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serg [7]
2 years ago
6

A cannonball and a marble roll smoothly from rest down an incline. Is the cannonball’s(a) time to the bottom and(b) translationa

l kinetic energy at the bottom more than, less than, or the same as the marble’s?
Physics
1 answer:
Salsk061 [2.6K]2 years ago
5 0

Answer:

Explained

Explanation:

a) Cannonball and marble will reach the bottom at the same time, as acceleration due to gravity is independent of the mass of the body.

b)Translational Kinetic energy of the cannonball will be more than that of the marble as the cannonball has more mass than the marble.

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A giant wall clock with diameter d rests vertically on the floor. The minute hand sticks out from the face of the clock, and its
Katyanochek1 [597]

Answer:

d_{x}(t)=(D/2)cos(\frac{\pi}{30}*t)

Explanation:

We can try writing the equation of the horizontal component of the length of the minute hand in terms of distance and the angle, that depends of time in this particular case.

The x-component of the length of the minute hand is:

d_{x}(t)=dcos(\theta (t)) (1)

  • d is the length of the minute hand (d=D/2)
  • D is the diameter of the clock
  • t is the time (min)

Now, using the angular kinematic equations we can express the angle in term of angular velocity and time. As we know, the minute hand moves with a constant angular velocity, so we can use this equation:

\theta (t)=\omega *t (2)

Also we know, that the minute hand moves 90 degrees or π/2 rad in 15 min, so using the definition of angular velocity, we have:

\omega=\frac{\Delta \theta}{\Delta t}=\frac{\theta_{f}-\theta_{i}}{t_{f}-t{i}}=\frac{\pi/2-0}{15-0}=\frac{\pi}{30}

Now, let's put this value on (2)

\theta (t)=\frac{\pi}{30}*t

Finally the length x(t) of the shadow of the minute hand as a function of time t, will be:

d_{x}(t)=(D/2)cos(\frac{\pi}{30}*t)

I hope it helps you!

6 0
2 years ago
The amount of kinetic energy an object has depends on its mass and its speed. Rank the following sets of oranges and cantaloupes
Elodia [21]

Explanation:

Below is an attachment containing the solution.

7 0
2 years ago
An electron moving at right angles to a 0.1 T magnetic field experiences an acceleration of 6 × 1015 m.s-2. What is the speed of
GaryK [48]

Explanation:

It is given that,

Magnetic field, B = 0.1 T

Acceleration, a=6\times 10^{15}\ m/s^2

Charge on electron, q=1.6\times 10^{-19}\ C    

Mass of electron, m=9.1\times 10^{-31}\ kg    

(a) The force acting on the electron when it is accelerated is, F = ma

The force acting on the electron when it is in magnetic field, F=qvB\ sin\theta

Here, \theta=90

So, ma=qvB

Where

v is the velocity of the electron

B is the magnetic field

v=\dfrac{ma}{qB}

v=\dfrac{9.1\times 10^{-31}\ kg\times 6\times 10^{15}\ m/s^2}{1.6\times 10^{-19}\ C\times 0.1\ T}

v = 341250  m/s

or

v=3.41\times 10^5\ m/s

So, the speed of the electron is 3.41\times 10^5\ m/s

(b) In 1 ns, the speed of the electron remains the same as the force is perpendicular to the cross product of velocity and the magnetic field.

7 0
2 years ago
The 1.5-in.-diameter shaft AB is made of a grade of steel with a 42-ksi tensile yield stress. Using the maximum-shearing-stress
PolarNik [594]

Answer:

T = 0.03 Nm.

Explanation:

d = 1.5 in = 0.04 m

r = d/2 = 0.02 m

P = 56 kips = 56 x 6.89 = 386.11 MPa

σ = 42-ksi = 42 x 6.89 = 289.58 MPa

Torque = T =?

<u>Solution:</u>

σ = (P x r) / T

T = (P x r) / σ

T = (386.11 x 0.02) / 289.58

T = 0.03 Nm.

7 0
2 years ago
Scientists and astronomers have found that in galaxies with central black holes, there are also large star formations near those
sweet-ann [11.9K]

Answer:

B

Explanation:

nothing to do with black holes creating star or related

5 0
2 years ago
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