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zzz [600]
2 years ago
5

What is the volume of a box with sides that measures 10 cm , 10 cm, 30 cm

Physics
2 answers:
Soloha48 [4]2 years ago
8 0
Volume of the box=
l×w×h
V=10×10×30
V=100×30
V=3000 cm. As a result, the volume of the box is 3000 cm. Hope it help!
timama [110]2 years ago
4 0
10 X 10 = 100 X 30 = 300
3000cm^3
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it is B

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A 48.0-kg skater is standing at rest in front of a wall. By pushing against the wall she propels herself backward with a velocit
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Answer:

F = 47.6 N

Explanation:

  • Newton's 2nd law can be expressed as the rate of change of the total momentum, respect of time, as follows:

       F = \frac{\Delta p}{\Delta t}

  • So, in order to find the average force exerted by the skater on the wall, we can find the change in momentum due to the force exerted by the wall (which is equal and opposite to the one exerted by the skater), and  divide it by the time interval , as follows:

       F_{wall} = \frac{\Delta p}{\Delta t} =\frac{(48.0 kg*(-1.06m/s)}{1.07s} = -47.6 N

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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

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2 years ago
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Answer:

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Explanation:

7 0
3 years ago
Read 2 more answers
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