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Mariulka [41]
2 years ago
8

1.is anyone else bored with school 2. is anyone getting out of school soon or at 3

Physics
1 answer:
storchak [24]2 years ago
3 0

Answer:

i am bored and i get out in 10 minuets

Explanation:

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a water-balloon launcher with mass 4 kg fires a 0.5 kg balloon with a velocity of 3 m/s to the east. what is the recoil velocity
kotykmax [81]
I think we will use the law of conservation of linear momentum;
M1V1 = M2V2
M1 =  4 kg (mass of the water balloon launcher)
V1=?
M2= 0.5 kg ( mass of the balloon)
V2 = 3 m/s

Therefore; 4 V1 = 0.5 × 3
                   4V1= 1.5
                     V1= 1.5/4
                          = 0.375 m/s










5 0
3 years ago
Read 2 more answers
Which of the following concepts helps explain the modern technology used in 3D films?
stealth61 [152]

Answer:

a.

Explanation:

6 0
3 years ago
A 4.0-m-diameter playground merry-go-round, with a moment of inertia of
HACTEHA [7]

Answer:

7.1 ms⁻¹

Explanation:

d = diameter of merry-go-round = 4 m

r = radius of merry-go-round = \frac{d}{2} =  \frac{4}{2} = 2 m

I = moment of inertia = 500 kgm²

w_{i} = angular velocity of merry-go-round before ryan jumps = 2.0 rad/s

w_{f} = angular velocity of merry-go-round after ryan jumps = 0 rad/s

v = velocity of ryan before jumping onto the merry-go-round

m = mass of ryan = 70 kg

Using conservation of angular momentum

Iw_{i} - m v r = (I + mr^{2})w_{f}

(500)(2.0) - (70) v (2) = (I + mr^{2})(0)

1000 = 140 v

v = 7.1 ms⁻¹

5 0
3 years ago
The tires on your truck have 0.35 m radius. In a straight line, you drive 2600 m. What is the angular displacement of the tire,
Travka [436]
1) In a circular motion, the angular displacement \theta is given by
\theta =  \frac{S}{r}
where S is the arc length and r is the radius. The problem says that the truck drove for 2600 m, so this corresponds to the total arc length covered by the tire: S=2600 m. Using the information about the radius, r=0.35 m, we find the total angular displacement:
\theta =  \frac{2600 m}{0.35 m} =7428 rad

2) If we put larger tires, with radius r=0.60 m, the angular displacement will be smaller. We can see this by using the same formula. In fact, this time we have:
\theta =  \frac{2600 m}{0.60 m}=4333 rad
8 0
3 years ago
A projectile is fired from the ground at a velocity of 30.0 m/s, 35.0 º from the horizontal. What is the maximum height the proj
Norma-Jean [14]

Answer:

Vy = V sin theta = 30 * ,574 = 17.2 m/s

t1 = 17.2 / 9.8 = 1.76 sec to reach max height

Max height = 17.2 * 1.76 - 1/2 * 4.9 * 1.76^2 = 15.1 m

H = V t - 1/2 g t^2 = 1.2 * 9.8 * 1.76^2 = 15.1 m

Time to fall from zero speed to ground = rise time = 1.76 sec

Vx = V cos 35 = 24.6 m / sec     horizontal speed

Time in air = 1.76 * 2 = 3.52 sec before returning to ground

S = 24.6 * 3.52 = 86.6 m

8 0
3 years ago
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