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frez [133]
3 years ago
13

Camels can run faster than horses in desert.Why

Physics
2 answers:
kobusy [5.1K]3 years ago
8 0
Because camels have thick and wide feet so they don't fall in the desert, also protect them from the heat of the desert
grigory [225]3 years ago
4 0

Most camels can outrun most horses, but the fastest racehorse would probably outrun the fastest camel (it would be a close race). But camels can carry more weight and can move at average speeds of about 12 kilometer per hour for as much as 18 hours.

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Im pregnaunt at the age 12 what shall i do now ??<br><br> i do live my belly bump&lt;3
Alex

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theoretically speaking I don't even wanna believe it's possible but if it does then then you should check for abortion

8 0
2 years ago
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Find the mass of a flying discus that has a net force of 1.05 newtons and accelerates at 3.5 m/s^2
Ilya [14]
F=ma
m=1.05/3.5= 0.3kg
8 0
3 years ago
two pulse waves a and b arrive at a point in a medium simustaneously. if the amplitude of wave a is 35 units the amplitude of wa
Shalnov [3]
It depends on the type of interference.
For constructive interference, add the amplitudes to get |35 + 41| = 76 units.
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3 0
3 years ago
A plant blossoms with violet-colored flowers. The flowers appear violet because they absorb all light rays except for____ rays.
jolli1 [7]
~Hello there! ^_^

Your question: A plant blossoms with violet-colored flowers. The flowers appear violet because they absorb all light rays except for____ rays.

Your answer: A plant blossoms with violet-colored flowers. The flowers appear violet because they absorb all light rays except for violet rays.


Hope this helps~

5 0
2 years ago
Read 2 more answers
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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