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2 years ago

(A) State the relation between acceleration and momentum (10 marks).

Physics

1 answer:

ohaa [14]2 years ago

4 0

Answer:

momentum is directly proportional to acceleration

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An emf is induced in response to a change in magnetic field inside a loop of wire. Which of the following changes would increase

goldenfox [79]

Answer:

changing the magnetic field more rapidly

Explanation:

According to Faraday's law, whenever there is a change in the magnetic lines of force, it leads the production of induced emf. The magnitude of induced emf is proportional to to the rate of change of flux.

Hence if the magnetic field inside a loop of wire is changed rapidly, the magnitude of induced emf increases in accordance with Faraday's law of electromagnetic induction stated above when the magnetic field is changed more rapidly, hence the answer.

8 0

2 years ago

A wheel moves in the xy plane in such a way that the location of its center is given by the equations xo = 12t3 and yo = R = 2,

Stella [2.4K]

Answer:

the velocity of the point P located on the horizontal diameter of the wheel at t = 1.4 s is $P = 104.04 \hat{i} -314.432 \hat{j}$

Explanation:

The free-body diagram below shows the interpretation of the question; from the diagram , the wheel that is rolling in a clockwise directio will have two velocities at point P;

the peripheral velocity that is directed downward $(-V_y)$ along the y-axis

the linear velocity $(V_x)$ that is directed along the x-axis

Now;

$V_x = \frac{d}{dt}(12t^3+2) = 36 t^2$

$V_x = 36(1.7)^2\\\\V_x = 104.04\ ft/s$

Also,

$-V_y = R* \omega$

where $\omega$ (angular velocity) = $\frac{d\theta}{dt} = \frac{d}{dt}(8t^4)$

$-V_y = 2*32t^3)\\\\\\-V_y = 2*32(1.7^3)\\\\-V_y = 314.432 \ ft/s$

∴ the velocity of the point P located on the horizontal diameter of the wheel at t = 1.4 s is $P = 104.04 \hat{i} -314.432 \hat{j}$

3 0

3 years ago

Eventually, the process approaches a steady state. In that steady state, the charge of the capacitor is not changing. What is th

devlian [24]

Answer:

The current in the circuit must be zero.

Explanation:

In a RC circuit, the steady state is reached when either the capacitor is fully charged or fully discharged. In either case, there must not be any current through the circuit because if it exists, it will deliver charge to the capacitor and thus change its charge, which is not a steady state.

6 0

2 years ago

An 20-cm-long Bicycle Crank Arm. With A Pedal At One End. Is Attached To A 25-cm-diameter Sprocket, The Toothed Disk Around Whic

malfutka [58]

To solve the problem, it is necessary to apply the concepts related to the kinematic equations of the description of angular movement.

The angular velocity can be described as

$\omega_f = \omega_0 + \alpha t$

Where,

$\omega_f =$ Final Angular Velocity

$\omega_0 =$ Initial Angular velocity

$\alpha =$ Angular acceleration

t = time

The relation between the tangential acceleration is given as,

$a = \alpha r$

where,

r = radius.

PART A ) Using our values and replacing at the previous equation we have that

$\omega_f = (94rpm)(\frac{2\pi rad}{60s})= 9.8436rad/s$

$\omega_0 = 63rpm(\frac{2\pi rad}{60s})= 6.5973rad/s$

$t = 11s$

Replacing the previous equation with our values we have,

$\omega_f = \omega_0 + \alpha t$

$9.8436 = 6.5973 + \alpha (11)$

$\alpha = \frac{9.8436- 6.5973}{11}$

$\alpha = 0.295rad/s^2$

The tangential velocity then would be,

$a = \alpha r$

$a = (0.295)(0.2)$

$a = 0.059m/s^2$

Part B) To find the displacement as a function of angular velocity and angular acceleration regardless of time, we would use the equation

$\omega_f^2=\omega_0^2+2\alpha\theta$

Replacing with our values and re-arrange to find $\theta,$

$\theta = \frac{\omega_f^2-\omega_0^2}{2\alpha}$

$\theta = \frac{9.8436^2-6.5973^2}{2*0.295}$

$\theta = 90.461rad$

That is equal in revolution to

$\theta = 90.461rad(\frac{1rev}{2\pi rad}) = 14.397rev$

The linear displacement of the system is,

$x = \theta*(2\pi*r)$

$x = 14.397*(2\pi*\frac{0.25}{2})$

$x = 11.3m$

5 0

3 years ago

2. Three super strong teenagers pull a heavy crate across the floor. Dion pulls with a force of 18.5 N towards 0°. Shirley pulls

kogti [31]

Answer:

A) The resultant force is 43.4 [N]

B) The movement of the heavy crate is going to the right and in the negative direction on the y-axis

Explanation:

We need to make a sketch of the different forces acting on the heavy crate.

In the attached image we can see the forces and the sum of the vector with their respective angles.

Forces in the X-axis

$Fdionx=18.5N\\\\Fshix=16.5*cos(30)=14.29N\\Fjoanx=19.5*cos(60)=9.75N\\\\Forcex= 18.5 + 14.29 + 9.75 = 42.54 N$

Forces in the y-axis

$FDiony=0[N]\\Fshirley= 16.5*sin(30)=8.25[N]\\Fjoany=19.5*sin(60)=16.88 [N]\\\\Forcesy=0+8.25-16.88= -8.63[N]$

Using the Pythagorean theorem

$Tforce=\sqrt{(42.54)^{2} +(8.63)^{2} } \\\\Tforce= 43.4N$

The movement of the heavy crate is going to the right and in the negative direction on the y-axis, this can be easily seen in the graphical sum of vectors.

8 0

2 years ago