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kiruha [24]
3 years ago
12

Please help me answer!! why can't you cry in space? thanks!! ​

Physics
2 answers:
Natalka [10]3 years ago
8 0
Your eyes make tears but they stick as a liquid ball.
Radda [10]3 years ago
6 0

Answer:

we can't cry in space like we do earth.... because there is no gravity... due to which the tears not fall down..instead they make a liquid clump...near eyes

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A 1000-kg car is moving at 30 m/s around a horizontal unbanked curve whose diameter is 0.20 km. What is the magnitude of the fri
omeli [17]

Answer:

4500 N

Explanation:

When a body is moving in a circular motion it will feel an acceleration directed towards the center of the circle, this acceleration is:

a = v^2/r

where v is the velocity of the body and r is the radius of the circumference:

Therefore, a body with mass m, will feel a force f:

f = m v^2/r

Therefore we need another force to keep the body(car) from sliding, this will be given by friction, remember that friction force is given a the normal times a constant of friction mu, that is:

fs = μN = μmg

The car will not slide if     f = fs,   i.e.

fs = μmg =  m v^2/r

That is, the magnitude of the friction force must be (at least) equal to the force due to the centripetal acceleration

fs = (1000 kg)  * (30m/s)^2 / (200 m) = 4500 N

7 0
4 years ago
Read 2 more answers
It takes 20 N of force to move a box a distance of 10 m. How much work is done on the box?
PIT_PIT [208]

Answer:

Ans is 200 J

Explanation:

Given:  Force = 20N

            Distance = 10m

Work done  = Force * displacement

                    =  20 * 10

                    =  200 J

3 0
4 years ago
Which type of force pulls objects toward one another
Klio2033 [76]
Gravity ALWAYS does that, and electrostatic force does it when two objects have opposite charges.
4 0
3 years ago
Read 2 more answers
On a straight road, a car speeds up at a constant rate from rest to 20 m/s over a 5 second interval and a truck slows at a const
IceJOKER [234]

Answer:

a)

Explanation:

  • Since the car speeds up at a constant rate, we can use the kinematic equation for distance (assuming that the initial position is x=0, and choosing t₀ =0), as follows:

        x_{fc} = v_{o}*t + \frac{1}{2}*a*t^{2}   (1)

  • Since the car starts from rest, v₀ =0.
  • We know the value of t = 5 sec., but we need to find the value of a.
  • Applying the definition of acceleration, as the rate of change of velocity with respect to time, and remembering that v₀ = 0 and t₀ =0, we can solve for a, as follows:

       a_{c} =\frac{v_{fc}}{t} = \frac{20m/s}{5s} = 4 m/s2  (2)

  • Replacing a and t in (1):

       x_{fc} = v_{o}*t + \frac{1}{2}*a*t^{2}  = \frac{1}{2}*a*t^{2} = \frac{1}{2}* 4 m/s2*(5s)^{2} = 50.0 m.  (3)

  • Now, if the truck slows down at a constant rate also, we can use (1) again, noting that v₀ is not equal to zero anymore.
  • Since we have the values of vf (it's zero because the truck stops), v₀, and t, we can find the new value of a, as follows:

       a_{t} =\frac{-v_{to}}{t} = \frac{-20m/s}{10s} = -2 m/s2  (4)

  • Replacing v₀, at and t in (1), we have:

       x_{ft} = 20m/s*10.0s + \frac{1}{2}*(-2 m/s2)*(10.0s)^{2} = 200m -100m = 100.0m   (5)

  • Therefore, as the truck travels twice as far as the car, the right answer is a).
7 0
3 years ago
A foot is 1/3 of a yard. What part of a meter is a millimeter?
N76 [4]

Answer:

Explanation:

1/1000

6 0
3 years ago
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