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Vaselesa [24]
4 years ago
7

A ball is moving across a level platform 1.6m above the floor. After rolling off the ball hits the floor 20 from the base of the

platform. What is the velocity of the ball as it left the platform? Remember that the platform is level and that the ball is moving horizontally when it leaves the platform.
Physics
2 answers:
irakobra [83]4 years ago
8 0

Answer:

The correct answer is

35.01 m/s

Explanation:

To solve the question, we note the given variables thus

Height of platform S = 1.6 m

Distance from the platform the ball landed = 20 m

S = ut + 0.5gt² therefore 1.6 = 0.5 × 9.81 × t² or t = 0.57 s

Distance = 20 m  = velocity × time

Therefore velocity = Distance / time = 20 m/0.57 s = 35 m/s

sergij07 [2.7K]4 years ago
6 0

Answer:

 

Explanation:

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A boy and a girl are on a spinning merry-go-round. The boy is at a radial distance of 1.2 m from the central axis; the girl is a
Licemer1 [7]

Answer:

E) True. The girl has a larger tangential acceleration than the boy.

Explanation:

In this exercise they do not ask us to say which statement is correct, for this we propose the solution to the problem.

Angular and linear quantities are related

          v = w r

          a = α r

the boy's radius is r₁ = 1.2m the girl's radius is r₂ = 1.8m

as the merry-go-round rotates at a constant angular velocity this is the same for both, but the tangential velocity is different

          v₁ = w 1,2 (boy)

          v₂ = w 1.8 (girl)

whereby

          v₂> v₁

reviewing the claims we have

          a₁ = α 1,2

          a₂ = α 1.8

          a₂> a₁

A) False. Tangential velocity is different from zero

B) False angular acceleration is the same for both

C) False. It is the opposite, according to the previous analysis

D) False. Angular acceleration is equal

E) True. You agree with the analysis above,

8 0
3 years ago
The predictions of Rutherford's scattering formula failed to correspond with experimental data when the energy of the incoming a
Vera_Pavlovna [14]

Answer:

The radius of the gold nucleus is 7.1x10⁻¹⁵m

Explanation:

The nearest distance is:

r=\frac{kze^{2} }{mpV^{2} } (eq. 1)

Where

z = atomic number of gold = 79

e = electron charge = 1.6x10⁻¹⁹C

k = electrostatic constant = 9x10⁹Nm²C²

energy of the particle = 32 MeV = 5.12x10⁻¹²J

At the potential energy is zero, all the energy will be kinetic energy:

E_{k} =\frac{1}{2} m V^{2}

Where

m = 4 mp = mass of proton

5.12x10^{-12} =\frac{1}{2} *4*m_{p}* V^{2} \\m_{p}* V^{2} = 2.56x10^{-12}

Replacing in equation 1

r=\frac{9x10^{9}*79*(1.6x10^{-19})^{2}   }{2.56x10^{-12} } =7.1x10^{-15} m

8 0
3 years ago
A mass on a spring with k=88.7 N/m oscillates 15 times in 9.24s. what is the objects mass? unit=kg?
sweet [91]

The mass on the spring is 0.86 kg

Explanation:

The period of a mass-spring system is given by the equation

T=2\pi \sqrt{\frac{m}{k}}

where

m is the mass

k is the spring constant

In this problem, we have:

k = 88.7 N/m is the spring constant

The system makes 15 oscillations in 9.24 s: therefore, the period of the system is

T=\frac{9.24}{15}=0.62 s

Now we can re-arrange the first equation  to solve for the mass:

m=k(\frac{T}{2\pi})^2=(88.7)(\frac{0.62}{2\pi})^2=0.86 kg

Learn more about period:

brainly.com/question/5438962

#LearnwithBrainly

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Jasmine is late to science class and misses the very beginning of notes for the day. These are Jasmine’s notes: –Round objects t
SpyIntel [72]

Answer:

d

Explanation:

8 0
3 years ago
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gavmur [86]
True! All sounds come from some type of vibrating object. Hopefully I helped!
3 0
3 years ago
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