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ahrayia [7]
3 years ago
9

Difference between weightlessness in space and weightlessness in earth​

Physics
1 answer:
nadya68 [22]3 years ago
8 0
In space we feel weightlessness because the earths gravity has less effect on us.The Earths gravitational attraction at those altitudes is only about 11% less than it is at the Earths surface. If you had a ladder that could reach as high as the the shuttles orbit, your weight would be 11% less at the top.
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Allison wants to calculate the speed of a sound wave.
Fittoniya [83]
D.) Distance travelled / time ...........
5 0
3 years ago
Imagine two billiard balls on a pool table. Ball A has a mass of 7 kilograms and ball B has a mass of 2 kilograms. The initial v
wlad13 [49]
1) In a perfectly inelastic collision, the two balls stick together after the collision. In this type of collision, the total kinetic energy of the system is not conserved, while the total momentum is conserved.
If we callv_f the final velocity of the two balls that stick together, the conservation of the total momentum before and after the collision can be written as
m_a v_{Ai} + m_b v_{Bi} = (m_A+m_B)v_f (1)
where
m_A=7 kg is the mass of ball A
m_B=2 kg is the mass of ball B
v_{Ai}=6 m/s is the initial velocity of ball A
v_{Bi}=-12 m/s is the initial velocity of ball B (taken with a negative sign, since it goes in the opposite direction of ball A)

If we solve (1) to find v_f, we find that the final velocity of the balls is
v_f= \frac{m_Av_{Ai}+m_Bv_{Bi}}{m_A+m_B}= \frac{(7\cdot 6)+(2 \cdot (- 12))}{7+2}= \frac{18}{9}=2 m/s
and the positive sign means the two balls are going to the right.


2) I assume here we are talking about an elastic collision. In this case, both total momentum and total kinetic energy are conserved:
m_A v_{Ai}+m_B v_{Bi} = m_A v_{fA} + m_B v_{fB}
\frac{1}{2}m_A v_{Ai}^2+ \frac{1}{2}m_B v_{Bi}^2= \frac{1}{2}m_Av_{fA}^2+ \frac{1}{2}m_B v_{fB}^2
where
v_{fA} is the final velocity of ball A
v_{fB} is the final velocity of ball B

If we solve simultaneously the two equations, we find:
v_{fA}= \frac{v_{Ai}(m_A-m_B)+2m_Bv_{Bi}}{m_A+m_B} = \frac{(6)(7-2)+2(2)(-12)}{7+2}=-2 m/s
v_{fB}= \frac{v_{Bi}(m_B-m_A)+2m_Av_{Ai}}{m_A+m_B} = \frac{(-12)(2-7)+2(7)(6)}{7+2}= \frac{144}{9}=16 m/s
So, after the collision, ball A moves to the left with velocity v=-2 m/s and ball B moves to the right with velocity v=16 m/s.

3) The total momentum before and after the collision is conserved.
In fact, the total momentum before the collision is:
p_i = m_A v_{A} + m_B v_{fB} = (7\cdot 6)+(2 \cdot (-12))=42-24=18 m/s
and the total momentum after the collision is:
p_f = m_A v_{A} + m_B v_{fB} = (7\cdot (-2))+(2 \cdot 16)=-14+32=18 m/s

3 0
3 years ago
Compressions and rarefactions are characteristic of
Korvikt [17]

Answer:

Of longitudinal waves

Explanation:

Depending on the direction of the oscillation, there are two types of waves:

- Transverse waves: in a transverse wave, the oscillations occur perpendicularly to the direction of propagation of the wave. Examples are electromagnetic waves.

- Longitudinal waves: in a longitudinal wave, the oscillations occur parallel to the direction of propagation of the wave. In such a wave, the oscillations are produced by alternating regions of higher density of particles, called compressions, and regions of lower density of particles, called rarefactions. Examples of longitudinal waves are sound waves.

6 0
3 years ago
Find equation tangent to a circle at given point
vlabodo [156]

the equation of the tangent line must be passed on a point A (a,b) and perpendicular to the radius of the circle. <span>
I will take an example for a clear explanation:
let x² + y² = 4 is the equation of the circle, its center is C(0,0). And we assume that the tangent line passes to the point A(2.3).

</span>since the tangent passes to the A(2,3), the line must be perpendicular to the radius of the circle. 

<span>Let's find the equation of the line parallel to the radius.</span>

<span>The line passes to the A(2,3) and C (0,0). y= ax+b is the standard form of the equation. AC(-2, -3) is a vector parallel to CM(x, y).</span>

det(AC, CM)= -2y +3x =0, is the equation of the line // to the radius.

let's find the equation of the line perpendicular to this previous line.

let M a point which lies on the line. so MA.AC=0 (scalar product), 

it is (2-x, 3-y) . (-2, -3)= -4+4x + -9+3y=4x +3y -13=0 is the equation of tangent



7 0
3 years ago
a swimmer experiences a total (absolute) pressure of 117,500 pa in a pool. how far below the surface are they located?
marta [7]

Answer:

Explanation:

We know that the pressure can be calculated in the following way:

p = d·g·h

with d being the density of the water, g the gravitational acceleration and h the depth.

Also d of the water = 1000 kg/m^3 circa and g = 9.8 m/s^2 circa

117,500 Pa = 1000kg/m³ · 9.8m/s² · h

Therefore h = 11,9 m

4 0
2 years ago
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