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3 years ago

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A boy starts from point A and moves 5 units toward the east, then turns back and moves 3 units toward the west. What is the disp

lacement in the position of the boy?

1 answer:

Flura [38]3 years ago

4 0

The displacement of the boy is 2 units

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1. An object on Earth and the same object on the Moon would have a difference in

Feliz [49]

Answers: (1) a. weight, (2)b. Force changes by 2/9, (3)b. movement, (4)a. 40,000 Joules, (5)c. the soil will be 5°C.

<h2>Answer 1: a. weight</h2>

Mass and weight are very different concepts.

Mass is the amount of matter that exists in a body, which only depends on the quantity and type of particles within it. This means mass is an intrinsic property of each body and remains the same regardless of where the body is located.

On the other hand, weight is a measure of the gravitational force acting on an object and is directly proportional to the product of the mass $m$ of the body by the acceleration of gravity $g$ :

$W=m.g$

Then, since the Earth and the Moon have different values of gravity, t<u>he weight of an object in each place will vary</u>, but its mass will not.

<h2>Answer 2: b. Force changes by 2/9</h2>

According to the law of universal gravitation, which is a classical physical law that describes the gravitational interaction between different bodies with mass:

$F=G\frac{m_{1}m_{2}}{r^2}$ (1)

Where:

$F$ is the module of the force exerted between both bodies

$G$ is the universal gravitation constant

$m_{1}$ and $m_{2}$ are the masses of both bodies.

$r$ is the distance between both bodies

If we double the mass of one object (for example $2m_{1}$ ) and triple the distance between both (for example $3r$ ). The equation (1) will be rewritten as:

$F=G\frac{2m_{1}m_{2}}{(3r)^2}$ (2)

$F=\frac{2}{9}G\frac{m_{1}m_{2}}{r^2}$ (3)

If we compare (1) and (2) we will be able to see the force changes by 2/9.

<h2>Answer 3: b. movement</h2>

The Work $W$ done by a Force $F$ refers to the release of potential energy from a body that is <u>moved</u> by the application of that force to overcome a resistance along a path.

When the applied force is constant and <u>the direction of the force and the direction of the movement are parallel,</u> the equation to calculate it is:

$W=(F)(d)$

Now, <u>when they are not parallel, both directions form an angle</u>, let's call it $\alpha$ . In that case the expression to calculate the Work is:

$W=Fdcos{\alpha}$

Therefore, pushing on a rock accomplishes no work unless there is movement (independently of the fact that movement is parallel to the applied force or not).

<h2>Answer 4: a. 40,000 Joules</h2>

The Kinetic Energy is given by:

$K=\frac{1}{2}mV^{2}$ (4)

Where $m$ is the mass of the body and $V$ its velocity

For the first case (kinetic energy $K_{1}=10000J$ for a car at $V_{1}=30 mph=13.4112m/s$ ):

$K_{1}=\frac{1}{2}mV_{1}^{2}$ (5)

Finding $m$ :

$m=\frac{2K_{1}}{V_{1}^{2}}$ (6)

$m=\frac{2(10000J)}{(13.4112m/s)^{2}}$ (7)

$m=111.197kg$ (8)

For the second case (unknown kinetic energy $K_{2}$ for a car with the same mass at $V_{2}=60 mph=26.8224m/s$ ):

$K_{2}=\frac{1}{2}mV_{2}^{2}$ (9)

$K_{2}=\frac{1}{2}(111.197kg)(26.8224m/s)^{2}$ (10)

$K_{2}=40000J$ (11)

<h2>Answer 5: c. the soil will be 5°C</h2>

The formula to calculate the amount of calories $Q$ is:

$Q=m. c. \Delta T$ (12)

Where:

$m$ is the mass

$c$ is the specific heat of the element. For water is $c_{w}=1 kcal/g\°C$ and for soil is $c_{s}=0.20 kcal/g\°C$

$\Delta T$ is the variation in temperature (the amount we want to find for both elements)

