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ruslelena [56]
3 years ago
13

Susan's 10.0kg baby brother Paul sits on a mat. Susan pulls the mat across the floor using a rope that is angled 30? above the f

loor. The tension is a constant 32.0N and the coefficient of friction is 0.190.
Use work and energy to find Paul's speed after being pulled 3.20m .

Solve for vf.
Physics
1 answer:
elena-s [515]3 years ago
8 0

Answer:

The speed after being pulled is 2.4123m/s

Explanation:

The work realize by the tension and the friction is equal to the change in the kinetic energy, so:

W_T+W_F=K_f-K_i (1)

Where:

W_T=T*x*cos(0)=32N*3.2m*cos(30)=88.6810J\\W_F=F_r*x*cos(180)=-0.190*mg*x =-0.190*10kg*9.8m/s^{2}*3.2m=59.584J\\ K_i=0\\K_f=\frac{1}{2}*m*v_f^{2}=5v_f^{2}

Because the work made by any force is equal to the multiplication of the force, the displacement and the cosine of the angle between them.

Additionally, the kinetic energy is equal to \frac{1}{2}mv^{2}, so if the initial velocity v_i is equal to zero, the initial kinetic energy K_i is equal to zero.

Then, replacing the values on the equation and solving for v_f, we get:

W_T+W_F=K_f-K_i\\88.6810-59.5840=5v_f^{2}\\29.097=5v_f^{2}

\frac{29.097}{5}=v_f^{2}\\\sqrt{5.8194}=v_f\\2.4123=v_f

So, the speed after being pulled 3.2m is 2.4123 m/s

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Calculate the linear acceleration (in m/s2) of a car, the 0.310 m radius tires of which have an angular acceleration of 15.0 rad
love history [14]

Answer:

a) The linear acceleration of the car is 4.65\,\frac{m}{s^{2}}, b) The tires did 7.46 revolutions in 2.50 seconds from rest.

Explanation:

a) A tire experiments a general plane motion, which is the sum of rotation and translation. The linear acceleration experimented by the car corresponds to the linear acceleration at the center of the tire with respect to the point of contact between tire and ground, whose magnitude is described by the following formula measured in meters per square second:

\| \vec a \| = \sqrt{a_{r}^{2} + a_{t}^{2}}

Where:

a_{r} - Magnitude of the radial acceleration, measured in meters per square second.

a_{t} - Magnitude of the tangent acceleration, measured in meters per square second.

Let suppose that tire is moving on a horizontal ground, since radius of curvature is too big, then radial acceleration tends to be zero. So that:

\| \vec a \| = a_{t}

\| \vec a \| = r \cdot \alpha

Where:

\alpha - Angular acceleration, measured in radians per square second.

r - Radius of rotation (Radius of a tire), measured in meters.

Given that \alpha = 15\,\frac{rad}{s^{2}} and r = 0.31\,m. The linear acceleration experimented by the car is:

\| \vec a \| = (0.31\,m)\cdot \left(15\,\frac{rad}{s^{2}} \right)

\| \vec a \| = 4.65\,\frac{m}{s^{2}}

The linear acceleration of the car is 4.65\,\frac{m}{s^{2}}.

b) Assuming that angular acceleration is constant, the following kinematic equation is used:

\theta = \theta_{o} + \omega_{o}\cdot t + \frac{1}{2}\cdot \alpha \cdot t^{2}

Where:

\theta - Final angular position, measured in radians.

\theta_{o} - Initial angular position, measured in radians.

\omega_{o} - Initial angular speed, measured in radians per second.

\alpha - Angular acceleration, measured in radians per square second.

t - Time, measured in seconds.

If \theta_{o} = 0\,rad, \omega_{o} = 0\,\frac{rad}{s}, \alpha = 15\,\frac{rad}{s^{2}}, the final angular position is:

\theta = 0\,rad + \left(0\,\frac{rad}{s}\right)\cdot (2.50\,s) + \frac{1}{2}\cdot \left(15\,\frac{rad}{s^{2}}\right)\cdot (2.50\,s)^{2}

\theta = 46.875\,rad

Let convert this outcome into revolutions: (1 revolution is equal to 2π radians)

\theta = 7.46\,rev

The tires did 7.46 revolutions in 2.50 seconds from rest.

3 0
3 years ago
Does the theory of relativity show that Newtonian mechanics is wrong?
valina [46]

Answer:

Einstein extended the rules of Newton for high speeds. For applications of mechanics at low speeds, Newtonian ideas are almost equal to reality. That is the reason we use Newtonian mechanics in practice at low speeds.

Explanation:

<em>But on a conceptual level, Einstein did prove Newtonian ideas quite wrong in some cases, e.g. the relativity of simultaneity. But again, in calculations, Newtonian ideas give pretty close to correct answer in low-speed regimes. So, the numerical validity of Newtonian laws in those regimes is something that no one can ever prove completely wrong - because they have been proven correct experimentally to a good approximation.</em>

4 0
3 years ago
The magnitude of the magnetic flux through the surface of a circular plate is 5.90 10-5 T · m2 when it is placed in a region of
valkas [14]

Answer:

The strength of the magnetic field is 3.5 x 10⁻³ T

Explanation:

Given;

magnitude of the magnetic flux , Φ = 5.90 x 10⁻⁵ T·m²

angle of inclination of the field, θ = 42.0°

radius of the circular plate, r = 8.50 cm = 0.085 m

Generally magnetic flux in a uniform magnetic field is given as;

Φ = BACosθ

where;

B is the strength of the magnetic field

A is the area of the circular plate

Area of the circular plate:

A = πr²

A = π (0.085)² = 0.0227 m²

The strength of the magnetic field:

B = Φ / ACosθ

B = ( 5.90 x 10⁻⁵) / ( 0.0227 x Cos42)

B = 3.5 x 10⁻³ T

Therefore, the strength of the magnetic field is 3.5 x 10⁻³ T

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A student increased the resister of the voltmeter year not the circuit
Flura [38]

Answer:

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3 0
3 years ago
A speaker vibrates at a frequency of 200 Hz. What is its period
Daniel [21]
Period,  T = 1/ f.
f = frequency =  200 Hz.

Period T =  1/200 =  0.005 seconds.
8 0
3 years ago
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