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Elis [28]
3 years ago
9

What is someone who leaves there country to settle in another?

Physics
2 answers:
Eddi Din [679]3 years ago
8 0

Answer:

an immigrant

Explanation:

gulaghasi [49]3 years ago
5 0

Answer:

it is an immigrant

Explanation:

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Travel north to south how would polaris be situated?
valentina_108 [34]

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idi

Explanation:

3 0
4 years ago
A cyclist is travelling at 4.0 m/s. She speeds up to 16 m/s in a time of 5.6 s. Calculate her acceleration.​
adell [148]

Answer:

Initial velocity, u=15m/s

Initial velocity, u=15m/sFinal velocity, v=0m/s

Initial velocity, u=15m/sFinal velocity, v=0m/sDistance, s=18m

Initial velocity, u=15m/sFinal velocity, v=0m/sDistance, s=18mAcceleration, a=?

Initial velocity, u=15m/sFinal velocity, v=0m/sDistance, s=18mAcceleration, a=?Using relation, v

Initial velocity, u=15m/sFinal velocity, v=0m/sDistance, s=18mAcceleration, a=?Using relation, v 2

Initial velocity, u=15m/sFinal velocity, v=0m/sDistance, s=18mAcceleration, a=?Using relation, v 2 −u

Initial velocity, u=15m/sFinal velocity, v=0m/sDistance, s=18mAcceleration, a=?Using relation, v 2 −u 2

Initial velocity, u=15m/sFinal velocity, v=0m/sDistance, s=18mAcceleration, a=?Using relation, v 2 −u 2 =2as

Initial velocity, u=15m/sFinal velocity, v=0m/sDistance, s=18mAcceleration, a=?Using relation, v 2 −u 2 =2as0

Initial velocity, u=15m/sFinal velocity, v=0m/sDistance, s=18mAcceleration, a=?Using relation, v 2 −u 2 =2as0 2

Initial velocity, u=15m/sFinal velocity, v=0m/sDistance, s=18mAcceleration, a=?Using relation, v 2 −u 2 =2as0 2 −(15)

Initial velocity, u=15m/sFinal velocity, v=0m/sDistance, s=18mAcceleration, a=?Using relation, v 2 −u 2 =2as0 2 −(15) 2

Initial velocity, u=15m/sFinal velocity, v=0m/sDistance, s=18mAcceleration, a=?Using relation, v 2 −u 2 =2as0 2 −(15) 2 =2a×18

Initial velocity, u=15m/sFinal velocity, v=0m/sDistance, s=18mAcceleration, a=?Using relation, v 2 −u 2 =2as0 2 −(15) 2 =2a×18−225=36a

Initial velocity, u=15m/sFinal velocity, v=0m/sDistance, s=18mAcceleration, a=?Using relation, v 2 −u 2 =2as0 2 −(15) 2 =2a×18−225=36aa=

Initial velocity, u=15m/sFinal velocity, v=0m/sDistance, s=18mAcceleration, a=?Using relation, v 2 −u 2 =2as0 2 −(15) 2 =2a×18−225=36aa= 36

Initial velocity, u=15m/sFinal velocity, v=0m/sDistance, s=18mAcceleration, a=?Using relation, v 2 −u 2 =2as0 2 −(15) 2 =2a×18−225=36aa= 36−225

Initial velocity, u=15m/sFinal velocity, v=0m/sDistance, s=18mAcceleration, a=?Using relation, v 2 −u 2 =2as0 2 −(15) 2 =2a×18−225=36aa= 36−225

Initial velocity, u=15m/sFinal velocity, v=0m/sDistance, s=18mAcceleration, a=?Using relation, v 2 −u 2 =2as0 2 −(15) 2 =2a×18−225=36aa= 36−225 =−6.25m/s

Initial velocity, u=15m/sFinal velocity, v=0m/sDistance, s=18mAcceleration, a=?Using relation, v 2 −u 2 =2as0 2 −(15) 2 =2a×18−225=36aa= 36−225 =−6.25m/s 2

Initial velocity, u=15m/sFinal velocity, v=0m/sDistance, s=18mAcceleration, a=?Using relation, v 2 −u 2 =2as0 2 −(15) 2 =2a×18−225=36aa= 36−225 =−6.25m/s 2

Initial velocity, u=15m/sFinal velocity, v=0m/sDistance, s=18mAcceleration, a=?Using relation, v 2 −u 2 =2as0 2 −(15) 2 =2a×18−225=36aa= 36−225 =−6.25m/s 2 So, deceleration is 6.25m/s

Initial velocity, u=15m/sFinal velocity, v=0m/sDistance, s=18mAcceleration, a=?Using relation, v 2 −u 2 =2as0 2 −(15) 2 =2a×18−225=36aa= 36−225 =−6.25m/s 2 So, deceleration is 6.25m/s 2

Initial velocity, u=15m/sFinal velocity, v=0m/sDistance, s=18mAcceleration, a=?Using relation, v 2 −u 2 =2as0 2 −(15) 2 =2a×18−225=36aa= 36−225 =−6.25m/s 2 So, deceleration is 6.25m/s 2 .

5 0
2 years ago
Read 2 more answers
The bending of light as it passes into a transparent material of different optical intensity is known as
defon
The bending of light as it passes into a transparent material of different optical intensity is known as refraction. The correct option among all the options given in the question is option "C".It can never be reflection as reflection is the bouncing of the light from a surface and so option “A” can easily be negated. The last option or option “D”<span> conversion and it mean converting something to another and that means a total change of form. So the last option should also be negated. The second option can also be negated.</span>
5 0
3 years ago
If a tube is placed vertically into a container of water, and the water level inside the tube is the same as the water level on
dybincka [34]
It's equal to d external atmospheric pressure
6 0
3 years ago
Two horizontal rods are each held up by vertical strings tied to their ends. Rod 1 has length L and mass M; rod 2 has length 2L
antiseptic1488 [7]

Answer:

Rod 1 has greater initial angular acceleration; The initial angular acceleration for rod 1 is greater than for rod 2.

Explanation:

For the rod 1 the angular acceleration is

\tau_1 = I_1\alpha _1 \\\\\alpha_1 = \dfrac{\tau_1}{I_1}

Similarly, for rod 2

\alpha_2 = \dfrac{\tau_2}{I_2}.

Now, the moment of inertia for rod 1 is

I_1 = \dfrac{1}{3}ML^2,

and the torque acting on it is (about the center of mass)

\tau_1 = Mg\dfrac{L}{2};

therefore, the angular acceleration of rod 1 is  

\alpha_1 = \dfrac{Mg\dfrac{L}{2}}{\dfrac{1}{3}ML^2},

\boxed{\alpha_1 = \dfrac{3g}{2L} }

Now, for rod 2 the moment of inertia is

I_2 = \dfrac{1}{3}(2M)(2L)^2

I_2 = \dfrac{8}{3} ML^2,

and the torque acting is (about the center of mass)

\tau _2 = (2M)g \dfrac{(2L)}{2}

\tau _2 = 2MgL;

therefore, the angular acceleration \alpha_2 is

\alpha_2 = \dfrac{2MgL;}{\dfrac{8}{3} ML^2,}.

\boxed{\alpha_2 = \dfrac{3g}{4L}}

We see here that

\dfrac{3g}{2L} > \dfrac{3g}{4L}

therefore

\boxed{\alpha_1 > \alpha_2.}

In other words , the initial angular acceleration for rod 1 is greater than for rod 2.

7 0
3 years ago
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