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Verdich [7]
2 years ago
6

A 1.0 kg object is attached to a string 0.50 m. It is twirled in a horizontal circle above the ground at a speed of 5.0 m/s. A b

all is moving on a circular path. The ball is currently at the top right position of the circle. Point a is above and to the left of the ball. Point b is above and to the right of the ball. Point c is down and to the right of the ball. Point d is down and to the left of the ball and inside the circle near the center. What is the magnitude of the centripetal force acting on the object? 2.5 N 10. N 25 N 50. N
Physics
1 answer:
aivan3 [116]2 years ago
6 0
<span>50 N The centripetal force upon an object is expressed as F = mv^2/r So let's substitute the known values and calculate F = mv^2/r F = 1.0 kg * (5.0 m/s)^2 / 0.5 m F = 1.0 kg * 25 m^2/s^2 / 0.5 m F = 25 kg*m^2/s^2 / 0.5 m F = 50 kg*m/s^2 F = 50 N So the answer is 50 N which matches one of the available choices.</span>
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The correct option is B.
Nuclear fission and fusion are two different types of nuclear reactions, through which energy may be obtained. Nuclear fission involves the splitting of a molecule into two different part in order to generate energy while nuclear fusion reaction involves the joining together of two elements in other to form one product. Nuclear fission generate much radioactive waste than nuclear fusion. The radioactive waste that is obtainable during nuclear fusion is less than 1% of that produce by nuclear fission.
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2 years ago
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In two separate double slit experiments, an interference pattern is observed on a screen. In the first experiment, violet light
sashaice [31]

Answer:

\lambda_3 = 4.72*10^{-7} m

Explanation:

given data:

wavelength \lambda = 708nm = 708*10^{-9} m

using the following relation:

y = \frac{mL\lambda}{d}

according to the given information

second and third dark fringe is at same location. so

y_2 = y_3

\frac{m_2L\lambda_2}{d} = \frac{m_3L\lambda_3}{d}

m_2\lambda_2 = m_3\lambda_3

2*708*10^{-9} = 3*\lambda_3

\lambda_3 = \frac{2*708*10^{-9}}{3}

\lambda_3 = 4.72*10^{-7} m

4 0
2 years ago
A pair of glasses is dropped from the top of a 32.0m stadium. A pen is dropped 2.Os later. How high above the ground is the pen
Svetllana [295]

Answer:

h_p = 30.46\ m

Explanation:

<u>Free Fall Motion</u>

A free-falling object refers to an object that is falling under the sole influence of gravity. If the object is dropped from a certain height h, it moves downwards until it reaches ground level.

The speed vf of the object when a time t has passed is given by:

v_f=g\cdot t

Where g = 9.8 m/s^2

Similarly, the distance y the object has traveled is calculated as follows:

\displaystyle y=\frac{g\cdot t^2}{2}

If we know the height h from which the object was dropped, we can solve the above equation for t:

\displaystyle t=\sqrt{\frac{2\cdot y}{g}}

The stadium is h=32 m high. A pair of glasses is dropped from the top and reaches the ground at a time:

\displaystyle t_1=\sqrt{\frac{2\cdot 32}{9.8}}=2.56\ sec

The pen is dropped 2 seconds after the glasses. When the glasses hit the ground, the pen has been falling for:

t_2=2.56 - 2 = 0.56\ sec

Therefore, it has traveled down a distance:

\displaystyle y=\frac{9.8\cdot 0.56^2}{2} = 1.54\ m

Thus, the height of the pen is:

h_p = 32 - 1.54\Rightarrow h_p=30.46\ m

8 0
2 years ago
The frequency of the applied RF signal used to excite spins is directly proportional to the magnitude of the static magnetic fie
Anni [7]

Answer:

The inverse frequency is \dfrac{3}{80}\ s

Explanation:

Given that,

Magnetic field = 20 T

Proportionality constant = 5 Hz/T

Change in magnetic field = 3 T

We know that,

B=\dfrac{K}{\dfrac{1}{\omega}}

We need to calculate the inverse frequency

Using formula of frequency

\Delta(\dfrac{1}{\omega})=\dfrac{\Delta B}{k\times(\dfrac{1}{\omega^2})}

\Delta(\dfrac{1}{\omega})=\dfrac{k\times\Delta B}{B^2}

Put the value into the formula

\Delta(\dfrac{1}{\omega})=\dfrac{3\times5}{(20)^2}

\Delta(\dfrac{1}{\omega})=\dfrac{3}{80}\ s

Hence, The inverse frequency is \dfrac{3}{80}\ s

5 0
2 years ago
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FrozenT [24]

Answer:

I = 2 kgm^2

Explanation:

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\tau=I\alpha     (1)

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α: angular acceleration of the door = 2.00 rad/s^2

τ: torque exerted on the door

You can calculate the torque by using the information about the Force exerted on the door, and the distance to the hinges. You use the following formula:

\tau=Fd        (2)

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You replace the equation (2) into the equation (1), and you solve for α:

Fd=I\alpha\\\\I=\frac{Fd}{\alpha}

Finally, you replace the values of all parameters in the previous equation for I:

I=\frac{(5.00N)(0.800m)}{2.00rad/s^2}=2kgm^2

The moment of inertia of the door around the hinges is 2 kgm^2

3 0
2 years ago
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