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mart [117]
2 years ago
15

What might Earth be like if it had never been hit by the theoretical protoplanet Orpheus?

Physics
1 answer:
Dafna1 [17]2 years ago
7 0

Answer:

If Earth hadn't been hit by Orpheus, it would be covered by ocean, with perhaps a few mountaintops emerging through the water. There would be no humans, but there could be other forms of life. Earth would rotate rapidly, as the moon would not be present to produce the tidal friction that slows Earth's rotation today

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A 60 kilogram student jumps down from a laboratory counter. At the instant he lands on the floor hus speed is 3 meters per secon
erastovalidia [21]

As per Newton's law rate of change in momentum is net force

so we can write it as

F = \frac{dP}{dt}

F = \frac{m(v_f - v_i)}{\Delta t}

now we know that

m = 60 kg

v_f = 3 m/s

v_i = 0

\Delta t= 0.2 s

from above equation

F = \frac{60(3 - 0)}{0.2} = 900 N

so he will experience 900 N force in above case

5 0
3 years ago
The exact speed of an object in a specific instant is?
GarryVolchara [31]
Im not so sure but it should be the
instantaneous speed 

3 0
3 years ago
Read 2 more answers
While in a car at 4.47 meters per second a passenger drops a ball from a height of 0.70 meters above the top of a bucket how far
viktelen [127]

Answer:

1.7 m

Explanation:

v_x = Velocity of ball in x direction = 4.47 m/s

u_y = Velocity of ball in y direction = 0

g = Acceleration due to gravity = 9.81\ \text{m/s}^2

t = Time taken

s_y = Vertical displacement = 0.7 m

s_y=u_yt+\dfrac{1}{2}gt^2\\\Rightarrow 0.7=0+\dfrac{1}{2}\times 9.81t^2\\\Rightarrow t=\sqrt{\dfrac{0.7\times 2}{9.81}}\\\Rightarrow t=0.38\ \text{s}

Horizontal displacement is given by

s_x=v_xt\\\Rightarrow s_x=4.47\times 0.38\\\Rightarrow s_x=1.7\ \text{m}

The passenger should throw the ball 1.7 m in front of the bucket.

5 0
3 years ago
A person suffering from anaemia gets tired after a short walk​
FrozenT [24]

Answer:

yes

Explanation:

you will feel weary after shorter times

5 0
3 years ago
Read 2 more answers
A 35.4 kg girl is riding on the back of a 15.23 kg cart. the cart and the kid are both moving eastward at 4.25 m/s when she step
Grace [21]

Answer:

The final  velocity of the cart is  v_c = 7.02 \  m/s

Explanation:

From the question we are told that

    The mass of the girl is  m_g  = 35.4 \ kg

     The mass of the cart is  m_c  = 15.23 \ kg

      The speed of the cart and  kid(girl) is  v = 4.25 \ m/s

     The final velocity of  the girl is v_g  = 3.06 \  m/s

Let assume that velocity eastward is  positive and velocity westward is negative (Note that if we assume vise versa it wouldn't affect the answer )

   The total momentum of the system before she steps off the back of the cart

is mathematically evaluated as

        p__{T1}} = (m_g + m_c) * v

substituting values

        p__{T1}} = (35.4 + 15.23) * 4.25

        p__{T1}} =215.17 \  kg m /s

The total momentum after she steps off the back of the cart is mathematically evaluated as

        p__{T2}} = (m_g * v_g ) +(  m_c * v_c )

Where  v_c  is the final velocity of the cart

substituting values    

      p__{T2}} = (35.4 * 3.06 ) +(  15.23 * v_c )

       p__{T2}} = 108. 324 + 15.23  v_c

Now according to the law of conservation of momentum

       p__{T1}} =p__{T2}}

So  

       215.17 \  kg m /s =  108. 324 + 15.23  v_c

=>      v_c = 7.02 \  m/s

Since the value is positive it implies that the cart moved eastward

7 0
3 years ago
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