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sp2606 [1]
3 years ago
12

A basketball is tossed up into the air, falls freely, and bounces from the wooden floor. from the moment after the player releas

es it until the ball reaches the top of its bounce, what is the smallest system for which momentum is conserved? the ball plus player the ball momentum is not conserved the ball plus earth
Physics
1 answer:
Lelu [443]3 years ago
7 0
I would say the smallest system for which the momentum will be preserved will be the ball plus the earth though in this case it would be the wooden floor since the last thing it does is bounce from the floor up in the air so that is the last system,
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Which is heavier , 1kg of steel or i kg of cork​
goblinko [34]

Answer:

Same weight so neither is heavier

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3 years ago
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Flies could be prey to predator plants, too what plant snacks on this fly?
sertanlavr [38]

Answer:

A Venus Fly trap

Explanation:

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3 years ago
In the diagram, q1 = +6.60*10^-9 C and q2 = +3.10*10^-9 C. Find the magnitude of the total electric field at point P. pls help?
kirza4 [7]

Answer:

|E(t)| = 1258,46 [N/C]

α = 67,5⁰  (angle with respect x-axis)

Explanation:

E(t)  Electric Field is a vector, so we need to determine module and direction

E(t)  =  E(q₁)  + E (q₂)  Where E(q₁) and E (q₂) are electric fields due to electric charge q₁ and q₂  respectively.

E(q₁) = K * q₁/ (d₁)²         K = 9 *10⁹   [N*m²/C²]    d₁ = 0,350 m

E(q₁) = 9 *10⁹ * 6,6*10⁻⁹ / 0,1225      [N*m²/C²] *C/m²

E(q₁) = 484,9 [N/C]

E(q₂) =  9 *10⁹ * 3,1*10⁻⁹ / 0,024025

E(q₂) = 1161,29

Then

|E(t)| = √ |Eq₁|² + |Eq₂|²

|E(t)| = √ ( 484,9)² +( 1161,29)²

|E(t)| = √ 235128 + 1348594,46

|E(t)| = 1258,46 [N/C]

And tanα = 1161,29/484,9        tanα =  2,395      α = 67,5⁰

The angle of the vector electric field with the x-axis

3 0
3 years ago
Read 2 more answers
5.
igor_vitrenko [27]

Answer:

a=40\ m/s^2

Explanation:

Given that,

Initial speed of a shuttlecock, u = 30 m/s

Final speed of the shuttlecock, v = 10 m/s

Time, t = 0.5 s

We need to find its average acceleration. The acceleration of an object is equal to the change in speed divided by time taken. It is given by :

a=\dfrac{v-u}{t}\\\\a=\dfrac{10-30}{0.5}\\\\a=-40\ m/s^2

So, the average acceleration of badminton shuttlecock is 40\ m/s^2.

3 0
3 years ago
A car’s 30.0-kg front tire is suspended by a spring with spring constant k=1.00x10^5 N/m. At what speed is the car moving if was
Taya2010 [7]

we know the equation for the period of oscillation in SHM is as follows:

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we know f = 1/T, so f = 1/(2 * pi) * sqrt(k/m).

since d = v*T, we can say v = d/t = d * f

the final equation, after combining everything, is as follows:

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by plugging everything in

v = .75/(2 * pi) * sqrt((1 * 10^5)/(30))

We find our velocity to be:

v = 6.89 m/s

6 0
3 years ago
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