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3 years ago

how your weight would change with time if you were on a space ship traveling away from,earth toward the moon

Physics

1 answer:

Mkey [24]3 years ago

4 0

U wouldn't get enough vitamins

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Given that the collision is elastic and glider 2 is initially at rest (v2,i =0), please use below Eqs. to explain why

Morgarella [4.7K]

Answer:

Explanation:

1 )

Put v2,i =0, in second equation

v2,f= (m2-m1)v2,i + 2m1v1,i/m1+m2

v2,f = 0 + 2m1v1,i/m1+m2

v2,f = 2m1v1,i/m1+m2

In this equation coefficient of v1,i is positive so v2,f and v1,i have the same sign.

2 )

Put m1 < m2 and v2,i =0 in first equation

v1,f= (m1-m2)v1,i + 2m2v2,1/m1+m2

v1,f= (m1-m2)v1,i

As m1-m2 is negative , v1f and v1i will have opposite sign.

3 )

Put m1 > m2 and v2,i =0 in first equation

v1,f= (m1-m2)v1,i + 2m2v2,1/m1+m2

v1,f= (m1-m2)v1,i

m1 - m2 is positive so v1f and v1i will have same sign.

4 )

Put m1 = m2 and v2,i =0 in first equation

v1,f= (m1-m2)v1,i

= 0 because m1 = m2

So glider 1 will stop because v1,f = 0 .

5 0

3 years ago

A car drives 24 meters to the left in three seconds what is the velocity of the car?

sineoko [7]

Answer:

8 m/s to the left.

Explanation:

Applying,

V = d/t...................... Equation 1

Where V = Velocity of the car, d = distance, t = time

From the question,

Given: d = 24 meters, t = 3 seconds

Substitute these values into equation 1

V = 24/3

V = 8 m/s to the left.

Hence the velocity of the car is 8 m/s to the left.

5 0

3 years ago

Two charges, one of 2.50μC and the other of -3.50μC, are placed on the x-axis, one at the origin and the other at x = 0.600 m

aev [14]

Answer:

Explanation:

Given

charge of first body $q_1=2.5\ mu C$

charge of second body $q_2=-3.5\ mu C$

Particle 1 is at origin and particle 2 is at $x=0.6\ m$

third Particle which charge +q must be placed left of $2.5\mu C$ because it will repel the q charge while $-3.5\mu C$ will attract it

suppose it is placed at a distance of x m

$F_{1q}=\frac{kq(2.5)}{x^2}$

$F_{2q}=\frac{kq(-3.5)}{(0.6+x)^2}$

$F_{1q}+F_{2q}=0$

$\frac{kq(2.5)}{x^2}+\frac{kq(-3.5)}{(0.6+x)^2}=0$

$\frac{kq(2.5)}{x^2}=\frac{kq(3.5)}{(0.6+x)^2}$

$\frac{0.6+x}{x}=(\frac{3.5}{2.5})^{0.5}$

$0.6+x=1.1832x$

$x=3.27\ m$

5 0

3 years ago

Find the velocity v=ds/dt at t=3

IRINA_888 [86]

Answer:

you need at least two out of the three to get any aenser

7 0

3 years ago

As one moves farther and farther from the Sun, the distance between adjacent planets is _____.

Alenkinab [10]

As one moves farther and farther from the Sun, the distance between adjacent planets is greater.

4 0

3 years ago