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nikklg [1K]
3 years ago
15

Exits must be separated from the rest of the building by:​

Physics
1 answer:
tigry1 [53]3 years ago
6 0
Fire resistant materials
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What is mean by the net displacement in transverse wave
natta225 [31]
In a transverse wave, the particles are disturbed in a direction perpendicular to the direction of wave propagation. Thus, waves travel through a medium with no net displacement of the distance between two successive particles of wave that are in wavelength. 
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3 years ago
A constant force is exerted for a short time interval on a cart that is initially at rest on an air track. This force gives the
Ilya [14]

Answer:

d) is the same as when it started from rest

Explanation:

using equation of motion

v = u + at

second law of momentum defines

F = ma

a = F /m

the equation becomes

v = u + (F/m)t

from hear

since v is directly proportional to the force and the force remain the same, the increase in the cart speed will also remain the same.

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3 years ago
a street light is mounted at the top of a 15 foot pole. A man 6 ft tall walks away from the pole wit a speed of 7 ft/s along a s
Mariulka [41]

Answer:

16.3 ft/s

Explanation:

Let d=distance

and

x = length of shadow.

Therfore,

x=(d + x)

 = 6/15

So,

    15x = 6x + 6d

     9x = 6d.

x = (2/3)d.

As we know that:

dx=dt

   = (2/3) (d/dt) 

Also,

Given:

d(d)=dt

     = 7 ft/s

Thus,

d(d + x)=dt

           = (7/3)d (d/dt)

Substitute, d= 7  

d(d + x) = 49/3 ft/s.

Hence,

d(d + x) = 16.3 ft/s.

4 0
3 years ago
Decision making worksheet answer key
mario62 [17]

Answer:

  1. <em><u>do </u></em><em><u>your</u></em><em><u> mom</u></em><em><u> hard</u></em><em><u> to</u></em><em><u> get</u></em><em><u> out</u></em><em><u> of</u></em><em><u> the</u></em><em><u> house</u></em><em><u> </u></em><em><u> you</u></em><em><u> doing</u></em><em><u> today</u></em><em><u> no</u></em><em><u> classes</u></em><em><u> are</u></em><em><u> you</u></em><em><u> doing</u></em><em><u> today</u></em><em><u> no</u></em><em><u> classes</u></em><em><u> are</u></em><em><u> you</u></em><em><u> doing</u></em><em><u> today</u></em><em><u> no</u></em><em><u> classes</u></em><em><u> are</u></em><em><u> not</u></em><em><u> sleeping</u></em><em><u> in</u></em><em><u> my</u></em><em><u> bed</u></em><em><u> you</u></em><em><u> have</u></em><em><u> any</u></em><em><u> best</u></em><em><u> to</u></em><em><u> get</u></em><em><u> the</u></em><em><u> meaning</u></em><em><u> do</u></em><em><u> you</u></em><em><u> want</u></em><em><u> for</u></em><em><u> it</u></em><em><u> is</u></em><em><u> my</u></em><em><u> friend</u></em><em><u> I</u></em><em><u> will</u></em><em><u> be</u></em><em><u> there</u></em><em><u> at</u></em><em><u> 12</u></em><em><u> I</u></em><em><u> have</u></em><em><u> send</u></em><em><u> me</u></em><em><u> a</u></em><em><u> </u></em><em><u>even</u></em><em><u> if</u></em><em><u> </u></em><em><u>we</u></em><em><u> </u></em><em><u>don't</u></em><em><u> ask</u></em><em><u> </u></em><em><u>everyone</u></em><em><u> </u></em><em><u>e</u></em><em><u> for</u></em><em><u> it</u></em><em><u> was</u></em><em><u> a</u></em><em><u> screenshot</u></em><em><u> I</u></em><em><u> am</u></em><em><u> </u></em><em><u>excited</u></em><em><u> </u></em><em><u>everything</u></em><em><u> </u></em><em><u>except</u></em><em><u> you</u></em><em><u> </u></em><em><u>eat</u></em><em><u> your</u></em><em><u> </u></em><em><u>eyes</u></em><em><u> </u></em><em><u>everything</u></em><em><u> </u></em><em><u>except</u></em><em><u> </u></em><em><u>everything</u></em><em><u> else</u></em><em><u> </u></em><em><u>ever</u></em><em><u> </u></em><em><u>even</u></em><em><u> </u></em><em><u>know</u></em><em><u> if</u></em><em><u> </u></em><em><u>everyone</u></em><em><u> e</u></em><em><u> </u></em><em><u>and</u></em><em><u> 10</u></em><em><u> </u></em><em><u>even</u></em><em><u> </u></em><em><u>though</u></em><em><u> it</u></em><em><u> </u></em><em><u>was</u></em><em><u> </u></em><em><u>kiss</u></em><em><u> emoji</u></em><em><u> </u></em><em><u>everything</u></em><em><u> </u></em><em><u>else</u></em><em><u> </u></em><em><u>ever</u></em><em><u> </u></em><em><u>eyebrows</u></em><em><u> </u></em><em><u>yet</u></em><em><u> I</u></em><em><u> </u></em><em><u>eat</u></em><em><u> </u></em><em><u>early</u></em><em><u> </u></em><em><u>to</u></em><em><u> get</u></em><em><u> </u></em><em><u>enough</u></em><em><u> </u></em><em><u>everything</u></em><em><u> </u></em><em><u>except</u></em><em><u> </u></em><em><u>everything</u></em><em><u> is</u></em><em><u> your</u></em><em><u> class</u></em><em><u> finish</u></em><em><u> up</u></em><em><u> </u></em><em><u>early</u></em><em><u> </u></em><em><u>enough</u></em><em><u> </u></em><em><u>even</u></em><em><u> know</u></em><em><u> what</u></em><em><u> is</u></em><em><u> going</u></em><em><u> on</u></em><em><u> with</u></em><em><u> out</u></em><em><u> hajab</u></em><em><u> I</u></em><em><u> have</u></em><em><u> send</u></em><em><u> it</u></em><em><u> out</u></em><em><u> to</u></em><em><u> </u></em><em><u>edit</u></em><em><u> a</u></em><em><u> video</u></em><em><u> on</u></em><em><u> </u></em><em><u>e</u></em><em><u> a</u></em><em><u> video</u></em><em><u> on</u></em><em><u> </u></em><em><u>each</u></em><em><u> </u></em><em><u>e</u></em>

