

Q1) We want to see how many significant figures have the measure:

0.0009

The number of significant figures is the number of known digits that are not the leading zeros.

Here we can see four leading zeros, and a single-digit different than zero, which is a 9.

Then we have only one significant figure, the 9.

Q2) Here we will use the measure that is the less exact, as the error of that measure may be larger than the smaller significant figures of the other measures.

Then:

31.2 lb + 38.02lb + 45 lb

The worst measure is 45lb, so the smallest significant figure that we should use is the first one at the left of the decimal point, then we need to round the other two measures to the next whole number, we will get:

31 lb + 38 lb + 45 lb = 114lbs

Q3) We know that the measure is 11.5 seconds and the uncertainty of 1.7%, then the uncertainty will be the 1.7% of the above measure:

(1.7%/100%)*11.5 s = 0.1955 s

Notice that our measure has one significant figure after the decimal point, so we need to round the error to the same significant figure.

0.1955 s ≈ 0.2s

Then the measure is:

11.5 s ± 0.20 s

Q4) We have the measure:

312.0 mph ± 3.9 mph.

The percent uncertainty will be the quotient between the error and the measure times 100%, or:

(3.9 mph/312.0 mph)*100% = 1.25%

This is a percent error, we do not need to round this.

If you want to learn more, you can read:

brainly.com/question/17339020

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3 years ago

Four capacitors with capacitance 3.0 pF, 2.0 pF, 5.0 pF and X pF are connected in series to each other. If the equivalent capaci

bagirrra123 [75]

Answer:

Explanation:

If unknown capacitance C1, C2, C3 and C4 are connected in series to one another, their equivalent capacitance of the circuit will be expressed as shown

$\frac{1}{C_t} = \frac{1}{C_1} +\frac{1}{C_2} +\frac{1}{C_3} +\frac{1}{C_4} \\$

Given the capacitance's 3.0 pF, 2.0 pF, 5.0 pF and X pF connected in series to each other. If the equivalent capacitance of the circuit is 0.83 pF, then to get X, we will apply the formula above;

$\frac{1}{0.83} = \frac{1}{3.0} +\frac{1}{2.0} +\frac{1}{5.0} +\frac{1}{C_4} \\\\\\1.205 = 0.333+0.5+0.2+\frac{1}{C_4} \\\\1.205 = 1.033 + \frac{1}{C_4} \\\\\frac{1}{C_4} = 1.205-1.033\\\\\frac{1}{C_4} = 0.172\\\\C_4 = \frac{1}{0.172}\\ \\C_4 = 5.8pF\\\\$

C₄ ≈ 6pF

Hence the value of the X capacitor is approximately 6pF

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3 years ago