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suter [353]
3 years ago
7

An athlete rotates a 1.00-kg discus along a circular path of radius 1.09 m. The maximum speed of the discus is 17.0 m/s. Determi

ne the magnitude of the maximum radial acceleration of the discus.
Physics
1 answer:
kaheart [24]3 years ago
8 0

Answer:

a= 289 m/s²  

Explanation:

Given that

mass , m =1 kg

Radius of the path ,R= 1.09 m

The maximum speed ,V= 17 m/s

Lets take the maximum acceleration = a

The radial acceleration is given as

a= \dfrac{V^2}{R}

V=Velocity

R=Radius of the circular path

Now by putting the values in the above equation we get

a= \dfrac{17^2}{1}\ m/s^2

a= 289 m/s²

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(This is a non-relativistic warm-up problem, to get you to think about reference frames.) A girl throws a baseball upwards at ti
____ [38]

Answer:

X(t) = 9.8 *t - 4.9 * t^2

Explanation:

We set a frame of reference with origin at the hand of the girl the moment she releases the ball. We assume her hand will be in the same position when she catches it again. The positive X axis point upwards.The ball will be subject to a constant gravitational acceleration of -9.81 m/s^2.

We use the equation for position under constant acceleration:

X(t) = X0 + V0 * t + 1/2 * a *t^2

X0 = 0 because it is at the origin of the coordinate system.

We know that at t = 2, the position will be zero.

X(2) = 0 = V0 * 2 + 1/2 * -9.81 * 2^2

0 = 2 * V0 - 4.9 * 4

2 * V0 = 19.6

V0 = 9.8 m/s

Then the position of the ball as a function of time is:

X(t) = 9.8 *t - 4.9 * t^2

5 0
2 years ago
At take off a plane flies 100 km north before turning east to fly 200 km east. How far is its destination from where the plane t
andrew-mc [135]

First the plane turns 100 km North, and than 200 km East. Since both the directions are perpendicular to each other, therefore we can apply the Pythagoras theorem to calculate the distance between the destination and the point where plane took off


=100^{2}+200^{2}

D=223.60 km=224 km

Therefore, The destination is 224 km from where the plane took off


8 0
3 years ago
A 1 kg mass is attached to a spring with spring constant 7 Nt/m. What is the frequency of the simple harmonic motion? What is th
Scorpion4ik [409]

1. 0.42 Hz

The frequency of a simple harmonic motion for a spring is given by:

f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}

where

k = 7 N/m is the spring constant

m = 1 kg is the mass attached to the spring

Substituting these numbers into the formula, we find

f=\frac{1}{2\pi}\sqrt{\frac{7 N/m}{1 kg}}=0.42 Hz

2. 2.38 s

The period of the harmonic motion is equal to the reciprocal of the frequency:

T=\frac{1}{f}

where f = 0.42 Hz is the frequency. Substituting into the formula, we find

T=\frac{1}{0.42 Hz}=2.38 s

3. 0.4 m

The amplitude in a simple harmonic motion corresponds to the maximum displacement of the mass-spring system. In this case, the mass is initially displaced by 0.4 m: this means that during its oscillation later, the displacement cannot be larger than this value (otherwise energy conservation would be violated). Therefore, this represents the maximum displacement of the mass-spring system, so it corresponds to the amplitude.

4. 0.19 m

We can solve this part of the problem by using the law of conservation of energy. In fact:

- When the mass is released from equilibrium position, the compression/stretching of the spring is zero: x=0, so the elastic potential energy is zero, and all the mechanical energy of the system is just equal to the kinetic energy of the mass:

E=K=\frac{1}{2}mv^2

where m = 1 kg and v = 0.5 m/s is the initial velocity of the mass

- When the spring reaches the maximum compression/stretching (x=A=amplitude), the velocity of the system is zero, so the kinetic energy is zero, and all the mechanical energy is just elastic potential energy:

E=U=\frac{1}{2}kA^2

Since the total energy must be conserved, we have:

\frac{1}{2}mv^2 = \frac{1}{2}kA^2\\A=\sqrt{\frac{m}{k}}v=\sqrt{\frac{1 kg}{7 N/m}}(0.5 m/s)=0.19 m

5. Amplitude of the motion: 0.44 m

We can use again the law of conservation of energy.

- E_i = \frac{1}{2}kx_0^2 + \frac{1}{2}mv_0^2 is the initial mechanical energy of the system, with x_0=0.4 m being the initial displacement of the mass and v_0=0.5 m/s being the initial velocity

- E_f = \frac{1}{2}kA^2 is the mechanical energy of the system when x=A (maximum displacement)

Equalizing the two expressions, we can solve to find A, the amplitude:

\frac{1}{2}kx_0^2 + \frac{1}{2}mv_0^2=\frac{1}{2}kA^2\\A=\sqrt{x_0^2+\frac{m}{k}v_0^2}=\sqrt{(0.4 m)^2+\frac{1 kg}{7 N/m}(0.5 m/s)^2}=0.44 m

6. Maximum velocity: 1.17 m/s

We can use again the law of conservation of energy.

- E_i = \frac{1}{2}kx_0^2 + \frac{1}{2}mv_0^2 is the initial mechanical energy of the system, with x_0=0.4 m being the initial displacement of the mass and v_0=0.5 m/s being the initial velocity

- E_f = \frac{1}{2}mv_{max}^2 is the mechanical energy of the system when x=0, which is when the system has maximum velocity, v_{max}

Equalizing the two expressions, we can solve to find v_{max}, the maximum velocity:

\frac{1}{2}kx_0^2 + \frac{1}{2}mv_0^2=\frac{1}{2}mv_{max}^2\\v_{max}=\sqrt{\frac{k}{m}x_0^2+v_0^2}=\sqrt{\frac{7 N/m}{1 kg}(0.4 m)^2+(0.5 m/s)^2}=1.17 m/s m

4 0
3 years ago
Read 2 more answers
How many electrons are in 204 C of charge?
zubka84 [21]

Answer:

The mass number 204 – 82 protons = 122 neutrons

Explanation:

Hope this helps!

5 0
3 years ago
1.In one challenge on the Titan Games, competitors have to lift 200 pounds up a long ramp. Angel is able to move the weight in 4
77julia77 [94]

#1).  Anthony does the same amount of work as Angel, with <em>more power</em>.

#2). Power = (Work)/(Time) = 41,000 J / 500 s  =  <em>82 watts .</em>

#3). Power = (Work) / (Time) = 83 J / 3 sec = <em>27.7 watts</em>

5 0
3 years ago
Read 2 more answers
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