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Temka [501]
4 years ago
7

To throw the discus, the thrower holds it with a fully outstretched arm. Starting from rest, he begins to turn with a constant a

ngular acceleration, releasing the discus after making one complete revolution. The diameter of the circle in which the discus moves is about 1.6 mm.
If the thrower takes 1.0 s to complete one revolution, starting from rest, what will be the speed of the discus at release?
Physics
1 answer:
yawa3891 [41]4 years ago
6 0

Answer:

Explanation:

Time taken to complete one revolution is called time period.

So, Time period, T = 1 s

Diameter = 1.6 mm

radius, r = 0.8 mm

Let the angular speed is ω.

The relation between angular velocity and the time period is

\omega =\frac{2\pi}{T}

ω = 2 x 3.14 = 6.28 rad/s

The relation between the linear velocity and the angular velocity is

v = r x ω

v = 0.8 x 10^-3 x 6.28

v = 0.005 m/s

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Answer:

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Explanation:

An object that is at rest stays at rest or stays in motion.

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3 years ago
The local church is hosting a carnival which includes a bumper car ride. Bumper car A and its driver have a mass of 300 kg; bump
lukranit [14]

Answer:

a. 20 s

b. 0 m/s  

c. right

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4 years ago
7.22 Ignoring reflection at the air–water boundary, if the amplitude of a 1 GHz incident wave in air is 20 V/m at the water surf
Serga [27]

Answer:

z = 0.8 (approx)

Explanation:

given,

Amplitude of 1 GHz incident wave in air = 20 V/m

Water has,

μr = 1

at 1 GHz, r = 80 and σ = 1 S/m.

depth of water when amplitude is down to  1 μV/m

Intrinsic impedance of air = 120 π  Ω

Intrinsic impedance of  water = \dfrac{120\pi}{\epsilon_r}

Using equation to solve the problem

  E(z) = E_0 e^{-\alpha\ z}

E(z) is the amplitude under water at z depth

E_o is the amplitude of wave on the surface of water

z is the depth under water

\alpha = \dfrac{\sigma}{2}\sqrt{\dfrac{(120\pi)^2}{\Epsilon_r}}

\alpha = \dfrac{1}{2}\sqrt{\dfrac{(120\pi)^2}{80}}

\alpha =21.07\ Np/m

now ,

  1 \times 10^{-6} = 20 e^{-21.07\times z}

  e^{21.07\times z}= 20\times 10^{6}

taking ln both side

21.07 x z = 16.81

z = 0.797

z = 0.8 (approx)

5 0
4 years ago
A student pushes a 2.85 kg cart causing it to accelerate at a rate of 4.9 m/s squared .What amount of force must the student hav
uysha [10]
In order to find the force (F), you would have to use the formula for it:
F=ma
where m is mass and a is acceleration.
In the problem, the mass is 2.85kg and the acceleration is 4.9m/s^2.
Therefore,
F=2.85kg(4.9m/s^2)
F=13.965kg(m/s^2)
Since N=kg(m/s^2)
F=13.965N
And because the problem requires that we use only 2 significant figures,
F=13N
Therefore, the student must exert 13N of force.

7 0
3 years ago
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