Explanation:

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7 0
3 years ago
A space probe is fired as a projectile from the Earth's surface with an initial speed of 2.05 104 m/s. What will its speed be wh
Elanso [62]

Answer:

The value is  v  =  2.3359 *10^{4} \ m/s

Explanation:

From the question we are told that

  The  initial speed is u =  2.05 *10^{4} \  m/s

 Generally the total energy possessed by the space probe when on earth is mathematically represented as

             T__{E}} =  KE__{i}} +  KE__{e}}

Here  KE_i is the kinetic energy of the space probe due to its initial speed which is mathematically represented as

          KE_i =   \frac{1}{2}  *  m  *  u^2

=>       KE_i =   \frac{1}{2}  *  m  *  (2.05 *10^{4})^2

=>       KE_i =  2.101 *10^{8} \ \ m \ \ J

And  KE_e is the kinetic energy that the space probe requires to escape the Earth's gravitational pull , this is mathematically represented as

       KE_e =  \frac{1}{2}  *  m *  v_e^2

Here v_e is the escape velocity from earth which has a value v_e =  11.2 *10^{3} \  m/s

=>    KE_e =  \frac{1}{2}  *  m *  (11.3 *10^{3})^2

=>    KE_e =  6.272 *10^{7} \  \  m  \ \   J

Generally given that at a position that is very far from the earth that the is Zero, the kinetic energy at that position is mathematically represented as

        KE_p =  \frac{1}{2}  *  m *  v^2

Generally from the law energy conservation we have that

        T__{E}} =  KE_p

So

       2.101 *10^{8}  m  +  6.272 *10^{7}  m  =   \frac{1}{2}  *  m *  v^2

=>     5.4564 *10^{8} =   v^2

=>     v =  \sqrt{5.4564 *10^{8}}

=>     v  =  2.3359 *10^{4} \ m/s

4 0
2 years ago
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