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qwelly [4]
4 years ago
9

Kent is twirling a ball attached to a string in a clockwise direction. If Kent lets go of the string, which of the following des

cribes the motion that the ball will take if there are no gravitational forces influencing the motion of the ball?

Physics
2 answers:
Serhud [2]4 years ago
7 0

It will continue in a straight line.

raketka [301]4 years ago
3 0
Kent's ball will go in the direction that the ball is going right when he lets go of it.

Imagine this like an analog clock. 12 is forwards, 6 is behind, 3 is right, and 9 is left relative to Kent.  If Kent is twirling the ball clockwise and lets go at 3, the, the ball will go behind him as the velocity vector exactly at 3 points behind Kent in a clockwise motion. Here are Kents velocity vectors according to my clock analogy. Hope this helps because this is how I learned it!

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A block of a plastic material floats in water with 42.9% of its volume under water. What is the density of the block in kg/m3?
adell [148]

To solve this problem we will apply the principle of buoyancy of Archimedes and the relationship given between density, mass and volume.

By balancing forces, the force of the weight must be counteracted by the buoyancy force, therefore

\sum F = 0

F_b -W = 0

F_b = W

F_b = mg

Here,

m = mass

g =Gravitational energy

The buoyancy force corresponds to that exerted by water, while the mass given there is that of the object, therefore

\rho_w V_{displaced} g = mg

Remember the expression for which you can determine the relationship between mass, volume and density, in which

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

In this case the density would be that of the object, replacing

\rho_w V_{displaced} g = V\rho g

Since the displaced volume of water is 0.429 we will have to

\rho_w (0.429V) = V \rho

0.429\rho_w= \rho

The density of water under normal conditions is 1000kg / m ^ 3, so

0.429(1000) = \rho

\rho = 429kg/m^3

The density of the object is 429kg / m ^ 3

7 0
3 years ago
An inductor has an inductance of 0.12 mH. Calculate the number of turns per unit length of this inductor if it has a total of 50
SSSSS [86.1K]

Answer:

oh                      

Explanation:

5 0
3 years ago
Which letter (A, B, or C) shows where you should apply the most effort to lift the stone?
sergij07 [2.7K]
A becuz its at da it dont got no wa
8 0
3 years ago
A skier starts from rest down a slope 500.0 M long, the skier accelerates at a constant rate of 2.00 m/s/s, what's the velocity
nevsk [136]
We can use the kinematic equation
(v_f)^2 = (v_i)^2 + 2*a*d
where Vf is what we are looking for
Vi is 0 since we start from rest
a is acceleration
and d is the distance

we get
(Vf)^2 = (0)^2 + 2*(2)*(500)
(Vf)^2 = 2000
Vf = about 44.721
or 44.7 m/s   [if you are rounding this by significant figures]
8 0
3 years ago
An alpha particle (the nucleus of a helium atom) consists of two protons and two neutrons, and has a mass of 6.64 * 10-27 kg. A
melamori03 [73]

Answer:

t = 4.21x10⁻⁷ s

Explanation:

The time (t) can be found using the angular velocity (ω):

\omega = \frac{\theta}{t}

<em>Where θ: is the angular displacement = π (since it moves halfway through a complete circle)</em>

We have:

t = \frac{\theta}{\omega} = \frac{\theta}{v/r}  

<u>Where</u>:      

<em>v: is the tangential speed </em>

<em>r: is the radius</em>

The radius can be found equaling the magnetic force with the centripetal force:

qvB = \frac{mv^{2}}{r} \rightarrow r = \frac{mv}{qB}

Where:

m: is the mass of the alpha particle = 6.64x10⁻²⁷ kg

q: is the charge of the alpha particle = 2*p (proton) = 2*1.6x10⁻¹⁹C

B: is the magnetic field = 0.155 T

Hence, the time is:

t = \frac{\theta*r}{v} = \frac{\theta}{v}*\frac{mv}{qB} = \frac{\theta m}{qB} = \frac{\pi * 6.64 \cdot 10^{-27} kg}{2*1.6 \cdot 10^{-19} C*0.155 T} = 4.21 \cdot 10^{-7} s

Therefore, the time that takes for an alpha particle to move halfway through a complete circle is 4.21x10⁻⁷ s.

I hope it helps you!    

4 0
3 years ago
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