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RideAnS [48]
3 years ago
8

How can a single crash of a car into a wall involve three separate collections?

Physics
1 answer:
kap26 [50]3 years ago
7 0

Answer:

because it can hit another one and git another one

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When all of the magnetic domains line up on their own, the material is called ____________________.
Gelneren [198K]

Probably C...:):):):):):):):):):):):):):):):):):):):):):):):):):):):):):):):):):):)


3 0
3 years ago
Read 2 more answers
In 2007, michael carter (u.s.) set a world record in the shot put with a throw of 24.77 m. what was the initial speed of the sho
Serga [27]

The initial speed of the shot is 15.02 m/s.

The Shot put is released at a height y<em> </em>from the ground with a speed u. It is released at an angle θ to the horizontal. In a time t, the shot put travels a distance <em>R</em> horizontally.

Pl refer to the attached diagram.

Resolve the velocity u into horizontal and vertical components, u ₓ=ucosθ and uy=u sinθ. The horizontal component remains constant in the absence of air resistance, while the vertical component varies due to the action of the gravitational force.

Write an expression for R.

R=u_xt=(ucos \theta)t

Therefore,

t=\frac{R}{ucos\theta} .......(1)

In the time t, the net displacement of the shotput is y in the downward direction.

Use the equation of motion,

y=u_yt-\frac{1}{2}gt^2=(usin\theta) t-\frac{1}{2}gt^2

Substitute the value of t from equation (1).

y=(ucos\theta)(\frac{R}{ucos\theta} )-\frac{1}{2} g(\frac{R}{ucos\theta} )^2\\ =Rtan\theta-(\frac{gR^2}{2u^2cos^2\theta} )

Substitute -2.10 m for y, 24.77 m for R and 38.0° for θ and solve for u.

y=Rtan\theta-(\frac{gR^2}{2u^2cos^2\theta} )\\ (-2.10m)=(24.77 m)(tan38.0^o)-\frac{(9.8 m/s^2)(24.77m)^2}{2u^2(cos38.0^o)^2} \\ u^2=225.71(m/s)^2\\ u=15.02m/s

The shot put was thrown with a speed 15.02 m/s.




7 0
3 years ago
To practice Problem-Solving Strategy 27.1: Magnetic Forces. A particle with mass 1.81×10−3 kgkg and a charge of 1.22×10−8 CC has
Roman55 [17]

Answer:

The magnitude of the acceleration  is a = 0.33 m/s^2

The direction is - \r k i.e the negative direction of the z-axis

Explanation:

 From  the question we are that

       The mass of the particle m = 1.8*10^{-3} kg

         The charge on the particle is q = 1.22*10^{-8}C

         The velocity is \= v = (3.0*10^4 m/s ) j

        The the magnetic field is  \= B = (1.63T)\r i + (0.980T) \r j

The charge experienced  a force which is mathematically represented as

         

                    F = q (\= v * \= B)

    Substituting value

         F = 1.22*10^{-8} (( 3*10^4 ) \r j \ \ X \ \  ( 1.63 \r i  + 0.980 \r j )T)

            = 1.22 *10^{-8} ((3*10^4 * 1.63)(\r j \ \  X \ \  \r i) + (3*10^4 * 0.980) (\r j \ \ X \ \  \r  j))

            = (-5.966*10^4 N) \r k

Note :

           i \ \ X \ \ j = k \\\\j \ \  X  \ \ k = i\\\\k  \ \ X \ \ i = j\\\\j \ \ X \ \ i = -k \\\\k \ \  X  \ \ j = -i\\\\i  \ \ X \ \ k = - j\\

Now force is also mathematically represented as

        F = ma

Making a the subject

      a = \frac{F}{m}

   Substituting values

     a =\frac{(-5.966*10^4) \r k}{1.81*10^{-3}}

        = (-0.33m/s^2)\r k

        = 0.33m/s^2 * (- \r k)

6 0
4 years ago
Which of the following units would need to be converted before being used for a calculation
Andreyy89

The units which would need to be converted before being used for a calculation are cm and 0 ks.

<h3>What is Unit?</h3>

This is referred to a standard which is used to make comparisons in the aspect of measurement.

The units cm and 0 ks aren't in their standard form which is m and s respectively.

Read more about Unit here brainly.com/question/4895463

#SPJ1

6 0
2 years ago
A pallet of bricks is to be suspended by attaching a rope to it and connecting the other end to a couple of heavy crates on the
SashulF [63]

Answer:

<h3><u>C</u><u>.</u><u>2</u><u>5</u><u>0</u><u> </u><u>l</u><u>b</u><u> </u><u>i</u><u>s</u><u> </u><u>t</u><u>h</u><u>e</u><u> </u><u>a</u><u>n</u><u>s</u><u>w</u><u>e</u><u>r</u><u>.</u><u>.</u><u>.</u></h3>
5 0
3 years ago
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