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Sergio [31]
3 years ago
7

Can someone help with these questions (1 , 3 , 4)

Physics
1 answer:
LenaWriter [7]3 years ago
8 0

Remark

4 is long (but not hard) and I am in a terrible time bind. I will take a short at doing it but it will be abbreviated.

One

Ax = 70*cos(51.7) = 43.38

Ay = 70*sin(51.7) = 54.93

That makes C the correct answer. To resolve these, Ax is always taken along the +x axis. and Ay is always associated with the y axis. A(x) always uses Cos(x) and Ay = sin(x)

Three

A would be true if Ax and Ay were vectors. They don't seem to be. They look like scalers. Magnitudes don't add that way.

B Theta = tan-1(Ay/Ax) The Statement given is simply not true.

C Tan(Theta) = Ay / Ax Not The given result

D is correct Cos(theta) = The magnitude along the +x axis divided by the hypotenuse which is sqrt(Ax^2 + Ay^2)

Four

The best way to do this is to set up x and y values in a table. I can't do this well, but I'll try.

<u>x</u>                                                            <u>y</u>

12*cos(53) = 7.22                             15*sin(53)= 9.58

15*cos(-40) = <u>11.49</u>                            15.sin(-40) = <u>-9.64</u>

Sum =            18.69                                             = - 0.08

As you can see most of the magnitude and direction will come from Ax

So the answer is 19 for the magnitude and 0o for the direction. The other choices are taken from theta = 40o. That minus sign makes all the difference.          

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3 years ago
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7. A beam of light converges at a point P. Now a lens is placed in the path of the convergent
Rudiy27

Answer:

Lens at a distance = 7.5 cm

Lens at a distance = 6.86 cm  (Approx)

Explanation:

Given:

Object distance u = 12 cm

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Computation:

a. 1/v = 1/u + 1/f

1/v = 1/20 + 1/12

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Lens at a distance = 6.86 cm  (Approx)

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inessss [21]

Answer:

at t=46/22, x=24 699/1210 ≈ 24.56m

Explanation:

The general equation for location is:

x(t) = x₀ + v₀·t + 1/2 a·t²

Where:

x(t) is the location at time t. Let's say this is the height above the base of the cliff.

x₀ is the starting position. At the base of the cliff we'll take x₀=0 and at the top x₀=46.0

v₀ is the initial velocity. For the ball it is 0, for the stone it is 22.0.

a is the standard gravity. In this example it is pointed downwards at -9.8 m/s².

Now that we have this formula, we have to write it two times, once for the ball and once for the stone, and then figure out for which t they are equal, which is the point of collision.

Ball: x(t) = 46.0 + 0 - 1/2*9.8 t²

Stone: x(t) = 0 + 22·t - 1/2*9.8 t²

Since both objects are subject to the same gravity, the 1/2 a·t² term cancels out on both side, and what we're left with is actually quite a simple equation:

46 = 22·t

so t = 46/22 ≈ 2.09

Put this t back into either original (i.e., with the quadratic term) equation and get:

x(46/22) = 46 - 1/2 * 9.806 * (46/22)² ≈ 24.56 m

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