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beks73 [17]
2 years ago
15

A puck of mass 0.5100.510kg is attached to the end of a cord 0.8270.827m long. The puck moves in a horizontal circle without fri

ction. If the cord can withstand a maximum tension of 126126N, what is the highest frequency at which the puck can go around the circle without the cord breaking?
Physics
1 answer:
yanalaym [24]2 years ago
6 0

Answer: 2.75 1/sec

Explanation:

The only external force (neglecting gravity) acting on the puck, is the centripetal force, which. in this case, is represented by the tension in the string, so we can say:

T = mv² / r (1)

Our unknown, is the frequency at which the puck can go around the circle, which is the inverse of the period Tp.

By definition, a period is the time needed by the puck to complete one entire circle.

By definition also , angular velocity is the rate of change of the angle advanced, so we can express this way:

ω = ∆θ / ∆t  

The angle advanced during one period, is exactly (by angle definition) 2 π radians.

So, we can always write the angular velocity, ω, as follows:

ω = 2π / Tp = 2πf

Now, there is a relationship between linear and angular velocity, that can be found applying simply the definition of velocity and of an angle too, as follows:

v = ∆s / ∆t = r ∆θ/∆t = ω r

Replacing in (1), we have:

T = mω2 r2 / r = m ω2r (2)

We have just found that ω= 2πf, so, replacing in (2) :

T = m (2π)2 f2 r  

Solving for f:

f = 1/2π√(T/mr) = 1/2π 17.28 1/sec = 2.75 1/sec

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Den pushes a desk 400 cm across the floor. He exerts a force of 10 N for 8 s to move the desk. What is his power output? (Power:
White raven [17]

Answer:

5 W

Explanation:

The formula of the power is:

● P = W/t

W is the work and t is the time needed to do it(in seconds)

Let's calculate first the work that the force exerced:

W = Vector F . Vector d

D is the distance ( here 400 cm wich is 4 m)

Make a representation to see how are the vectors F and V.(picture below)

The vector F and d are colinear since Den is pushing the desk on the ground.

● W = 4 × 10 = 40 J

J is Joule

■■■■■■■■■■■■■■■■■■■■■■■■■■

● P = W / t

● P = 40/ 8

● P = 5 W

7 0
3 years ago
Electric field lines always begin at _______ charges (or at infinity) and end at _______ charges (or at infinity). One could als
lesantik [10]

Answer:

b)

Explanation:

By convention, the electric field lines (which are tangent to the direction of the electric field at a given point) always begin at positive charges, and finish at negative charges.

This is a consequence of the convention that states that the electric field has the direction of the trajectory of a positive test charge when released from rest in an electric field.

(As the positive charge would move away from positive charges and would  be attracted by negative ones).

So, the combination of answers that is true is b) (positive, negative, positive).

3 0
3 years ago
Please help me with this physics prooblem
zaharov [31]

Take the missile's starting position to be the origin. Assuming the angles given are taken to be counterclockwise from the positive horizontal axis, the missile has position vector with components

x=v_0\cos20.0^\circ t+\dfrac12a_xt^2

y=v_0\sin20.0^\circ t+\dfrac12a_yt^2

The missile's final position after 9.20 s has to be a vector whose distance from the origin is 19,500 m and situated 32.0 deg relative the positive horizontal axis. This means the final position should have components

x_{9.20\,\mathrm s}=(19,500\,\mathrm m)\cos32.0^\circ

y_{9.20\,\mathrm s}=(19,500\,\mathrm m)\sin32.0^\circ

So we have enough information to solve for the components of the acceleration vector, a_x and a_y:

x_{9.20\,\mathrm s}=\left(1810\,\dfrac{\mathrm m}{\mathrm s}\right)\cos20.0^\circ(9.20\,\mathrm s)+\dfrac12a_x(9.20\,\mathrm s)^2\implies a_x=21.0\,\dfrac{\mathrm m}{\mathrm s^2}

y_{9.20\,\mathrm s}=\left(1810\,\dfrac{\mathrm m}{\mathrm s}\right)\sin20.0^\circ(9.20\,\mathrm s)+\dfrac12a_y(9.20\,\mathrm s)^2\implies a_y=110\,\dfrac{\mathrm m}{\mathrm s^2}

The acceleration vector then has direction \theta where

\tan\theta=\dfrac{a_y}{a_x}\implies\theta=79.2^\circ

5 0
2 years ago
One mole of magnesium (6 × 1023 atoms) has a mass of 24 grams, as shown in the periodic table on the inside front cover of the t
natka813 [3]

This question involves the concepts of density, volume, and mass.

The approximate diameter of a magnesium atom is "3.55 x 10⁻¹⁰ m".

<h3>STEP 1 (FINDING MASS OF INDIVIDUAL ATOM)</h3>

It is given that:

Mass of one mole = 24 grams

Mass of 6 x 10²³ atoms = 24 grams

Mass of 1 atom = \frac{24\ grams}{6\ x\ 10^{23}\ atoms} = 4 x 10⁻²³ grams

<h3>STEP 2 (FINDING VOLUME OF A SINGLE ATOM)</h3>

\rho = \frac{m}{V}\\\\V=\frac{m}{\rho}

where,

  • \rho = density = 1.7 grams/cm³
  • m = mass of single atom = 4 x 10⁻²³ grams
  • V = volume of single atom = ?

Therefore,

V=\frac{4\ x\ 10^{-23}\ grams}{1.7\ grams/cm^3}

V = 2.35 x 10⁻²³ cm³

<h3>STEP 3 (FINDING DIAMETER OF ATOM)</h3>

The atom is in a spherical shape. Hence, its Volume can be given as follows:

V =\frac{\pi d^3}{6}\\\\d=\sqrt[3]{ \frac{6V}{\pi}}\\\\d=\sqrt[3]{ \frac{6(2.35\ x\ 10^{-23}\ cm^3)}{\pi}}

d = 0.355 x 10⁻⁷ cm = 3.55 x 10⁻¹⁰ m

Learn more about density here:

brainly.com/question/952755

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2 years ago
Why is the big bang called a theory?<br>​
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There is no scientific way to prove that it happened it’s like a hypothesis without being able to test the hypothesis
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3 years ago
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