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3 years ago

6.

Physics

1 answer:

yaroslaw [1]3 years ago

6 0

Answer:

12 J

Explanation:

From the question given above, the following data were obtained:

Mass (m) = 7.6 kg

Distance (d) = 6 m

Velocity (v) = 5 m/s

Force (F) = 2 N

Workdone (Wd) =.?

Workdone can be defined as the product of force and distance moved in the direction of the force. Mathematically, it is expressed as:

Workdone = Force × distance

Wd = F × d

With the above formula, we can obtain the workdone as follow:

Distance (d) = 6 m

Force (F) = 2 N

Workdone (Wd) =.?

Wd = F × d

Wd = 2 × 6

Wd = 12 J

Thus, the workdone is 12 J

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A 42.0-cm diameter disk rotates with a constant angular acceleration of 2.70 rad/s2. It starts from rest at t = 0, and a line dr

Zielflug [23.3K]

Answer:

6.21 rad/s

1.3041 m/s, 0.567 m/s²

$106.4778\ ^{\circ}$

Explanation:

$\omega_f$ = Final angular velocity

$\omega_i$ = Initial angular velocity = 0

$\alpha$ = Angular acceleration = 2.3 rad/s²

$\theta$ = Angle of rotation

t = Time taken = 2.3 s

Equation of rotational motion

$\omega_f=\omega_i+\alpha t\\\Rightarrow \omega_f=0+2.7\times 2.3\\\Rightarrow \omega_f=6.21\ rad/s$

The angular speed is 6.21 rad/s

Linear velocity is given by

$v=r\omega\\\Rightarrow v=0.21\times 6.21\\\Rightarrow v=1.3041\ m/s$

Linear velocity is 1.3041 m/s

Tangential acceleration is given by

$a_t=r\alpha\\\Rightarrow a_t=0.21\times 2.7\\\Rightarrow a_t=0.567\ m/s^2$

Tangential acceleration is 0.567 m/s²

$\theta=\omega_it+\dfrac{1}{2}\alpha t^2\\\Rightarrow \theta=0\times t+\dfrac{1}{2}\times 2.7\times 2.3^2\\\Rightarrow \theta=7.1415\ rad$

In degress the angle would be

$57.3+7.1415\times \dfrac{180}{\pi}=466.47780\ ^{\circ}$

From x axis it would be

$466.47780-360=106.4778\ ^{\circ}$

The angle is $106.4778\ ^{\circ}$ from x axis

7 0

4 years ago

A coil is wrapped with 300 turns of wire on the perimeter of a circular frame (radius = 8.0 cm). Each turn has the same area, eq

MAVERICK [17]

Answer:

Approximately 18 volts when the magnetic field strength increases from $\rm 20\; mT$ to $\rm 80\;mT$ at a constant rate.

Explanation:

By the Faraday's Law of Induction, the EMF $\epsilon$ that a changing magnetic flux induces in a coil is:

$\displaystyle \epsilon = N \cdot \frac{d\phi}{dt}$ ,

where

$N$ is the number of turns in the coil, and
$\displaystyle \frac{d\phi}{dt}$ is the rate of change in magnetic flux through this coil.

However, for a coil the magnetic flux $\phi$ is equal to

$\phi = B \cdot A\cdot \cos{\theta}$ ,

where

$B$ is the magnetic field strength at the coil, and
$A\cdot \cos{\theta}$ is the area of the coil perpendicular to the magnetic field.

For this coil, the magnetic field is perpendicular to coil, so $\theta = 0$ and $A\cdot \cos{\theta} = A$ . The area of this circular coil is equal to $\pi\cdot r^{2} = \pi\times 8.0\times 10^{-2}\approx \rm 0.0201062\; m^{2}$ .

$A\cdot \cos{\theta} = A$ doesn't change, so the rate of change in the magnetic flux $\phi$ through the coil depends only on the rate of change in the magnetic field strength $B$ . The size of the magnetic field at the instant that $B = \rm 50\; mT$ will not matter as long as the rate of change in $B$ is constant.

$\displaystyle \begin{aligned} \frac{d\phi}{dt} &= \frac{\Delta B}{\Delta t}\times A \\&= \rm \frac{80\times 10^{-3}\; T- 20\times 10^{-3}\; T}{20\times 10^{-3}\; s}\times 0.0201062\;m^{2}\\&= \rm 0.0603186\; T\cdot m^{2}\cdot s^{-1}\end{aligned}$ .