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3 years ago

65 POINTS! PLEASE ANSWER EVERY QUESTION! NEED HELP ASAP!

1 answer:

3 0

Maybe you can split up the questions. I will try to answer your first question.

1. In an elastic collision, momentum is conserved. The momentum before the collision is equal to the momentum after the collision. This is a consequence of Newton's 3rd law. (Action = Reaction)

2. Momentum: p = m₁v₁ + m₂v₂

m₁ mass of ball A
v₁ velocity of ball A
m₂ mass of ball B
v₂ velocity of ball B

Momentum before the collision:
p = 2*9 + 3*(-6) = 18 - 18 = 0

Momentum after the collision:
p = 2*(-9) + 3*6 = -18 + 18 = 0

3: mv + m(-v) = m(-v) + m(v)
the velocities would reverse.

4.This question is not factual since the energy of an elastic collision must also be conserved. The final velocities should be: v₁ = -1 m/s and v₂ = 5 m/s. That said assuming the given velocities were correct:
before collision
p = 10*3 + 5*(-3) = 30 - 15 = 15
after collision:
p = 10*(-2) + 5 * v₂ = 15
v₂ = 7

5.You figure out.

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A coin is resting on the bottom of an empty container. The container is then filled to the brim three times, each time with a di

Virty [35]

Answer:

Refractive index of liquid C > Refractive index of liquid B > Refractive index of liquid A

Explanation:

Let the depth of each section is h.

That means the real depth for each section is h.

Apparent depth is liquid A is 7 cm.

Apparent depth in liquid B is 6 cm.

Apparent depth in liquid C is 5 cm.

by the formula of the refractive index

n = real depth / apparent depth

where, n is the refractive index of the liquid.

For liquid A:

$n_{A}=\frac{h}{7}$ .... (1)

For liquid B:

$n_{B}=\frac{h}{6}$ ..... (2)

For liquid C:

$n_{C}=\frac{h}{5}$ ..... (3)

By comparing all the three equations

nc > nB > nA

Refractive index of liquid C > Refractive index of liquid B > Refractive index of liquid A

5 0

3 years ago

A concrete slab shown in Figure 5 is being lifted by using three cables connected to the slab at points A, B and C. The slab is

xxMikexx [17]

Answer:

Fad = 28.8 kN

Fbd = 16.4 kN

Fcd = 28.1 kN

Explanation:

First, find the length of each cable.

AD = √((2 m)² + (0.5 m)² + (2.5 m)²)

AD = √10.5 m

AD ≈ 3.24 m

BD = √((1.5 m)² + (1 m)² + (2.5 m)²)

BD = √9.5 m

BD ≈ 3.08 m

CD = √((1 m)² + (1 m)² + (2.5 m)²)

CD = √8.25 m

CD ≈ 2.87 m

Next, use similar triangles to find the x, y, and z components of each tension force.

Fadx = 2/3.24 Fad = 0.617 Fad

Fady = 0.5/3.24 Fad = 0.154 Fad

Fadz = 2.5/3.24 Fad = 0.772 Fad

Fbdx = 1.5/3.08 Fbd = 0.487 Fbd

Fbdy = 1/3.08 Fbd = 0.324 Fbd

Fbdz = 2.5 / 3.08 Fbd = 0.811 Fbd

Fcdx = 1/2.87 Fcd = 0.348 Fcd

Fcdy = 1/2.87 Fcd = 0.348 Fcd

Fcdz = 2.5/2.87 Fcd = 0.870 Fcd

Now sum the forces in the x, y, and z directions:

∑Fx = ma

-0.617 Fad + 0.487 Fbd + 0.348 Fcd = 0

∑Fy = ma

-0.154 Fad − 0.324 Fbd + 0.348 Fcd = 0

∑Fz = ma

60 kN − 0.772 Fad − 0.811 Fbd − 0.870 Fcd = 0

To solve this system of equations algebraically, start by subtracting the first two equations, eliminating Fcd.

-0.463 Fad + 0.811 Fbd = 0

0.811 Fbd = 0.463 Fad

Fbd = 0.571 Fad

Substitute into either of the first two equations:

-0.617 Fad + 0.487 (0.571 Fad) + 0.348 Fcd = 0

-0.617 Fad + 0.278 Fad + 0.348 Fcd = 0

-0.339 Fad + 0.348 Fcd = 0

0.348 Fcd = 0.339 Fad

Fcd = 0.975 Fad

Now substituting into the third equation:

60 kN − 0.772 Fad − 0.811 Fbd − 0.870 Fcd = 0

60 kN − 0.772 Fad − 0.811 (0.571 Fad) − 0.870 (0.975 Fad) = 0

60 kN − 0.772 Fad − 0.463 Fad − 0.849 Fad = 0

60 kN − 2.083 Fad = 0

Fad = 28.8 kN

Solving for the other two tension forces:

Fbd = 0.571 Fad = 16.4 kN

Fcd = 0.975 Fad = 28.1 kN

3 0

3 years ago

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Calculate the force of gravity on the 1.2-kg mass if it were 1.9×107 m above earth's surface (that is, if it were four earth rad

Anettt [7]

Answer:

Force of gravity, F = 0.74 N

Mass of an object, m = 1.2 kg

Distance above earth's surface, $d=1.9\times 10^7\ m$

Mass of Earth, $M=5.97\times 10^{24}\ kg$

Radius of Earth, $r=6.37\times 10^6\ m$

We need to find the force of gravity above the surface of Earth. It is given by :

$F=G\dfrac{mM}{R^2}$

R = r + d

R = 25370000 m

$F=6.67\times 10^{-11}\times \dfrac{5.97\times 10^{24}\ kg\times 1.2\ kg}{(25370000\ m)^2}$

F = 0.74 N

So, the force of gravity on the object is 0.74 N. Hence, this is the required solution.

5 0

4 years ago

A man can swim with a speed of 3 m/s in still water. He crosses a river of width 100 m by swimming in the direction that will ta

Elis [28]

Explanation:

The given data is as follows.

width = 100 m, speed = 3 m/s

So, minimum time required to cross the stream is calculated as follows.

t = $\frac{100 m}{3 m/s}$

Also, in the same time the man has moved downstream by 50 meter.

Hence, distance traveled downstream is as follows.

d = velocity of river × t

$v_{river} = \frac{d}{t}$

= $\frac{50}{\frac{100}{3}}$

= $\frac{3}{2}$

= 1.5 m/s

Thus, we can conclude that the river is flowing by 1.5 m/s.

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3 years ago

Pleaswe help me ewwfea

Talja [164]

Answer:

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6 0

3 years ago

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