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2 years ago

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A puppy finds a rawhide bone and begins to pull it with a force, Ft. The free-body diagram is shown. Which describes what happen

s to the bone?
It begins to move toward the right.
It stays in static equilibrium.
It stays in dynamic equilibrium.
It begins to move toward the left.

2 answers:

Anastasy [175]2 years ago

8 0

It begins to move towards the right

GalinKa [24]2 years ago

4 0

It begins to move towards the right

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A +5.0-μC point charge is placed at the 0 cm mark of a meter stick and a -4.0-μC charge is placed at the 50 cm mark. What is the

algol13

Answer:

1.4 *10^6 N/C

Explanation:

The electric field caused by a charge at a certain point is given by the equation:

$E = k \frac{q}{r} \^r$

where k is the Coulomb constant equal to 8.99 *10^9 Nm^2/C^2, q the charge of the particle in coulombs, r is the distance from the point to the charge in meters.

$\^r$ is the unitary vector that goes from the charge to the point. This vector will give us the direction of the Electric Field vector.

The unitary vector of the +5.0-μC charge will go to the right (+i), as the point is to the right of the charge. Then, the electric field caused by the charge will be:

$E_1 = k \frac{q}{r^2} \^r = 8.99*10^9 Nm^2/C^2 \frac{5.0 *10^{-6}C}{(0.3m - 0m)^2}(+\^i) = +0.5*10^6 N/C$

The unitary vector of the -4.0-μC charge will go to the left (-i), as the point is to the left of the charge. Then, the electric field caused by the charge will be:

$E_2 = k \frac{q}{r^2} \^r = 8.99*10^9 Nm^2/C^2 \frac{-4.0 *10^{-6}C}{(0.5m - 0.3m)^2}(-\^i) = +0.9*10^6 N/C$

The electric field at the 30 cm mark will be the addition of both electric field:

$E_{total} = E_1 +E_2 = 0.5 *10^6 N/C + 0.9*10^6 N/C = 1.4 *10^6 N/C$

3 0

2 years ago

A turntable rotates counterclockwise at 76 rpm . A speck of dust on the turntable is at 0.47 rad at t=0. What is the angle of th

icang [17]

To solve this exercise it is necessary to apply the kinematic equations of angular motion.

By definition we know that the displacement when there is constant angular velocity is

$\theta= \theta_0 +\omega t$

From our given data we know that,

$\omega = 76\frac{rev}{min}$

$\omega = 76\frac{rev}{min}(\frac{2\pi rad}{1rev})(\frac{1 min}{60s})$

$\omega = 7.958rad/s$

Moreover we know that

$\theta_0 = 0.47 rad$

Therefore for time t=8.1s we have,

$\theta= \theta_0+ \omega t$

$\theta= 0.47+(7.958)(8.1)$

$\theta = 64.9298rad$

That number in revolution is:

$\theta = 64.9298rad(\frac{1rev}{2\pi})$

$\theta = 15.108 Revolutions$

Here, we see that there are 15 complete revolutions

And 0.108 revolutions i not complete, so the tunable rotation is

$\theta_{net} = 0.108*2\pi=0.216\pi$

Therefore the angle of the speck at a time 8.1s is $0.216\pi$

4 0

2 years ago

A 0.500-kg ball traveling horizontally on a frictionless surface approaches a very massive stone at 20.0 m/s perpendicular to wa

gregori [183]

The magnitude of the change in momentum of the stone is about 18.4 kg.m/s

$\texttt{ }$

<h3>Further explanation</h3>

Let's recall Impulse formula as follows:

$\boxed {I = \Sigma F \times t}$

<em>where:</em>

<em>I = impulse on the object ( kg m/s )</em>

<em>∑F = net force acting on object ( kg m /s² = Newton )</em>

<em>t = elapsed time ( s )</em>

Let us now tackle the problem!

$\texttt{ }$

<u>Given:</u>

mass of ball = m = 0.500 kg

initial speed of ball = vo = 20.0 m/s

final kinetic energy = Ek = 70% Eko

<u>Asked:</u>

magnitude of the change of momentum of the stone = Δp = ?

<u>Solution:</u>

<em>Firstly, we will calculate the final speed of the ball as follows:</em>

$Ek = 70\% \ Ek_o$

$\frac{1}{2} m v^2 = 70\% \ ( \frac{1}{2} m (v_o)^2 )$

$v^2 = 70 \% \ (v_o)^2$

$v = - v_o \sqrt{70 \%}$ → <em>negative sign due to ball rebounds</em>

$v = - v_o \sqrt{0.7} \texttt{ m/s}$

$\texttt{ }$

<em>Next, we could find the magnitude of the change of momentum of the stone as follows:</em>

$\Delta p_{stone} = - \Delta p_{ball}$

$\Delta p_{stone} = - [ mv - mv_o ]$

$\Delta p_{stone} = m[ v_o - v ]$

$\Delta p_{stone} = m[ v_o + v_o\sqrt{0.7} ]$

$\Delta p_{stone} = mv_o [ 1 + \sqrt{0.7} ]$

$\Delta p_{stone} = 0.500 ( 20.0 ) [ 1 + \sqrt{0.7} ]$

$\Delta p_{stone} \approx 18.4 \texttt{ kg.m/s}$

$\texttt{ }$

<h3>Learn more</h3>

Velocity of Runner : brainly.com/question/3813437

Kinetic Energy : brainly.com/question/692781

Acceleration : brainly.com/question/2283922

The Speed of Car : brainly.com/question/568302
Average Speed of Plane : brainly.com/question/12826372
Impulse : brainly.com/question/12855855
Gravity : brainly.com/question/1724648

$\texttt{ }$

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Dynamics

8 0

2 years ago

A 1.0 kg object moving at 4.5 m/s has a wavelength of:

Lisa [10]

By wave particle duality.

Wavelength , λ = h / mv

where h = Planck's constant = 6.63 * 10⁻³⁴ Js, m = mass in kg, v = velocity in m/s.
m = 1kg, v = 4.5 m/s

λ = h / mv

λ = (6.63 * 10⁻³⁴) /(1*4.5)

λ ≈ 1.473 * 10⁻³⁴ m

Option D.

7 0

2 years ago

Determine whether each substance will sink or float in corn syrup, which has a density of 1.36 g/cm3. Write “sink” or “float” in

Tatiana [17]

Explanation:

A substance will floats if it is having lower density than the density of the liquid in which it is placed.
A substance will sink if it is having density greater than the density of the liquid in which it is kept.

Density of corn syrup = $1.36 g/cm^3$

1) Density of gasoline = $0.748 g/cm^3$

Density of the gasoline is less than the the density of corn syrup which means it will float in corn syrup.

2) Density of water = $1 g/cm^3$

Density of the water is less than the the density of corn syrup which means it will float in corn syrup.

3) Density of honey = $1.45 g/cm^3$

Density of the gasoline is more than the the density of corn syrup which means it will sink in corn syrup.

4) Density of titanium = $4.506 g/cm^3$

Density of the titanium is more than the the density of corn syrup which means it will sink in corn syrup.

3 0

2 years ago

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