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3 years ago

How many (whole number of) 91 kg people

2 answers:

6 0

We are given that the maximum mass capacity is 1 metric ton or 1000 kg. Therefore the amount of people this can hold is simply the ratio of the maximum mass capacity over the mass of the person, that is:

people = 1000 kg / 91 kg

people = 10.98

So the elevator can safely occupy 10 people. We will not exceed to 11.

Vesna [10]3 years ago

6 0

Answer:

<h2>10 people.</h2>

Explanation:

Gives

The elevator can hold a maximum weight of 1 metric ton.
1 metric ton is equivalen to 1000 kilograms, or which is the same $1.000 \times 10^{3} \ kg$ .
Each person weighs 91 kg.

To know the number of people that can safely occupy the elevator, we just need to divide

$\frac{1000 \ kg}{91 \ kg} \approx 10$

Therefore, around 10 people can safely occupy the elevator.

Notice that we can round the number to 11, because we cannot do that in real life with people. The decimal number is 10.98, but we there can't ve 10.98 people, because we can't have there a part of a person.

Therefore, the answer is 10 people.

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Suppose a small planet is discovered that is 16 times as far from the Sun as the Earth's distance is from the Sun. Use Kepler's

mamaluj [8]

Answer:

23376 days

Explanation:

The problem can be solved using Kepler's third law of planetary motion which states that the square of the period T of a planet round the sun is directly proportional to the cube of its mean distance R from the sun.

$T^2\alpha R^3\\T^2=kR^3.......................(1)$

where k is a constant.

From equation (1) we can deduce that the ratio of the square of the period of a planet to the cube of its mean distance from the sun is a constant.

$\frac{T^2}{R^3}=k.......................(2)$

Let the orbital period of the earth be $T_e$ and its mean distance of from the sun be $R_e$ .

Also let the orbital period of the planet be $T_p$ and its mean distance from the sun be $R_p$ .

Equation (2) therefore implies the following;

$\frac{T_e^2}{R_e^3}=\frac{T_p^2}{R_p^3}....................(3)$

We make the period of the planet $T_p$ the subject of formula as follows;

$T_p^2=\frac{T_e^2R_p^3}{R_e^3}\\T_p=\sqrt{\frac{T_e^2R_p^3}{R_e^3}\\}................(4)$

But recall that from the problem stated, the mean distance of the planet from the sun is 16 times that of the earth, so therefore

$R_p=16R_e...............(5)$

Substituting equation (5) into (4), we obtain the following;

$T_p=\sqrt{\frac{T_e^2(16R_e)^3}{(R_e^3}\\}\\T_p=\sqrt{\frac{T_e^24096R_e^3}{R_e^3}\\}$

$R_e^3$ cancels out and we are left with the following;

$T_p=\sqrt{4096T_e^2}\\T_p=64T_e..............(6)$

Recall that the orbital period of the earth is about 365.25 days, hence;

$T_p=64*365.25\\T_p=23376days$

4 0

3 years ago

A light rope is attached to a block with mass 3.60 kg that rests on a frictionless, horizontal surface. The horizontal rope pass

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Answer and Explanation:

(a) The fre-body diagrams for each block is shown below. In the block of mass 3.60 kg, there are 3 forces acting on it: horizontal force due to the rope ( $F_{t}$ ), vertical gravitational force ( $F_{g}$ ) and vertical normal force ( $F_{n}$ ), due to the surface. Since there is no vertical movement, $F_{g}$ and $F_{n}$ cancels it out. So, for this block, net force is horizontal due to the rope $F_{t}$ .

The block of mass m is hanging from the pulley, so there is the force of the rope ( $F_{t}$ ) and the gravitational force ( $F_{g}$ ). Both are vertical, because there is no surface "holding" block m.

(b) Since both blocks are attached to each other, the acceleration will be the same. To calculate it, we use the Second Law of Motion:

$F_{r}=m.a$

$a=\frac{F_{r}}{m}$

$a=\frac{18.8}{3.6}$

a = 5.22

The acceleration of either block is 5.22 m/s².

(c) Block m has 2 forces acting on it: tension and gravitational force. Gravitational force is the force of attraction the Earth does over an object. It is calculated as the product of mass and gravitational acceleration, which has magnitude g = 9.8 m/s².

Suppose positive referential is going up. To determine mass:

$F_{r}=m.a$

$F_{t}-F_{g}=m.a$

$F_{t}-m.g=m.a$

$18.8-9.8m=5.22m$

$15.02m=18.8$

m = 1.25

Block m has 1.25 kg.

(d) Gravitational force is also called weight. So, as described above: $F_{g}=m.g$ .

The weight for the hanging block is

$F_{g}=1.25*9.8$

$F_{g}=$ 12.25 N

Comparing tension and weight:

$\frac{12.25}{18.8}$ ≈ 0.65

We can see that, weight of the hanging block is almost 0.65 times smaller than the tension on the rope.

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3 years ago

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3 years ago

What are challenges of securing a crime scene

kotegsom [21]

At any crime scene, the two greatest challenges to the physical evidence are contamination and loss of continuity.

<h3>What is the meaning of physical evidence?</h3>

In evidence law, physical evidence (also called real evidence or material evidence) is any material object that plays some role in the matter that gave rise to the litigation, introduced as evidence in a judicial proceeding (such as a trial) to prove a fact in issue based on the object's physical characteristics.

The two types of evidence at crime scenes:

Biological evidence (e.g., blood, body fluids, hair and other tissues)

Latent print evidence (e.g., fingerprints, palm prints, footprints)

The biggest impediment to an investigation is the removal or loss of a piece of evidence from the scene of a crime.

Hence, at any crime scene, the two greatest challenges to the physical evidence are contamination and loss of continuity.

Learn more about the physical evidence here:

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2 years ago

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3 years ago

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