This means we have to clear $\Delta T$ from (12) :

$\Delta T=\frac{Q}{m.c}$ (13)

For Water:

$\Delta T_{w}=\frac{Q_{w}}{m_{w}.c_{w}}$ (14)

$\Delta T_{w}=\frac{1kcal}{(1kg)(1 kcal/g\°C)}$ (15)

$\Delta T_{w}=1\°C)}$ (16)

For Soil:

$\Delta T_{s}=\frac{Q_{s}}{m_{s.c_{s}}$ (17)

$\Delta T_{s}=\frac{1kcal}{(1kg)(0.20 kcal/g\°C)}$ (18)

$\Delta T_{s}=5\°C)}$ (19)

Hence the correct option is c.

5 0

3 years ago

Read 2 more answers

Two forces F1 and F2 act on an object at a point P in the indicated directions. The magnitude of F1 is 15 lb and the magnitude o

adoni [48]

Answer

given,

F₁ = 15 lb

F₂ = 8 lb

θ₁ = 45°

θ₂ = 25°

Assuming the question's diagram is attached below.

now,

computing the horizontal component of the forces.

F_h = F₁ cos θ₁ - F₂ cos θ₂

F_h = 15 cos 45° - 8 cos 25°

F_h = 3.36 lb

now, vertical component of the forces

F_v = F₁ sin θ₁ + F₂ sin θ₂

F_v = 15 sin 45° + 8 sin 25°

F_v = 13.98 lb

resultant force would be equal to

$F = \sqrt{F_h^2+F_v^2}$

$F = \sqrt{3.36^2+13.98^2}$

F = 14.38 lb

the magnitude of resultant force is equal to 14.38 lb

direction of forces

$\theta =tan^{-1}(\dfrac{F_v}{F_h})$

$\theta =tan^{-1}(\dfrac{13.98}{3.36})$

θ = 76.48°

3 0

3 years ago

When the government sets a price for wheat that is above the equilibrium price, it is imposing a _____.

nalin [4]

<h3><u>Answer</u>;</h3>

Price floor

When the government sets a price for wheat that is above the equilibrium price, it is imposing a<u> price floor</u>.

<h3><u>Explanation</u>;</h3>

<u>Price floors prevent a price from falling below a certain level. When a price floor is set above the equilibrium price, quantity supplied will exceed quantity demanded, and excess supply. </u>

<u>Price control </u>occurs when the government laws regulate prices instead of letting market forces determine prices.

<u>Price ceilings</u> prevent a price from rising above a certain level.

5 0

3 years ago

Read 2 more answers

Two carts, one of mass 2m and one of mass m, approach each other with the same speed, v. When the carts collide, they hook toget

Nikitich [7]

Answer:

<em>Second option</em>

Explanation:

<u>Linear Momentum</u>

The linear momentum of an object of mass m and speed v is

P=mv

If two or more objects are interacting in the same axis, the total momentum is

$P_t=m_1v_1+m_2v_2+...$

Where the speeds must be signed according to a fixed reference

The images show a cart of mass 2m moves to the left with speed v since our reference is positive to the right

$P_1=-2mv$

The second cart of mass m goes to the right at a speed v

$P_2=mv$

The total momentum before the impact is

$P_t=-2mv+mv=-mv$

The total momentum after the collision is negative, both carts will join and go to the left side

The first option shows both carts with the same momentum before the collision and therefore, zero momentum after. It's not correct as we have already proven

The third option shows the 2m cart has a positive greater momentum than the other one. We have proven the 2m car has negative momentum. This option is not correct either

The fourth option shows the two carts keep separated after the collision, which contradicts the condition of the question regarding "they hook together".

The second option is the correct one because the mass $m_2$ has a negative momentum and then the sum of both masses keeps being negative

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3 years ago

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Se define como la masa por unidad de volume

kogti [31]

Answer:

is it density

Explanation:

my bad if u get it wrong

7 0

3 years ago