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vesna_86 [32]
3 years ago
9

9) Cart 1 has a mass of 4 kg and an initial speed of 4 m/s. It eventually elastically collides with cart 2, whose mass is 6 kg,

and which moves at an initial speed of 4 m/s toward cart 1. How fast are the carts moving after their collision? [Enter cart 1's final speed in answer box 1 and cart 2's final speed in answer box 2.]

Physics
1 answer:
postnew [5]3 years ago
3 0

Answer:

Cart 1 = 4 m/s

Cart 2 = 4 m/s

Explanation:

See attachment

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An airplane is flying at a speed of 200 m/s in level flight at an altitude of 800 m. A package is to be dropped from the airplan
MArishka [77]

Answer:

2560m or 2.56km (rounded to 3 significant figures)

Explanation:

First, list all known and desired values/variables (initial vertical velocity is 0 as the plane is kept level and vertical acceleration is just gravity):

Vertical \ velocity \ (\frac{m}{s} ) =  u_{v} = 0 \\\\ Horizontal \ velocity \ (\frac{m}{s} ) =  u_{h} = 200\\\\ Vertical \ acceleration \ (\frac{m}{s^{2} } ) =  a_{v} =  9.8 \\\\ Horizontal \ acceleration \ (\frac{m}{s^{2} } ) =  a_{h} =  0 \\\\ Vertical \ displacement \ (m) = s_{v} = 800 \\\\ Horizontal \ displacement \ (m) = s_{h}

The horizontal displacement is going to be the distance travelled, horizontally of course, once the package is released;

First thing to understand is that the vertical and horizontal components are to be dealt with separately because they don't affect each other;

Since there is no horizontal acceleration (ignoring air resistance), we simply require a velocity and time to find the horizontal displacement, using the formula v = d/t (or speed = distance/time);

What we have is the horizontal velocity but we don't have the time taken;

One thing we know is that the time elapsed for the vertical fall of 800m and for the horizontal displacement must be the same;

What we do, therefore, is find the time taken for the vertical displacement using the formula, s = ut + ¹/₂·at², since we know the vertical velocity, height and acceleration:

800 = (0)t + ¹/₂·(9.8)t²

800 = 4.9t²

t² = 163.26...

t = 12.77...

We now have the time taken for the vertical fall and the horizontal displacement, we can use this with the horizontal velocity we know already and get the horizontal displacement:

u_{h} = \frac{s_{h} }{t} \\\\ 200 = \frac{s_{h} }{12.77...} \\\\ s_{h} = 200(12.77...) \\\\ s_{h} = 2555.5...

7 0
2 years ago
Why is friction like applied force but different from gravity?
kirill115 [55]

Answer:

Friction is when a force is applied or done by weight dragging onto something.

Explanation:

Gravity is when an object is getting pulled toward the center of what is attracting it. And applied force is when someone/sommething is applying force.

3 0
3 years ago
Read 2 more answers
Can y’all please help me quickkkkk just give me answers it wouldn’t even take a minute from ur life pleaseee
shusha [124]

There's only one question there.

The answer is "Greater amplitude".

6 0
3 years ago
The left end of a long glass rod 8.00 cm in diameter, with an index of refraction 1.60, is ground to a concave hemispherical sur
sp2606 [1]

Answer:

a) q = -9.23 cm, b)  h’= 0.577 mm , c) image is right and virtual

Explanation:

This is an optical exercise, where the constructor equation should be used

        1 / f = 1 / p + 1 / q

Where f is the focal length, p the distance to the object and q the distance to the image

A) The cocal distance is framed with the relationship

       1 / f = (n₂-1) (1 /R₁ -1 /R₂)

In this case we have a rod whereby the first surface is flat R1 =∞ and the second surface R2 = -4 cm, the sign is for being concave

       1 / f = (1.60 -1) (1 /∞ - 1 / (-4))

       1 / f = 0.6 / 4 = 0.15

        f = 6.67 cm

We have the distance to the object p = 24.0 cm, let's calculate

       1 / q = 1 / f - 1 / p

       1 / q = 1 / 6.67 - 1/24

       1 / q = 0.15 - 0.04167 = 0.10833

       q = -9.23 cm

distance to the negative image is before the lens

B) the magnification of the lenses is given by

       M = h ’/ h = - q / p

        h’= - q / p h

        h’= - (-9.23) / 24.0 0.150

        h’= 0.05759 cm

        h’= 0.577 mm

C) the object is after the focal length, therefore, the image is right and virtual

6 0
3 years ago
What type of relationship exists between the length of a wire and the resistance, if all other factors remain the same? A. Resis
ZanzabumX [31]
<h3>Answer</h3>

(A) Resistance is directly related to length.

<h3>Explanation</h3>

Formula for resistance

R = p(length) / A

where R = resistance

           p = resistivity(material of wire)

           A = cross sectional area

So it can be seen that resistance depends upon 3 factors that are length of wire , resistivity of wire and the cross sectional area of the wire.

If two of the factors, resistivity and cross sectional area, are kept constant then the resistance is directly proportional to the length of wire.

<h3> R ∝ length</h3>

This means that the resistance of the wire increases with the increase in length of the wire and decreases with the decrease of length of the wire.

5 0
3 years ago
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