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AVprozaik [17]
2 years ago
10

Sarah and Maisie are analysing data from their school sports day. Looking at the 1500 m results for Stephen, Maisie believes tha

t Stephen’s displacement from the start line is 1500 m. Sarah says that she is incorrect and that his displacement from the start is actually 0 m. Which of the students is correct? Give reasoning for your answer.
Physics
1 answer:
Dahasolnce [82]2 years ago
4 0

Answer:

Sarah is right

Explanation:

This is an exercise that differentiates between scalars and vectors.

A scalar is a number, instead a vector is a number that represents the module in addition to direction and sense.

In this case, the distance (scalar) traveled is a number, which is why it is worth 1500m, but the displacement is a vector and since the point where it leaves is the same point where the vector's modulus arrives is zero, so the DISPLACEMENT VECTOR is zero

consequently Sarah is right

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A jogger runs 5.0 km on a straight trail at an angle of 60° south of west. What is the southern component of the run rounded to
geniusboy [140]

Answer:

4.3 km

Explanation:

5 0
2 years ago
You stand on a frictional platform that is rotating at 1.1 rev/s. Your arms are outstretched, and you hold a heavy weight in eac
bezimeni [28]

Answer:

a) The resulting angular speed of platform is 1.38 rev/sec

b) The change in kinetic energy of the system is 53 J.

Explanation:

This question is incomplete. The complete question will be:

You stand on a frictional platform that is rotating at 1.1 rev/s. Your arms are outstretched, and you hold a heavy weight in each hand. The moment of inertia of you, the extended weights, and the platform is 8.8 kg · m2. When you pull the weights in toward your body, the moment of inertia decreases to  7.0 k g .m 2  

a) What is the resulting angular speed of the platform? Answer in units of r e v / s .

b)What is the change in kinetic energy of the system? Answer in units of J.

<h3>ANSWER:</h3>

a)

we know that:

Angular Momentum = L = Iω

From conservation of momentum:

Lo = Lf

(Io) (ωo) = (If) (ωf)

ωf = (Io) (ωo)/(If)

ωf = (8.8 kg.m²)(1.1 rev/s)/(7.0 kg.m²)

<u>ωf = 1.38 rev/sec =</u>

b)

ωf = (1.38 rev/sec)(2π rad/ 1 rev) = 8.67 rad/sec

ωo = (1.1 rev/sec)(2π rad/ 1 rev) = 6.91 rad/sec

The kinetic energy for rotational motion is given as:

K.E = (1/2)Iω²

Thus, the change in kinetic energy will be:

ΔK.E = (K.E)f - (K.E)o

ΔK.E = (1/2)Ifωf² - (1/2)Ioωo²

ΔK.E = (1/2)(Ifωf² - Ioωo²)

ΔK.E = (1/2)[(7 kg.m²)(8.67 rad/sec)² - (8.8 kg.m²)(6.91 rad/sec)²

<u>ΔK.E = 53 J</u>

5 0
3 years ago
During which month is the bacteria population just under 1 million?
Thepotemich [5.8K]
D. March because it is just below the 1 million marker on the graph and it is the only one that low.
5 0
3 years ago
A centrifuge was used to separate into it’s components. What conclusions out the sample can be formed based on the technique use
Tatiana [17]

-- The sample was a fluid.

-- It was a mixture or a suspension ... NOT a solution.

4 0
3 years ago
As shown above an adhesive has been applied to contacting faces of two blocks so that the blocks interact with an adhesive force
rusak2 [61]

The magnitude of the adhesive force allows the top block to remain

attached to the bottom when the blocks and the wire are balanced.

The option that must be true is; \mathbf{W_{bottom} = F_{adhesion}}

Reason:

The tension exerted by the wire attached to the top block = T

Magnitude of the adhesive = F_{adhesion}

Weight of the top block = W_{top}

Weight of the bottom block = W_{bottom}

Given that with the exertion of the tension, the two blocks remain at rest, we have;

  • T =  W_{top} + W_{bottom}

The adhesive causes the bottom block to remain attached to the top block, we have;

Therefore, the magnitude of the adhesive force adds the bottom weight, to the top, weight, which gives;

The magnitude of the adhesive force = The weight of the bottom block

Therefore;

  • W_{bottom} = F_{adhesion}

Learn more here:

brainly.com/question/18907970

8 0
3 years ago
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