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3 years ago

Air has weight. a. True b. False

Physics

2 answers:

Ganezh [65]3 years ago

8 0

Air is a combination of gases, gases do have a mass/weight.

padilas [110]3 years ago

7 0

So basically yes air does have weight

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A 500g cart moving at 0.25 m/s collides and sticks to a stationary 750g cart. How fast do the two carts

Tresset [83]

Answer:

0.1 m/s

Explanation:

Please see attached photo for explanation.

Mass of 1st cart (m₁) = 500 g

Initial velocity of 1st cart (u₁) = 0.25 m/s

Mass of 2nd cart (m₂) = 750 g

Initial velocity of 2nd cart (u₂) = 0 m/s

Velocity (v) after collision =.?

m₁u₁ + m₂u₂ = v(m₁ + m₂)

(500 × 0.25) + (750 × 0) = v(500 + 750)

125 + 0 = v(1250)

125 = 1250v

Divide both side by 1250

v = 125 / 1250

v = 0.1 m/s

Thus, the two cart will move with a velocity of 0.1 m/s after collision.

3 0

3 years ago

A) Julie is running to the right at 5 m/s, as shown in the 1st figure. Balls 1 and 2 are thrown toward her at 10 m/s by friends

Svetach [21]

Part 1)

here Julie is running at speed 5 m/s

So here two balls are thrown at speed 10 m/s towards Julie with respect to her friend standing on the ground.

So here this all speed is real speed of all.

Now as per Anita (let say she is one of her friend standing on ground) the speed of two balls will be same as the given speed as she is observing from ground or stationary frame

As per the frame of Julie

speed of ball 1

$v_{1j} = v_1 - v_j$

$v_{1j} = 10 - 5 = 5 m/s$ towards her in same side

speed of ball 2

$v_{2j} = v_2 - v_j$

$v_{2j} = -10 - 5 = -15 m/s$ towards her from opposite side

Part b)

Now in this case the speed of two balls is given with respect to Julie

so we can say

for ball 1

$v_{1j} = v_1 - v_j$

$10 = v_1 - 5$

so in ground frame speed of ball 1 is

$v_1 = 15 m/s$

Similarly for ball 2

$v_{2j} = v_2 - v_j$

$-10 = v_2 - 5$

So speed in ground frame of ball 2 is

$v_2 = - 5 m/s$

so its 5 m/s from opposite side

6 0

3 years ago

What is center of mass?

yuradex [85]

“a point representing the mean position of the matter in a body or system.”

3 0

3 years ago

Two strings are adjusted to vibrate at exactly 202 Hz. Then the tension in one string isincreased slightly. Afterward, three bea

Semenov [28]

The concepts necessary to solve this problem are framed in the expression of string vibration frequency as well as the expression of the number of beats per second conditioned at two frequencies.

Mathematically, the frequency of the vibration of a string can be expressed as

$f = \frac{1}{2L}\sqrt{\frac{T}{\mu}}$

Where,

L = Vibrating length string

T = Tension in the string

$\mu =$ Linear mass density

At the same time we have the expression for the number of beats described as

$n = |f_1-f_2|$

Where

$f_1$ = First frequency

$f_2$ = Second frequency

From the previously given data we can directly observe that the frequency is directly proportional to the root of the mechanical Tension:

$f \propto \sqrt{T}$

If we analyze carefully we can realize that when there is an increase in the frequency ratio on the tight string it increases. Therefore, the beats will be constituted under two waves; one from the first string and the second as a residue of the tight wave, as well

$n = f_2-f_1$