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RUDIKE [14]
1 year ago
11

A green ball has a mass of 0.525 kg and a blue ball has a mass of 0.482 kg. A croquet player strikes the green ball and it gains

an initial velocity of 2.26 m/s. It then strikes the blue ball, which is initially at rest. After the collision, the green ball has a velocity of 1.14 m/s in the same direction. If the balls roll on a frictionless surface and the collision is head-on, what is the final velocity of the blue ball? (Round your answer to the nearest hundredths place.)
Physics
1 answer:
Gelneren [198K]1 year ago
4 0

Answer:

v' = 1.21 m/s

Explanation:

Mass of a green ball, m = 0.525 kg

Mass of a blue ball, m' = 0.482 kg

Initial velocity of green ball, u = 2.26 m/s

Initial velocity of blue ball, u' = 0 (at rest)

After the collision,

The final velocity of the green ball, v = 1.14 m/s

We need to find the final velocity of the blue ball after the collision if the collision is head on. Let v' is the final velcity of the blue ball. Using the conservation of momentum to find it :

mu+m'u'=mv+m'v'\\\\0.525 (2.26)+0=0.525 (1.14)+0.482v'\\\\0.588=0.482v'\\\\v'=\dfrac{0.588}{0.482}\\\\v'=1.21\ m/s

So, the final velocity of the blue ball is 1.21 m/s.

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Answer: F = 1391 N

Explanation:

The information given to you are:

Mass M = 1300 kg

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The magnitude of the force striking the building will be

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Substitute mass M and acceleration a into the formula

F = 1300 × 1.07

F = 1391 N

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A small mass m is tied to a string of length L and is whirled in vertical circular motion. The speed of the mass v is such that
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A 6-in-wide polyamide F-1 flat belt is used to connect a 2-in-diameter pulley to drive a larger pulley with an angular velocity
Likurg_2 [28]

Answer:

a) Fc = 4.15 N, Fi = 435.65 N, (F1)a = 640 N, and F2  = 239.6 N,

b) Ha = 1863.75 N, nfs = 1 , length = 11.8 mm

Explanation:

Given that:

γ= 9.5 kN/m³ = 9500N/m3

b = 6 inches = 0.1524 m

t = 0.0013 mm

d = 2 inches  = 0.0508 m

n = 1750 rpm

H_{nom}=2hp=1491.4W

L = 9 ft = 2.7432 m

Ks = 1.25

g = 9.81 m/s²

a)

w=\gamma b t = 9500* 0.1524*0.0013=1.88N/m

V=\frac{\pi d n}{60} =\pi *0.0508*1750/60=4.65 m/s

F_c=\frac{wV^2}{g}=1.88*4.65^2/9.81=4.15N

(F_1)_a=bF_aC_pC_v=0.1524*6000*0.7*1=640N

T=\frac{H_{nom}n_dK_s}{2\pi n}= \frac{1491*1.25*1}{2*\pi*1750/60}=10.17Nm

F_2=(F_1)_a-\frac{2T}{D}= 640-\frac{2*10.17}{0.0508} =239.6N

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b)

H_a=1491*1.25=1863.75W

n_f_s=\frac{H_a}{H_{nom}K_S }=1

dip = \frac{L^2w}{8F_i} =\frac{2.7432*1.88}{435.65}=11.8mm

7 0
1 year ago
If Emily throws the ball at an angle of 30∘ below the horizontal with a speed of 14m/s, how far from the base of the dorm should
liubo4ka [24]

Emily throws the ball at 30 degree below the horizontal

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