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STatiana [176]
2 years ago
11

balls a, with a mass of 20 kg, is moving to the right at 20 m/s. At what velocity should Ball B, with a mass of 40 kg, move so t

hat they both come to standstill upon collision?
Physics
1 answer:
Leya [2.2K]2 years ago
3 0
In that case, their momentum must be equal. 
So, m1v1 = m2v2
20 * 20 = 40 * v2
v2 = 400 / 40
v2 = 10

In short, Your Answer would be: 10 m/s

Hope this helps!
You might be interested in
If a 10. m3 volume of air (acting as an ideal gas) is at a pressure of 760 mm and a temperature of 27 degrees Celsius is taken t
kow [346]
We know, the ideal gas equation, 
P1V1 / T1 = P2V2 / T2

Here, P1 = 760 mm
V1 = 10 m3
T1 = 27 + 273 = 300 K

P2 = 400 mm Hg
T2 = -23 + 273 = 250 K

Substitute their values, 
760*10 / 300 = 400 * V2 / 250
25.33 * 250 = 400 * V2
V2 = 6333.333/ 400
V2 = 15.83

In short, Your Answer would be approx. 15.83 m3

Hope this helps!
7 0
3 years ago
Learning Task 2: Write the words that can be associated with the ''Music During Classical Era''. You may ask help from the membe
Natasha_Volkova [10]

Answer:

classical

concert

exposition

orchesta

devolopment

recapitulation

sympathy

sonata

sinfonia

soloist

Explanation:

5 0
3 years ago
A car turns a certain curve of radius 24.98 m with constant linear speed of
Anastaziya [24]

Answer:

3525.19 kg

Explanation:

The computation of the mass of the car is shown below:

As we know that

Fc = m × V^2 ÷ R

m = Fc × R ÷ V^2

Provided that:

Fc = 34.652 kN = 34652 N

R = Radius = 24.98 m

V = speed = 15.67 m/s

So,

m = 34652 × 24.98 ÷ 15.67^2

 = 3525.19 kg

7 0
3 years ago
A ball is projected with an initial velocity of 40 meter per second and reached maximum height of 160 meters calculate tge angle
Andru [333]

There's a problem with the question as given. Even with a maximum projection angle of <em>θ</em> = 90°, the initial velocity is not large enough to get the ball up in the air 160 m. With angle 90°, the ball's height <em>y</em> at time <em>t</em> would be

<em>y</em> = (40 m/s) <em>t</em> - 1/2 <em>g t</em> ²

Set <em>y</em> = 160 m, and you'll find that there is no (real) solution for<em> t</em>, so the ball never attains the given maximum height.

From another perspective: recall that

<em>v </em>² - <em>v</em>₀² = 2<em>a </em>∆<em>y</em>

where

• <em>v</em>₀ = initial velocity

• <em>v</em> = final velocity

• <em>a</em> = acceleration

• ∆<em>y</em> = displacement

At its maximum height, the ball has zero vertical velocity, and ∆<em>y</em> = maximum height = 160 m. The ball is in free fall once it's launched, so <em>a</em> = -<em>g</em>.

So we have

0² - (40 m/s)² = -2<em>g </em>(160 m)

but this reduces to

(40 m/s)² = 2 (9.8 m/s²) (160 m)

1600 m²/s² ≠ 3136 m²/s²

7 0
3 years ago
When lamps with wattages greater than the rating of the luminaire are installed, a fire could occur because the luminaire is bei
sergejj [24]

Answer:

true

Explanation:

Yes, it is true.

As the wattage is more than the prescribed wattage, it becomes overheated.

6 0
3 years ago
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