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2 years ago

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A box is being pulled by two ropes. Eduardo pulls to the left with a force of 500 N, and Clara pulls to the right with a force o

f 200 N. The box moves because of the two forces applied to it. Leon records the forces and direction of the forces acting on the box in his lab notebook. A 2 column table with 5 rows. Column 1 is labeled Force with entries Normal, Tension by Eduardo, Tension by Clara, Kinetic friction, Gravity. Column 2 is labeled Direction of Force Vector with entries Upward, Left, Right, Left, Downward. In the table, which force has the wrong direction? Tension by Eduardo Tension by Clara Kinetic friction Gravity

1 answer:

sergejj [24]2 years ago

8 0

Answer:

kinetic friction or something else

Explanation:

Draw a diagram to illustrate the problem, as shown below.

m = mass of the box, kg

mg = weight of the box, N, acting downward

Because of Newton's 3rd law, a normal force of N = mg acts upward on the box.

C = 200 N, the force applied by Clara, horizontally to the right

E = 500 N, the force applied by Eduardo, horizontally to the left.

Because E is greater than C, the box will move left.

The frictional force, F, will resist the initial motion. Its magnitude is

F = μN = μmg, where μ = kinetic coefficient of friction.

F acts horizontally to the right.

From the given table, we can conclude that

Normal force acting upward is correct

Tension by Eduardoacting left is correct.

Tension by Clara acting right is correct.

Kinetic friction acting left is Incorrect.

Gravity acting downward is correct.correct

Answer:

The direction of the kinetic friction force is incorrect

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Answer:

v = [√(g/2h)]L

Explanation:

Let v be the initial horizontal velocity, t be the time James Bond uses to jump over the ledge of length, L.

So, vt = L and t = L/v

Also, since James Bond has no initial horizontal velocity, he falls freely through the distance, h so we use the equation y - y' = ut - 1/2gt², where y = 0 (at the top of the cliff) and y' = -h, u = 0 (initial vertical velocity), g = acceleration due to gravity = 9.8 m/s² and t = the time it takes to jump off the cliff = L/v.

Substituting these values into the equation, we have

y' - y = ut - 1/2gt²

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taking square root of both sides, we have

v = [√(g/2h)]L

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2 years ago

Two astronauts on opposite ends of a spaceship are comparing lunches. One has an apple, the other has an orange. They decide to

vesna_86 [32]

Answer:

The speed and direction of the apple is 1.448 m/s and 66.65°.

Explanation:

Given that,

Mass of apple = 0.110 kg

Speed = 1.13 m/s

Mass of orange = 0.150 kg

Speed = 1.25 m/s

Suppose we find the final speed and direction of the apple in this case

Using conservation of momentum:

Before:

In x direction,

$P_{b}=m_{p}v_{p}-m_{o}v_{o}$

$P_{b}=0.110\times1.13-0.150\times1.25$

$P_{b}=−0.0632\ kg-m/s$

In y direction = 0

After:

$v_{ay}$ is velocity of the apple in the y direction

$v_{ax}$ is the velocity of the apple in the x direction

Momentum again:

In x direction,

$0.110\times v_{ax}+0=−0.0632$

$v_{x}=\dfrac{−0.0632}{0.110}$

$v_{x}=−0.574\ m/s$

In y-direction,

$0.110\times v_{ay}-0.150\times0.977=0$

$v_{ay}=\dfrac{0.150\times0.977}{0.110}$

$v_{ay}=1.33\ m/s$

We need to calculate the speed of apple

$v_{a}=\sqrt{(v_{x})^2+(v_{y})^2}$

Put the value into the formula

$v_{a}=\sqrt{(−0.574)^2+(1.33)^2}$

$v_{a}=1.448\ m/s$

We need to calculate the direction of the apple

Using formula of angle

$\tan\theta=\dfrac{v_{ay}}{v_{ax}}$

Put the value into the formula

$\theta=\tan^{-1}(\dfrac{1.33}{0.574})$

$\theta=66.65^{\circ}$

Hence, The speed and direction of the apple is 1.448 m/s and 66.65°.

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2 years ago

A jogger runs 10.0 blocks do east, 5.0 blocks due South, and another two. Zero blocks do east. Assume all blocks are equal size,

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Jogger moves in three displacements

d1 = 10 blocks East

d2 = 5 blocks South

d3 = 2 blocks East

now we can say

total displacement towards East direction will be

$d_x = 10 + 2= 12 blocks$

Total displacement towards South

$d_y = 5 block$

now to find the net displacement we can use vector addition

$d = \sqrt{d_x^2 + d_y^2}$

$d = \sqrt{12^2 + 5^2}$

$d = 13 blocks$

<em>so magnitude of net displacement will be equal to 13 blocks</em>

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2 years ago

A 50-kg person stands 1.5 m away from one end of a uniform 6.0-m-long scaffold of mass 70.0 kg.

babymother [125]

Answer

given,

mass of the person, m = 50 Kg

length of scaffold = 6 m

mass of scaffold, M= 70 Kg

distance of person standing from one end = 1.5 m

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now equating all the vertical forces acting in the system.

T₁ + T₂ = m g + M g

T₁ + T₂ = 50 x 9.8 + 70 x 9.8

T₁ + T₂ = 1176...........(1)

system is equilibrium so, the moment along the system will also be zero.

taking moment about rope with tension T₂.

now,

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'3 m' is used because the weight of the scaffold pass through center of gravity.

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from equation (1)

T₂ = 1176 - 710.5

T₂ = 465.5 N

hence, T₁ = 710.5 N and T₂ = 465.5 N

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algol [13]

Answer :

The number of vacancies (per meter cube) = 5.778 × 10^22/m^3.

Explanation:

Given,

Atomic mass of silver = 107.87 g/mol

Density of silver = 10.35 g/cm^3

Converting to g/m^3,

= 10.35 g/cm^3 × 10^6cm^3/m^3

= 10.35 × 10^6 g/m^3

Avogadro's number = 6.022 × 10^23 atoms/mol

Fraction of lattice sites that are vacant in silver = 1 × 10^-6

Nag = (Na * Da)/Aag

Where,

Nag = Total number of lattice sites in Ag

Na = Avogadro's number

Da = Density of silver

Aag = Atomic weight of silver

= (6.022 × 10^23 × (10.35 × 10^6)/107.87

= 5.778 × 10^28 atoms/m^3

The number of vacancies (per meter cube) = 5.778 × 10^28 × 1 × 10^-6

= 5.778 × 10^22/m^3.

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2 years ago