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3 years ago

8

A student leaves a sidewalk corner walks 3.0 blocks north then walks 3.0 blocks west.what is the students displacement from the

starting position

1 answer:

Sliva [168]3 years ago

3 0

Answer:

The students displacement is 3·√3 blocks, 45° North of West

Explanation:

From the question, we have;

The distance North in which the student walks = 3.0 blocks

The next walking distance West by the student = 3.0 blocks

The displacement of the student = The shortest distance between the student start and final point

By representing the motion of the student in vector format, we have;

$\mathbf{d} = 3\cdot \hat{i} + 3\cdot \hat{{j}}$

Therefore the magnitude of d is given as follows;

$\left | d\right | = \sqrt{3^2 + 3^2} = 3\cdot \sqrt{3} \ blocks$

The direction of the student = tan⁻¹(3/3) = 45° West of North

The students displacement = 3·√3 blocks, 45° North of West.

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A large piece of jewelry has a mass of 132.6 g. a graduated cylinder initially contains 48.6 ml water. when the jewelry is subme

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The density of the substance is<u> 10.5 g/cm³.</u>

The jewelry is made out of <u>Silver.</u>

Density ρ is defined as the ratio of mass <em>m</em> of the substance to its volume V<em>. </em> The cylinder contains a volume <em>V₁ of water</em> and when the jewelry is immersed in it, the total volume of water and the jewelry is found to be V₂.

The volume <em>V</em> of the jewelry is given by,

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Substitute 48.6 ml for <em>V₁ </em>and 61.2 ml for V₂.

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calculate the density ρ of the jewelry using the expression,

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

Substitute 132.6 g for <em>m</em> and 12.6 ml for <em>V</em>.

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3 years ago

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3 years ago

7. When will an object's displacement and distance traveled be different?

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If an object changes direction while travelling will an object's displacement and distance travelled be different.

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The magnitude of the displacement is always less than or equal to the distance because it is measured along the shortest path between two points.

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2 years ago

A torque of 36.5 N · m is applied to an initially motionless wheel which rotates around a fixed axis. This torque is the result

vivado [14]

Answer:

$21.6\ \text{kg m}^2$

$3.672\ \text{Nm}$

$54.66\ \text{revolutions}$

Explanation:

$\tau$ = Torque = 36.5 Nm

$\omega_i$ = Initial angular velocity = 0

$\omega_f$ = Final angular velocity = 10.3 rad/s

t = Time = 6.1 s

I = Moment of inertia

From the kinematic equations of linear motion we have

$\omega_f=\omega_i+\alpha_1 t\\\Rightarrow \alpha_1=\dfrac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha_1=\dfrac{10.3-0}{6.1}\\\Rightarrow \alpha_1=1.69\ \text{rad/s}^2$

Torque is given by

$\tau=I\alpha_1\\\Rightarrow I=\dfrac{\tau}{\alpha_1}\\\Rightarrow I=\dfrac{36.5}{1.69}\\\Rightarrow I=21.6\ \text{kg m}^2$

The wheel's moment of inertia is $21.6\ \text{kg m}^2$

t = 60.6 s

$\omega_i$ = 10.3 rad/s

$\omega_f$ = 0

$\alpha_2=\dfrac{0-10.3}{60.6}\\\Rightarrow \alpha_1=-0.17\ \text{rad/s}^2$

Frictional torque is given by

$\tau_f=I\alpha_2\\\Rightarrow \tau_f=21.6\times -0.17\\\Rightarrow \tau=-3.672\ \text{Nm}$

The magnitude of the torque caused by friction is $3.672\ \text{Nm}$

Speeding up

$\theta_1=0\times t+\dfrac{1}{2}\times 1.69\times 6.1^2\\\Rightarrow \theta_1=31.44\ \text{rad}$

Slowing down

$\theta_2=10.3\times 60.6+\dfrac{1}{2}\times (-0.17)\times 60.6^2\\\Rightarrow \theta_2=312.03\ \text{rad}$

Total number of revolutions

$\theta=\theta_1+\theta_2\\\Rightarrow \theta=31.44+312.03=343.47\ \text{rad}$

$\dfrac{343.47}{2\pi}=54.66\ \text{revolutions}$

The total number of revolutions the wheel goes through is $54.66\ \text{revolutions}$ .

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3 years ago